wrong_submission_id
stringlengths 10
10
| problem_id
stringlengths 6
6
| user_id
stringlengths 10
10
| time_limit
float64 1k
8k
| memory_limit
float64 131k
1.05M
| wrong_status
stringclasses 2
values | wrong_cpu_time
float64 10
40k
| wrong_memory
float64 2.94k
3.37M
| wrong_code_size
int64 1
15.5k
| problem_description
stringlengths 1
4.75k
| wrong_code
stringlengths 1
6.92k
| acc_submission_id
stringlengths 10
10
| acc_status
stringclasses 1
value | acc_cpu_time
float64 10
27.8k
| acc_memory
float64 2.94k
960k
| acc_code_size
int64 19
14.9k
| acc_code
stringlengths 19
14.9k
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s724272650
|
p02263
|
u535719732
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 680 |
An expression is given in a line. Two consequtive symbols (operand or operator) are separated by a space character. You can assume that +, - and * are given as the operator and an operand is a positive integer less than 106
|
data = list(map(str,input().split()))
stack = []
for i in data:
if(i == "+" or i == "-" or i == "*" or i == "/"):
stack.append(i)
else:
stack.append(int(i))
op1 = 0
op2 = 0
n = len(stack)
while(len(stack) != 1):
i = stack.pop()
if(i == "+"):
stack.append(op1+op2)
op1,op2 = op1+op2,0
elif(i == "-"):
stack.append(op1-op2)
op1,op2 = op1-op2,0
elif(i == "*"):
stack.append(op1*op2)
op1,op2 = op1*op2,0
elif(i == "/"):
stack.append(op1/op2)
op1,op2 = op1/op2,0
else:
if(op2 == 0):
op2 = i
else:
op1 = i
print(stack.pop())
|
s531504515
|
Accepted
| 20 | 5,596 | 427 |
data = list(map(str,input().split()))
stack = []
op1 = 0
op2 = 0
k = 0
for i in data:
# print(i,type(i))
if(i == "+" or i == "-" or i == "*" or i == "/"):
op1 = stack.pop()
op2 = stack.pop()
if(i == "+"): stack.append(op2+op1)
elif(i == "-"): stack.append(op2-op1)
elif(i == "*"): stack.append(op2*op1)
elif(i == "/"): stack.append(int(op2/op1))
else:
stack.append(int(i))
print(stack.pop())
|
s380227514
|
p02267
|
u394181453
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,540 | 87 |
You are given a sequence of _n_ integers S and a sequence of different _q_ integers T. Write a program which outputs C, the number of integers in T which are also in the set S.
|
input()
n1 = set(input().split())
input()
n2 = set(input().split())
print(len(n1-n2))
|
s949052768
|
Accepted
| 20 | 6,196 | 87 |
input()
n1 = set(input().split())
input()
n2 = set(input().split())
print(len(n1&n2))
|
s837654402
|
p03478
|
u893063840
| 2,000 | 262,144 |
Wrong Answer
| 32 | 2,940 | 192 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n, a, b = map(int, input().split())
cnt = 0
for num in range(1, n + 1):
s = str(num)
sm = 0
for let in s:
sm += int(let)
if a <= sm <= b:
cnt += 1
print(cnt)
|
s431171684
|
Accepted
| 33 | 2,940 | 194 |
n, a, b = map(int, input().split())
ans = 0
for num in range(1, n + 1):
s = str(num)
sm = 0
for let in s:
sm += int(let)
if a <= sm <= b:
ans += num
print(ans)
|
s296159868
|
p03565
|
u586577600
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 494 |
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
import sys
s_d = input()
t = input()
s = ''
for i in range(len(s_d)):
find_num = t.find(s_d[i])
if not find_num == -1:
if len(s_d[:i]) >= len(t[:find_num]) or len(s_d[i:]) >= len(t[find_num:]):
s_d = s_d[:i-find_num] + t + s_d[i+len(t)-1:]
break
elif i == len(s_d)-1:
print('UNRESTORABLE');
sys.exit()
for i in range(len(s_d)):
if s_d[i] == '?':
s = s + 'a'
else:
s = s + s_d[i]
print(s)
|
s435793845
|
Accepted
| 17 | 3,064 | 457 |
s = input()
t = input()
def check(s, t):
for i in range(len(s)):
for j in range(min(len(t), len(s[i:]))):
if s[i:][j] != '?' and s[i:][j] != t[j]:
break
if j == len(t)-1:
return True, i
return False, 0
s = s[::-1]
t = t[::-1]
c, i = check(s, t)
ans = ''
if c:
ans = s[:i] + t + s[(i+len(t)):]
ans = ans.replace('?', 'a')
print(ans[::-1])
else:
print('UNRESTORABLE')
|
s128931658
|
p03227
|
u517447467
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 28 |
You are given a string S of length 2 or 3 consisting of lowercase English letters. If the length of the string is 2, print it as is; if the length is 3, print the string after reversing it.
|
print(input()[::-1].upper())
|
s076857854
|
Accepted
| 18 | 2,940 | 61 |
S = input()
if len(S) == 2:
print(S)
else:
print(S[::-1])
|
s637731346
|
p03545
|
u364862909
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,136 | 277 |
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
A,B,C,D = map(int,list(input()))
dic = { 1:"+" , -1:"-"}
import itertools
for AB,BC,CD in itertools.product([1,-1],repeat=3):
tmp = A + B*AB + C*BC + D*CD
if tmp ==7:
ans = str(A)+dic[AB]+str(B)+dic[BC]+str(C)+dic[CD]+str(D)
print(ans)
exit()
|
s006365081
|
Accepted
| 31 | 9,200 | 282 |
A,B,C,D = map(int,list(input()))
dic = { 1:"+" , -1:"-"}
import itertools
for AB,BC,CD in itertools.product([1,-1],repeat=3):
tmp = A + B*AB + C*BC + D*CD
if tmp ==7:
ans = str(A)+dic[AB]+str(B)+dic[BC]+str(C)+dic[CD]+str(D)+"=7"
print(ans)
exit()
|
s592260562
|
p02406
|
u498511622
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,652 | 177 |
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
lst=""
x=int(input())
for i in range(1,x+1):
if i%3==0:
lst += " "+str(i)
else:
for s in range(i):
if s=="3":
lst += " "+str(s)
break
print(lst)
|
s083617868
|
Accepted
| 20 | 7,652 | 233 |
n = int(input())
result = ""
for s in range(1, n+1):
if s % 3 == 0:
result += " "+str(s)
else:
for i in str(s):
if i == "3":
result += " "+str(s)
break
print(result)
|
s514236602
|
p03493
|
u049355439
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 97 |
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
s = input()
number = 0
for i in range(len(s)):
if s[i] == 1:
number += 1
print(number)
|
s657443832
|
Accepted
| 17 | 2,940 | 99 |
s = input()
number = 0
for i in range(len(s)):
if s[i] == "1":
number += 1
print(number)
|
s682706516
|
p03359
|
u290187182
| 2,000 | 262,144 |
Wrong Answer
| 26 | 3,828 | 357 |
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
import sys
import copy
import math
import bisect
import pprint
import bisect
from functools import reduce
from copy import deepcopy
from collections import deque
def lcm(x, y):
return (x * y) // math.gcd(x, y)
if __name__ == '__main__':
a = [int(i) for i in input().split()]
if a[0] < a[1]:
print(a[0])
else:
print(a[0]-1)
|
s914513256
|
Accepted
| 27 | 3,828 | 358 |
import sys
import copy
import math
import bisect
import pprint
import bisect
from functools import reduce
from copy import deepcopy
from collections import deque
def lcm(x, y):
return (x * y) // math.gcd(x, y)
if __name__ == '__main__':
a = [int(i) for i in input().split()]
if a[0] <= a[1]:
print(a[0])
else:
print(a[0]-1)
|
s124033218
|
p02260
|
u432957752
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,596 | 354 |
Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini.
|
def selection(A,n):
count = 0
for i in range(0,n-1):
min = i
for j in range(i,n):
if(A[j] < A[min]):
min = j
if(min!=i):
A[i], A[min] = A[min], A[i]
count+=1
print(count)
def printList(A):
print(" ".join(str(x) for x in A))
n = int(input())
A = [int(x) for x in input().split()]
selection(A,n)
printList(A)
|
s272566380
|
Accepted
| 20 | 7,760 | 356 |
def selection(A,n):
count = 0
for i in range(0,n-1):
min = i
for j in range(i,n):
if(A[j] < A[min]):
min = j
if(min!=i):
A[i], A[min] = A[min], A[i]
count+=1
printList(A)
print(count)
def printList(A):
print(" ".join(str(x) for x in A))
n = int(input())
A = [int(x) for x in input().split()]
selection(A,n)
|
s944898761
|
p03455
|
u223060578
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,144 | 394 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
def answer(a:int, b:int) -> str:
if (a * b) % 2 == 0:
return 'Even'
else:
return 'Odd'
def test_積が偶数ならEvenとなる_入力例1():
assert answer(3, 4) == 'Even'
def test_積が奇数ならOddとなる_入力例2():
assert answer(1, 21) == 'Odd'
def test_積が偶数ならEvenとなる_最大値の場合():
assert answer(10000,10000) == 'Even'
|
s263925456
|
Accepted
| 29 | 9,116 | 91 |
a,b = map(int,input().split())
if a * b % 2 == 0:
print("Even")
else:
print("Odd")
|
s326538432
|
p03860
|
u696048490
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 58 |
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
a, s, c = map(str,input().split())
print('a' + s[0] + 'c')
|
s450385793
|
Accepted
| 17 | 2,940 | 58 |
a, s, c = map(str,input().split())
print('A' + s[0] + 'C')
|
s951141925
|
p03503
|
u126232616
| 2,000 | 262,144 |
Wrong Answer
| 85 | 3,064 | 441 |
Joisino is planning to open a shop in a shopping street. Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods. There are already N stores in the street, numbered 1 through N. You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2. Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}. Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
|
n = int(input())
f = [int(input().replace(" ","")) for i in range(n)]
p = [list(map(int,input().split())) for i in range(n)]
ans = -10**12
for i in range(1,1024):
#print("i=",i)
temp = 0
for j in range(n):
fi = f[j]
ci = bin(fi&j).count("1")
temp += p[j][ci]
#print(temp)
ans = max(ans,temp)
print(ans)
|
s673503963
|
Accepted
| 86 | 3,064 | 484 |
n = int(input())
f = [int(input().replace(" ",""),2) for i in range(n)]
p = [list(map(int,input().split())) for i in range(n)]
ans = -10**12
for i in range(1,1024):
#print("i=",i)
temp = 0
for j in range(n):
fi = f[j]
ci = bin(fi&i).count("1")
#print("ci=",ci)
temp += p[j][ci]
#print(temp)
#print(temp)
ans = max(ans,temp)
print(ans)
|
s762430443
|
p03377
|
u201660334
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,316 | 106 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
a = list(map(int, input().split()))
if a[0] <= a[2] <= a[0] + a[1]:
print("Yes")
else:
print("No")
|
s458797287
|
Accepted
| 17 | 2,940 | 115 |
a = list(map(int, input().split()))
if a[0] <= a[2] and a[2] <= a[0] + a[1]:
print("YES")
else:
print("NO")
|
s622068078
|
p03408
|
u153729035
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 182 |
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
d={}
for _ in [0]*int(input()):
s=input()
d[s]=d.get(s,0)+1
for _ in [0]*int(input()):
s=input()
d[s]=d.get(s,0)-1
print(sorted(d.items(), key=lambda x: x[1])[-1][0])
|
s577842880
|
Accepted
| 17 | 3,060 | 189 |
d={}
for _ in [0]*int(input()):
s=input()
d[s]=d.get(s,0)+1
for _ in [0]*int(input()):
s=input()
d[s]=d.get(s,0)-1
print(max(sorted(d.items(), key=lambda x: x[1])[-1][1],0))
|
s651405587
|
p00002
|
u234052535
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,640 | 149 |
Write a program which computes the digit number of sum of two integers a and b.
|
while True:
try:
line = input().split(" ")
a = int(line[0])
b = int(line[1])
print(a+b)
except:
break
|
s598403116
|
Accepted
| 20 | 7,656 | 152 |
import math
while True:
try:
line = input().split(" ")
print(int(math.log10(int(line[0])+int(line[1])))+1)
except:
break
|
s622476177
|
p00001
|
u602903519
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,468 | 119 |
There is a data which provides heights (in meter) of mountains. The data is only for ten mountains. Write a program which prints heights of the top three mountains in descending order.
|
list=list()
for i in range(10):
t=input()
list.append(int(t))
list.sort()
for i in range(3):
print(list[i])
|
s052113091
|
Accepted
| 30 | 7,536 | 131 |
list=list()
for i in range(10):
t=input()
list.append(int(t))
list.sort(reverse=True)
for i in range(3):
print(list[i])
|
s030781389
|
p03007
|
u706414019
| 2,000 | 1,048,576 |
Wrong Answer
| 2,314 | 1,681,420 | 407 |
There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.
|
import sys,math,collections,itertools,bisect
input = sys.stdin.readline
N=int(input())
A=sorted(list(map(int,input().split())))
q=collections.deque(A)
process = []
for i in range(N-2):
maxA = q.pop()
minA = q.popleft()
tmp = minA-maxA
q.insert(bisect.bisect_left(A,tmp),tmp)
process.append(list(q))
maxA = q.pop()
minA = q.popleft()
print(maxA-minA)
for pro in process:
print(*pro)
|
s860176615
|
Accepted
| 206 | 25,076 | 419 |
import sys,math,collections,itertools,bisect
input = sys.stdin.readline
N=int(input())
A=sorted(list(map(int,input().split())))
process = []
for i in range(N-2):
tmp = A.pop()
y=A.pop()
A.append(tmp)
if y>=0:
process.append([A[0],y])
A[0]-=y
else:
process.append([A[-1],y])
A[-1]-=y
process.append([A[-1],A[0]])
print(A[-1]-A[0])
for pro in process:
print(*pro)
|
s650370338
|
p03369
|
u902024288
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 92 |
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
lista = list(input())
lista2 = lista.count('x')
lista3 = 700 + (lista2 * 100 )
print(lista3)
|
s375848338
|
Accepted
| 17 | 2,940 | 92 |
lista = list(input())
lista2 = lista.count('o')
lista3 = 700 + (lista2 * 100 )
print(lista3)
|
s038300607
|
p03565
|
u566159623
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,064 | 475 |
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
sp = list(input())
t = list(input())
if len(sp)-len(t)>=0:
for i in range(len(sp)-len(t)+1):
for j in range(len(t)):
if t[j] != sp[i+j] and sp[i+j] != "?":
break
else:
for j in range(len(t)):
sp[i+j] = t[j]
print(sp)
continue
for i in range(len(sp)):
if sp[i] =="?":
sp[i] = "a"
print("".join(map(str, sp)))
else:
print("UNRESORABLE")
|
s474635781
|
Accepted
| 17 | 3,064 | 610 |
sp = list(input())
t = list(input())
boo = False
if len(sp)-len(t)>=0:
for i in reversed(range(len(sp)-len(t)+1)):
for j in reversed(range(len(t))):
if t[j] != sp[i+j] and sp[i+j] != "?":
break
else:
for j in reversed(range(len(t))):
sp[i+j] = t[j]
boo = True
break
continue
if boo:
for i in range(len(sp)):
if sp[i] =="?":
sp[i] = "a"
print("".join(map(str, sp)))
else:
print("UNRESTORABLE")
else:
print("UNRESTORABLE")
|
s153672840
|
p03679
|
u548834738
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 131 |
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
X, A, B = map(int, input().split())
if A - B >= 0:
print('delicious')
elif B - A <= X:
print('sage')
else:
print('dangerous')
|
s237809905
|
Accepted
| 17 | 2,940 | 131 |
X, A, B = map(int, input().split())
if B - A <= 0:
print('delicious')
elif B - A <= X:
print('safe')
else:
print('dangerous')
|
s570775342
|
p02261
|
u841567836
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 764 |
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def selection(l):
N = len(l)
for i in range(N):
minj = i
for j in range(i, N):
if int(l[j][1]) < int(l[minj][1]):
minj = j
if i == minj:
continue
else:
l[i], l[minj] = l[minj], l[i]
return l
def bubble(l):
N = len(l)
flag = 1
while flag:
flag = 0
for j in range(N - 1, 1 - 1, -1):
if int(l[j][1]) < int(l[j - 1][1]):
l[j], l[j + 1] = l[j + 1], l[j]
flag = 0
return l
def output(lists):
count = 0
leng = len(l)
for i in l:
count += 1
if count < leng:
print(i, end =' ')
else:
print(i)
if __name__ == '__main__':
n = int(input())
l = input().split()
l_1 = bubble(l)
l_2 = selection(l)
output(l_1)
print("Stable")
output(l_2)
if l_1 == l_2:
print("Stable")
else:
prin("Not stable")
|
s163220162
|
Accepted
| 20 | 5,616 | 854 |
def selection(l):
lists = l[:]
N = len(lists)
for i in range(N):
minj = i
for j in range(i, N):
if int(lists[j][1]) < int(lists[minj][1]):
minj = j
if i == minj:
continue
else:
lists[i], lists[minj] = lists[minj], lists[i]
return lists
def bubble(l):
lists = l[:]
N = len(lists)
flag = 1
while flag:
flag = 0
for j in range(N - 1, 1 - 1, -1):
if int(lists[j][1]) < int(lists[j - 1][1]):
lists[j], lists[j - 1] = lists[j - 1], lists[j]
flag = 1
return lists
def output(l):
count = 0
leng = len(l)
for i in l:
count += 1
if count < leng:
print(i, end =' ')
else:
print(i)
if __name__ == '__main__':
n = int(input())
l = input().split()
l_1 = bubble(l)
l_2 = selection(l)
output(l_1)
print("Stable")
output(l_2)
if l_1 == l_2:
print("Stable")
else:
print("Not stable")
|
s681720392
|
p03369
|
u482157295
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 64 |
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
a = input()
ans = 700
for i in a:
if i == "o":
ans += 100
|
s703737590
|
Accepted
| 19 | 2,940 | 74 |
a = input()
ans = 700
for i in a:
if i == "o":
ans += 100
print(ans)
|
s987980522
|
p03699
|
u497952650
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 240 |
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
N = int(input())
s = [int(input()) for _ in range(N)]
if sum(s)%10 != 0:
print(sum(s))
else:
ans = sum(s)
s.sort()
for i in s:
ans -= i
if ans%10 != 0:
print(ans)
break
print(ans)
|
s758881939
|
Accepted
| 19 | 3,060 | 204 |
N = int(input())
s = [int(input()) for _ in range(N)]
if sum(s)%10 != 0:
print(sum(s))
exit()
ans = 0
for i in s:
tmp = sum(s) - i
if tmp%10 != 0:
ans = max(ans,tmp)
print(ans)
|
s994875284
|
p02259
|
u663910047
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,704 | 265 |
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode.
|
n = int(input())
k = [int(i) for i in input().split()]
for i in range(n-1):
for l in range(n-i-1):
if k[n-2-l] > k[n-1-l]:
v = k[n-1-l]
k[n - 1 - l]=k[n-2-l]
k[n - 2 - l]=v
else:
pass
print(k)
|
s635987016
|
Accepted
| 20 | 7,720 | 314 |
n = int(input())
k = [int(i) for i in input().split()]
M = 0
for i in range(n-1):
for l in range(n-i-1):
if k[n-2-l] > k[n-1-l]:
v = k[n-1-l]
k[n - 1 - l]=k[n-2-l]
k[n - 2 - l]=v
M+=1
else:
pass
print(' '.join(map(str, k)))
print(M)
|
s180142809
|
p03474
|
u056358163
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 202 |
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
A, B = map(int, input().split())
S = input()
a = S[:A]
c = S[A]
b = S[A:]
try:
a = int(a)
b = int(b)
if c == '-':
print('yes')
else:
print('no')
except:
print('no')
|
s813017021
|
Accepted
| 18 | 3,060 | 202 |
A, B = map(int, input().split())
S = input()
a = S[:A]
c = S[A]
b = S[A:]
try:
a = int(a)
b = int(b)
if c == '-':
print('Yes')
else:
print('No')
except:
print('No')
|
s099057954
|
p03502
|
u088553842
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,084 | 93 |
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
n = int(input())
dv = 0
for s in str(n):
dv += int(s)
print('YES' if n % dv == 0 else 'NO')
|
s129765597
|
Accepted
| 25 | 9,104 | 93 |
n = int(input())
dv = 0
for s in str(n):
dv += int(s)
print('Yes' if n % dv == 0 else 'No')
|
s494593026
|
p03679
|
u033524082
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 133 |
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
X,A,B=map(int,input().split())
if A-B<0:
print("delicious")
elif X+1>B-A:
print("safe")
elif X+1<=B-A:
print("dangerous")
|
s492156288
|
Accepted
| 18 | 2,940 | 133 |
X,A,B=map(int,input().split())
if A-B>=0:
print("delicious")
elif B-A<=X:
print("safe")
elif X+1<=B-A:
print("dangerous")
|
s612569731
|
p03737
|
u389910364
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,572 | 1,164 |
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
import functools
import os
INF = float('inf')
def inp():
return int(input())
def inpf():
return float(input())
def inps():
return input()
def inl():
return list(map(int, input().split()))
def inlf():
return list(map(float, input().split()))
def inls():
return input().split()
def inpm(line):
return [inp() for _ in range(line)]
def inpfm(line):
return [inpf() for _ in range(line)]
def inpsm(line):
return [inps() for _ in range(line)]
def inlm(line):
return [inl() for _ in range(line)]
def inlfm(line):
return [inlf() for _ in range(line)]
def inlsm(line):
return [inls() for _ in range(line)]
def debug(fn):
if not os.getenv('LOCAL'):
return fn
@functools.wraps(fn)
def wrapper(*args, **kwargs):
print('DEBUG: {}({}) -> '.format(
fn.__name__,
', '.join(
list(map(str, args)) +
['{}={}'.format(k, str(v)) for k, v in kwargs.items()]
)
), end='')
ret = fn(*args, **kwargs)
print(ret)
return ret
return wrapper
a, b, c = inls()
print(a[0] + b[0]+c[0])
|
s155791892
|
Accepted
| 23 | 3,572 | 1,175 |
import functools
import os
INF = float('inf')
def inp():
return int(input())
def inpf():
return float(input())
def inps():
return input()
def inl():
return list(map(int, input().split()))
def inlf():
return list(map(float, input().split()))
def inls():
return input().split()
def inpm(line):
return [inp() for _ in range(line)]
def inpfm(line):
return [inpf() for _ in range(line)]
def inpsm(line):
return [inps() for _ in range(line)]
def inlm(line):
return [inl() for _ in range(line)]
def inlfm(line):
return [inlf() for _ in range(line)]
def inlsm(line):
return [inls() for _ in range(line)]
def debug(fn):
if not os.getenv('LOCAL'):
return fn
@functools.wraps(fn)
def wrapper(*args, **kwargs):
print('DEBUG: {}({}) -> '.format(
fn.__name__,
', '.join(
list(map(str, args)) +
['{}={}'.format(k, str(v)) for k, v in kwargs.items()]
)
), end='')
ret = fn(*args, **kwargs)
print(ret)
return ret
return wrapper
a, b, c = inls()
print((a[0] + b[0]+c[0]).upper())
|
s206610450
|
p02257
|
u299231628
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,608 | 242 |
A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.
|
def isprime(n):
if n < 2: return 0
elif n == 2: return 1
if n % 2 == 0: return 0
for i in range(3, n, 2):
if n % i == 0 : return 0
return 1
n = str([isprime(int(i)) for i in input().split()].count(1))
print(n)
|
s436709859
|
Accepted
| 380 | 6,000 | 304 |
def isprime(n):
if n < 2: return 0
elif n == 2: return 1
if n % 2 == 0: return 0
for i in range(3, n, 2):
if i > n/i: return 1
if n % i == 0 : return 0
return 1
N = int(input())
n = [int(input()) for i in range(N)]
a = [i for i in n if isprime(i)]
print(len(a))
|
s114998730
|
p02392
|
u656153606
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,632 | 141 |
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
a, b, c = [int(i) for i in input().split()]
if a > b:
if b > c:
print('Yes')
else:
print('No')
else:
print('No')
|
s445898318
|
Accepted
| 20 | 7,640 | 141 |
a, b, c = [int(i) for i in input().split()]
if a < b:
if b < c:
print('Yes')
else:
print('No')
else:
print('No')
|
s421948765
|
p03227
|
u240793404
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 52 |
You are given a string S of length 2 or 3 consisting of lowercase English letters. If the length of the string is 2, print it as is; if the length is 3, print the string after reversing it.
|
s = input()
if len(s) == 2:
s = s[::-1]
print(s)
|
s826347722
|
Accepted
| 17 | 2,940 | 52 |
s = input()
if len(s) == 3:
s = s[::-1]
print(s)
|
s200397201
|
p03971
|
u777283665
| 2,000 | 262,144 |
Wrong Answer
| 101 | 4,016 | 434 |
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
n, a, b = map(int, input().split())
s = input()
capacity = 0
f_capacity = 0
for c in s:
if c == 'a':
if capacity < a+b:
capacity += 1
print('YES')
else:
print('NO')
elif c == 'b':
if capacity < a+b and f_capacity < b:
capacity += 1
f_capacity += 1
print('YES')
else:
print('NO')
else:
print('NO')
|
s459495308
|
Accepted
| 104 | 4,016 | 381 |
n, A, B = map(int, input().split())
s = input()
ab = 0
f = 0
for t in s:
if t == "a":
if ab < A + B:
print("Yes")
ab += 1
else:
print("No")
elif t == "b":
if ab < A + B and f < B:
print("Yes")
ab += 1
f += 1
else:
print("No")
else:
print("No")
|
s500975767
|
p04029
|
u403986473
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 41 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
sum([a for a in range(1,int(input())+1)])
|
s929699691
|
Accepted
| 17 | 2,940 | 54 |
n = int(input())
print(sum([a for a in range(1,n+1)]))
|
s692248222
|
p02692
|
u608088992
| 2,000 | 1,048,576 |
Wrong Answer
| 256 | 104,516 | 1,132 |
There is a game that involves three variables, denoted A, B, and C. As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or C then subtract 1 from the other; if s_i is `BC`, you must add 1 to B or C then subtract 1 from the other. After each choice, none of A, B, and C should be negative. Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.
|
import sys
sys.setrecursionlimit(10000000)
input = sys.stdin.readline
N, A, B, C = map(int, input().split())
S = [input().strip("\n").replace("A", "0").replace("B", "1").replace("C", "2") for _ in range(N)]
Ans = [None] * N
def search(i, num):
if i == N - 1:
if num[int(S[N-1][0])] > 0:
Ans[N-1] = S[N-1][0]
return True
elif num[int(S[N-1][1])] > 0:
Ans[N-1] = S[N-1][1]
return True
else: return False
f = S[i][0]
s = S[i][1]
num[int(f)] += 1
num[int(s)] -= 1
if num[int(s)] >= 0:
if search(i+1, num):
Ans[i] = f
return True
num[int(f)] -= 2
num[int(s)] += 2
if num[int(f)] >= 0:
if search(i+1, num):
Ans[i] = s
return True
num[int(f)] += 1
num[int(s)] -= 1
return False
def solve():
if search(0, [A, B, C]):
print("Yes")
for i, a in enumerate(Ans):
if a == 0: print("A")
elif a == 1: print("B")
else: print("C")
else: print("No")
return 0
if __name__ == "__main__":
solve()
|
s655499199
|
Accepted
| 289 | 120,232 | 935 |
import sys
sys.setrecursionlimit(10000000)
input = sys.stdin.readline
N, A, B, C = map(int, input().split())
S = [input().strip("\n").replace("A", "0").replace("B", "1").replace("C", "2") for _ in range(N)]
Ans = [None] * N
def search(i, num):
if i == N:
if min(num) >= 0: return True
else: return False
if min(num) < 0: return False
L = [num[0], num[1], num[2]]
f = S[i][0]
s = S[i][1]
L[int(f)] += 1
L[int(s)] -= 1
if search(i+1, L):
Ans[i] = f
return True
L[int(f)] -= 2
L[int(s)] += 2
if search(i+1, L):
Ans[i] = s
return True
return False
def solve():
if search(0, [A, B, C]):
print("Yes")
for i, a in enumerate(Ans):
if a == "0": print("A")
elif a == "1": print("B")
else: print("C")
else: print("No")
#print(Ans)
return 0
if __name__ == "__main__":
solve()
|
s367096676
|
p02743
|
u173494138
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 305 |
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
#!/user/bin/env python3
# -*- conding: utf-8 -*-
def ans(a, b, c):
aa = a * a
bb = b * b
cc = c * c
if aa + bb < cc:
return 'Yes'
else:
return 'No'
def main():
a, b, c = map(int, input().split())
print(ans(a, b, c))
if __name__ == '__main__':
main()
|
s790766960
|
Accepted
| 17 | 3,064 | 365 |
#!/user/bin/env python3
# -*- conding: utf-8 -*-
import math
def ans(a, b, c):
rab = math.sqrt(a * b)
if c - a - b <= 0:
return 'No'
elif (c - a - b) ** 2 > 4 * a * b:
return 'Yes'
else:
return 'No'
def main():
a, b, c = map(int, input().split())
print(ans(a, b, c))
if __name__ == '__main__':
main()
|
s923720733
|
p00065
|
u043254318
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,308 | 833 |
取引先の顧客番号と取引日を月ごとに記録したデータがあります。今月のデータと先月のデータを読み込んで、先月から２ヶ月連続で取引のある会社の顧客番号と取引のあった回数を出力するプログラムを作成してください。ただし、月々の取引先数は 1,000 社以内です。
|
def get_input():
while True:
try:
yield ''.join(input())
except EOFError:
break
table = [[False for i in range(32)] for j in range(1001)]
table2 = [[False for i in range(32)] for j in range(1001)]
C = [False for i in range(1001)]
C2 = [False for i in range(1001)]
while True:
N = input()
if len(N) <= 1:
break
c,d = [int(i) for i in N.split(",")]
table[c][d] = True
print(c,d)
C[c] = True
M = list(get_input())
for l in range(len(M)):
c,d = [int(i) for i in M[l].split(",")]
table2[c][d] = True
print(c,d)
C2[c] = True
for i in range(1001):
if C[i] and C2[i]:
cnt = 0
for j in range(32):
if table[i][j]:
cnt += 1
if table2[i][j]:
cnt += 1
print(i,cnt)
|
s077011629
|
Accepted
| 30 | 6,300 | 806 |
def get_input():
while True:
try:
yield ''.join(input())
except EOFError:
break
table = [[False for i in range(32)] for j in range(1001)]
table2 = [[False for i in range(32)] for j in range(1001)]
C = [False for i in range(1001)]
C2 = [False for i in range(1001)]
while True:
N = input()
if len(N) <= 1:
break
c, d = [int(i) for i in N.split(",")]
table[c][d] = True
C[c] = True
M = list(get_input())
for l in range(len(M)):
c, d = [int(i) for i in M[l].split(",")]
table2[c][d] = True
C2[c] = True
for i in range(1001):
if C[i] and C2[i]:
cnt = 0
for j in range(32):
if table[i][j]:
cnt += 1
if table2[i][j]:
cnt += 1
print(i, cnt)
|
s745841782
|
p03963
|
u993268357
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 53 |
There are N balls placed in a row. AtCoDeer the deer is painting each of these in one of the K colors of his paint cans. For aesthetic reasons, any two adjacent balls must be painted in different colors. Find the number of the possible ways to paint the balls.
|
a, b = map(int, input().split())
print(b+(a-1)*(b-1))
|
s254419927
|
Accepted
| 17 | 2,940 | 54 |
a, b = map(int, input().split())
print(b*(b-1)**(a-1))
|
s914480384
|
p03861
|
u143903328
| 2,000 | 262,144 |
Wrong Answer
| 34 | 5,076 | 94 |
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
from decimal import *
a,b,x = map(int,input().split())
print((Decimal(b//x)-Decimal((a-1)/x)))
|
s574124486
|
Accepted
| 17 | 2,940 | 65 |
a,b,x = map(int,input().split())
print((int(b//x)-int((a-1)//x)))
|
s496344752
|
p02409
|
u513411598
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,628 | 517 |
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
n = int(input())
count = list()
for b in range(4):
count.append(list())
for f in range(3):
count[b].append(list())
for r in range(10):
count[b][f].append(0)
# count[b][f][r] = 0
read = 0
while read < n:
b,f,r,v = [int(x) for x in input().split(" ")]
count[b-1][f-1][r-1] = v
read += 1
for b in range(4):
for f in range(3):
for r in range(10):
print("", count[b][f][r], end="")
print()
print("####################")
|
s264820335
|
Accepted
| 40 | 7,696 | 546 |
n = int(input())
count = list()
for b in range(4):
count.append(list())
for f in range(3):
count[b].append(list())
for r in range(10):
count[b][f].append(0)
read = 0
while read < n:
b,f,r,v = [int(x) for x in input().split(" ")]
count[b-1][f-1][r-1] += v
read += 1
for b in range(len(count)):
for f in range(len(count[b])):
for r in range(len(count[b][f])):
print("", count[b][f][r], end="")
print()
if b < len(count) - 1:
print("####################")
|
s047430296
|
p02731
|
u580404776
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 82 |
Given is a positive integer L. Find the maximum possible volume of a rectangular cuboid whose sum of the dimensions (not necessarily integers) is L.
|
L=int(input())
ans=0
if L%3==0:
print((L/3)**3)
else:
print(((L-1)/2)**2)
|
s038145146
|
Accepted
| 17 | 2,940 | 47 |
L=int(input())
print("{:.12f}".format(L**3/27))
|
s345709401
|
p02614
|
u921617614
| 1,000 | 1,048,576 |
Wrong Answer
| 170 | 27,000 | 685 |
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
import itertools
import numpy as np
H,W,K=map(int,input().split())
m=[]
for i in range(H):
m.append(list(input()))
print(m)
ans=0
for i in range(2**H):
for j in range(2**W):
i_s=format(i, "0{0}b".format(H))
j_s=format(j, "0{0}b".format(W))
M=[[[] for _ in range(W)] for _ in range(H)]
for h in range(H):
for w in range(W):
if i_s[h]=="1" or j_s[w]=="1":
M[h][w]="$"
else:
M[h][w]=m[h][w]
M_1=list(itertools.chain.from_iterable(M))
if M_1.count("#")==K:
ans+=1
print(ans)
|
s232032547
|
Accepted
| 168 | 27,188 | 677 |
import itertools
import numpy as np
H,W,K=map(int,input().split())
m=[]
for i in range(H):
m.append(list(input()))
ans=0
for i in range(2**H):
for j in range(2**W):
i_s=format(i, "0{0}b".format(H))
j_s=format(j, "0{0}b".format(W))
M=[[[] for _ in range(W)] for _ in range(H)]
for h in range(H):
for w in range(W):
if i_s[h]=="1" or j_s[w]=="1":
M[h][w]="$"
else:
M[h][w]=m[h][w]
M_1=list(itertools.chain.from_iterable(M))
if M_1.count("#")==K:
ans+=1
print(ans)
|
s678393814
|
p03657
|
u870297120
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 119 |
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a, b = map(int, input().split())
if a%3==0 or b%3==0 or (a+b)%3==0:
print('possible')
else:
print('Impossible')
|
s492358018
|
Accepted
| 19 | 2,940 | 119 |
a, b = map(int, input().split())
if a%3==0 or b%3==0 or (a+b)%3==0:
print('Possible')
else:
print('Impossible')
|
s059036323
|
p03379
|
u780962115
| 2,000 | 262,144 |
Wrong Answer
| 216 | 25,220 | 326 |
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
#many medians
n=int(input())
B=list(map(int,input().split()))
C=B.copy()
B=sorted(B)
median1=B[n//2]
median2=B[n//2-1]
if median1==median2:
for _ in range(n):
print(median1)
elif median1<median2:
for i in range(n):
if i>=n//2:
print(median1)
elif i<n//2:
print(median2)
|
s596794607
|
Accepted
| 318 | 26,772 | 324 |
#many medians
n=int(input())
B=list(map(int,input().split()))
C=B.copy()
B=sorted(B)
median1=B[n//2]
median2=B[n//2-1]
if median1==median2:
for _ in range(n):
print(median1)
elif median1>median2:
for i in range(n):
if C[i]>=median1:
print(median2)
else:
print(median1)
|
s330168210
|
p03339
|
u121161758
| 2,000 | 1,048,576 |
Wrong Answer
| 307 | 72,380 | 757 |
There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions.
|
N = int(input())
S = list(input())
L = {}
L[0] = -1 if S[0] == "W" else -1
for i in range(1,N):
L[i] = L[i-1] - 1 if S[i] == "W" else L[i-1] + 1
R = {}
R[N-1] = -1 if S[N-1] == "W" else -1
for i in range(N-2, -1 , -1):
R[i] = R[i+1] -1 if S[i] == "W" else R[i+1] + 1
#print(L)
#print(R)
ans_key = 0
for i in range(N-1):
if L[i] - R[i + 1] >= ans_key:
ans_key = i
if L[N-1] > ans_key:
ans_key = L[N-1]
elif R[0] > ans_key:
ans_key = R[0]
ans = 0
l_num = (ans_key - (-1 * L[ans_key-1]) if L[i-1] < 0 else 0) //2 + (-1 * L[ans_key-1]) if L[i-1] < 0 else 0
r_num = ((N - ans_key -1 + R[ans_key + 1] if R[ans_key + 1] > 0 else 0 )) // 2 + R[ans_key + 1] if R[ans_key + 1] > 0 else 0
#print(ans_key)
print(l_num + r_num)
|
s968395928
|
Accepted
| 473 | 90,144 | 722 |
N = int(input())
S = list(input())
L_count = {}
L_count[0] = 0
for i in range(1,N):
if S[i-1] == "W":
L_count[i] = L_count[i-1] + 1
else:
L_count[i] = L_count[i-1]
R_count = {}
R_count[N-1] = 0
for i in range(N-2,-1,-1):
if S[i+1] == "E":
R_count[i] = R_count[i+1] + 1
else:
R_count[i] = R_count[i+1]
leader_i = 0
turn_num = N+1
for i in range(N):
#print(L_count[i] + R_count[i],"i is",i,"leader_i is",leader_i)
if L_count[i] + R_count[i] < turn_num:
leader_i = i
turn_num = L_count[i] + R_count[i]
R_sort = sorted(R_count.items(), key=lambda x: x[0])
#print(L_count)
#print(R_sort)
#print(leader_i)
print(L_count[leader_i] + R_count[leader_i])
|
s554755387
|
p02972
|
u625729943
| 2,000 | 1,048,576 |
Wrong Answer
| 285 | 9,812 | 404 |
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
N = int(input())
A = list(map(int, input().split()))
chk = [0]*N
print(N, A)
A.reverse()
for i, a in enumerate(A):
print('i', i)
sum_chk = sum(chk[N-i-1::N-i])
if sum_chk == a-1:
chk[N-i-1] = 1
continue
elif sum_chk == a:
continue
else:
print(-1)
exit()
M = sum(chk)
b = [i+1 for i, v in enumerate(chk) if v==1]
print(M)
if M != 0: print(*b)
|
s270161630
|
Accepted
| 253 | 14,516 | 525 |
import sys
def input():
return sys.stdin.readline()[:-1]
N = int(input())
A = list(map(int, input().split()))
chk = [0]*N
A.reverse()
for i, a in enumerate(A):
if i+1 <= N//2:
if a:
chk[N-i-1] = 1
else:
sum_chk = sum(chk[N-i-1::N-i])
s2 = sum_chk%2
if a:
if s2 == 0:
chk[N-i-1] = 1
else:
if s2 == 1:
chk[N-i-1] = 1
M = sum(chk)
b = [i+1 for i, v in enumerate(chk) if v==1]
print(M)
if M != 0: print(*b)
|
s614841299
|
p04011
|
u976225138
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 157 |
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
n, k, x, y = map(int, open(0).read().split())
ans = 0
while n < 0:
if n <= k:
ans += x
else:
ans += y
n -= 1
else:
print(ans)
|
s324749992
|
Accepted
| 19 | 2,940 | 153 |
n, k, x, y = [int(input()) for _ in range(4)]
ans = 0
while n:
if n <= k:
ans += x
else:
ans += y
n -= 1
else:
print(ans)
|
s369284941
|
p03855
|
u994064513
| 2,000 | 262,144 |
Wrong Answer
| 1,310 | 80,136 | 1,654 |
There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities. We will say city A and B are _connected by roads_ if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define _connectivity by railways_ similarly. For each city, find the number of the cities connected to that city by both roads and railways.
|
from collections import deque
N, K, L = [int(s) for s in input().split()]
edge_road = [[int(s) - 1 for s in input().split()] for _ in range(K)]
edge_train = [[int(s) - 1 for s in input().split()] for _ in range(L)]
graph_road = [[] for _ in range(N)]
graph_train = [[] for _ in range(N)]
for i, j in edge_road:
graph_road[i].append(j)
graph_road[j].append(i)
for i, j in edge_train:
graph_train[i].append(j)
graph_train[j].append(i)
completed = [False] * N
road = {}
train = {}
group_road = [0] * N
group_num = 0
visited = [False]*N
for i in range(N):
if not visited[i]:
continue
dq = deque([i])
while(dq):
search_obj = dq.pop()
for candidate_obj in graph_road[search_obj]:
if not visited[candidate_obj]:
group_road[candidate_obj] = group_num
dq.appendleft(search_obj)
visited[candidate_obj] = True
group_num += 1
group_train = [0] * N
group_num = 0
visited = [False]*N
for i in range(N):
if not visited[i]:
continue
dq = deque([i])
while(dq):
search_obj = dq.pop()
for candidate_obj in graph_train[search_obj]:
if not visited[candidate_obj]:
group_train[candidate_obj] = group_num
dq.appendleft(search_obj)
visited[candidate_obj] = True
group_num += 1
group_count = {}
for gr, gt in zip(group_road, group_train):
if (gr, gt) in group_count.keys():
group_count[(gr, gt)] += 1
else:
group_count[(gr, gt)] = 1
for gr, gt in zip(group_road, group_train):
print(group_count[(gr, gt)] - 1, end=' ')
|
s505966732
|
Accepted
| 1,415 | 127,784 | 1,330 |
N, M, L = [int(s) for s in input().split()]
parent_road = list(range(N))
parent_train = list(range(N))
rank_road = [0] * N
rank_train = [0] * N
class UnionFind:
def __init__(self, n):
self.parent = list(range(n))
self.rank = [0] * N
def find(self, x):
if x == self.parent[x]:
return x
else:
self.parent[x] = self.find(self.parent[x])
return self.parent[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if self.rank[x] < self.rank[y]:
self.parent[x] = y
else:
self.parent[y] = x
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
edge_road = [[int(s) - 1 for s in input().split()] for _ in range(M)]
edge_train = [[int(s) - 1 for s in input().split()] for _ in range(L)]
group_road = UnionFind(N)
group_train = UnionFind(N)
for x, y in edge_road:
group_road.union(x, y)
for x, y in edge_train:
group_train.union(x, y)
group_count = {}
group_list = []
for i in range(N):
gr = group_road.find(i)
gt = group_train.find(i)
group_list.append((gr, gt))
if (gr, gt) in group_count.keys():
group_count[(gr, gt)] += 1
else:
group_count[(gr, gt)] = 1
print(' '.join([str(group_count[g]) for g in group_list]))
|
s491107289
|
p03161
|
u690037900
| 2,000 | 1,048,576 |
Wrong Answer
| 2,109 | 23,036 | 523 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
import sys
import numpy as np
input = sys.stdin.readline
n, k = map(int, input().split())
h = [int(i) for i in input().split()]
dp = np.zeros(n, dtype=int)
h = np.array(h)
for i in range(1, n):
start = max(0, i-k)
print(start)
dp[i] = min(dp[start: i] + np.abs(h[i] - h[start: i]))
print(dp)
print(dp[n-1])
|
s895600183
|
Accepted
| 1,890 | 23,056 | 492 |
import sys
import numpy as np
input = sys.stdin.readline
n, k = map(int, input().split())
h = [int(i) for i in input().split()]
dp = np.zeros(n, dtype=int)
h = np.array(h)
for i in range(1, n):
start = max(0, i-k)
dp[i] = min(dp[start: i] + np.abs(h[i] - h[start: i]))
print(dp[n-1])
|
s387926294
|
p03471
|
u075303794
| 2,000 | 262,144 |
Wrong Answer
| 293 | 9,204 | 302 |
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
import sys
N,Y=map(int,input().split())
if 10_000*N < Y:
print(-1,-1,-1)
sys.exit()
elif 1_000*N > Y:
print(-1,-1,-1)
sys.exit()
else:
Y-=1_000*N
for i in range(1,2001):
for j in range(2000-i+1):
if 4_000*i+10_000*j==Y:
print(N-i-j,i,j)
sys.exit()
print(-1,-1,-1)
|
s282527302
|
Accepted
| 621 | 9,200 | 352 |
import sys
N,Y=map(int,input().split())
if 10_000*N < Y:
print(-1,-1,-1)
sys.exit()
elif 1_000*N > Y:
print(-1,-1,-1)
sys.exit()
else:
Y-=1_000*N
for i in range(2001):
for j in range(2000-i+1):
if 4_000*i+9_000*j>Y:
break
if i+j<=N and 4_000*i+9_000*j==Y:
print(j,i,N-i-j)
sys.exit()
print(-1,-1,-1)
|
s042294718
|
p03697
|
u357630630
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 81 |
You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead.
|
A, B = map(int, input().split())
print(A + B) if A + B >= 10 else print("error")
|
s954948060
|
Accepted
| 17 | 2,940 | 80 |
A, B = map(int, input().split())
print(A + B) if A + B < 10 else print("error")
|
s888055721
|
p02255
|
u150984829
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 141 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
n=int(input())
a=input().split()
for i in range(n):
k=int(a[i])
j=i-1
while j>=0 and int(a[j])>k:
a[i]=a[j]
j-=1
a[j+1]=k
print(*a)
|
s810955077
|
Accepted
| 20 | 5,976 | 119 |
N = int(input())
A = list(map(int, input().split()))
for i in range(1, N + 1):
A[:i] = sorted(A[:i])
print(*A)
|
s913815664
|
p03455
|
u847033024
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,028 | 90 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
if (a + b) % 2 == 0:
print('Even')
else:
print('Odd')
|
s024498132
|
Accepted
| 29 | 9,028 | 89 |
a, b = map(int, input().split())
if a * b % 2 == 0:
print('Even')
else:
print('Odd')
|
s718053634
|
p03407
|
u992759582
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 87 |
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
a,b,c = map(int,input().split())
if a + c <= c:
print('Yes')
else:
print('No')
|
s057902689
|
Accepted
| 17 | 2,940 | 87 |
a,b,c = map(int,input().split())
if a + b >= c:
print('Yes')
else:
print('No')
|
s506477135
|
p03795
|
u928784113
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 94 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
# -*- coding: utf-8 -*-
N = int(input())
M = N * 800
L = int(N / 15)
print("{}".format(M - L))
|
s601003237
|
Accepted
| 18 | 2,940 | 228 |
# -*- coding: utf-8 -*-
N = int(input())
M = N * 800
L = int(N // 15)
print("{}".format(M - 200*L))
|
s833983397
|
p03644
|
u934442292
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 109 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
N = int(input())
ans = 0
i = 0
while True:
if 2**i <= N:
ans = i
else:
break
i += 1
print(ans)
|
s412647150
|
Accepted
| 17 | 2,940 | 112 |
N = int(input())
ans = 0
i = 0
while True:
if 2**i <= N:
ans = 2**i
else:
break
i += 1
print(ans)
|
s001898788
|
p03434
|
u105302073
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 215 |
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
N = int(input())
a = list(map(int, input().split()))
sorted_a = sorted(a)
alice = 0
bob = 0
for i in range(N):
if i % 2 == 0:
alice += sorted_a[i]
else:
bob += sorted_a[i]
print(alice - bob)
|
s816629918
|
Accepted
| 17 | 3,060 | 230 |
N = int(input())
a = list(map(int, input().split()))
sorted_a = list(reversed(sorted(a)))
alice = 0
bob = 0
for i in range(N):
if i % 2 == 0:
alice += sorted_a[i]
else:
bob += sorted_a[i]
print(alice - bob)
|
s358909032
|
p02613
|
u033893324
| 2,000 | 1,048,576 |
Wrong Answer
| 142 | 16,184 | 200 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n=int(input())
a = [input() for i in range(n)]
A=str(a.count("AC"))
B=str(a.count("WA"))
C=str(a.count("TLE"))
D=str(a.count("RE"))
print("AC x"+ A)
print("WA x"+ B)
print("TLE x"+ C)
print("RE x"+ D)
|
s540008426
|
Accepted
| 142 | 16,300 | 204 |
n=int(input())
a = [input() for i in range(n)]
A=str(a.count("AC"))
B=str(a.count("WA"))
C=str(a.count("TLE"))
D=str(a.count("RE"))
print("AC x "+ A)
print("WA x "+ B)
print("TLE x "+ C)
print("RE x "+ D)
|
s766159671
|
p03068
|
u497952650
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,156 | 132 |
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
N = int(input())
S = input()
s = S[int(input())-1]
ans = ""
for i in s:
if i == s:
ans += s
else:
ans += "*"
print(ans)
|
s353130382
|
Accepted
| 23 | 9,152 | 133 |
N = int(input())
S = input()
s = S[int(input())-1]
ans = ""
for i in S:
if i == s:
ans += s
else:
ans += "*"
print(ans)
|
s003049762
|
p03434
|
u316095188
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 155 |
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
N =int(input())
a = list(map(int, input().split()))
b = sorted(a,reverse=True)
Alice = sum(b[::2])
Bob = sum(b[1::2])
ans = Alice - Bob
print(ans)
print(b)
|
s020694770
|
Accepted
| 17 | 3,060 | 146 |
N =int(input())
a = list(map(int, input().split()))
b = sorted(a,reverse=True)
Alice = sum(b[::2])
Bob = sum(b[1::2])
ans = Alice - Bob
print(ans)
|
s481399309
|
p03129
|
u504562455
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 92 |
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
n, k = [int(i) for i in input().split()]
if n//k > 1:
print("YES")
else:
print("NO")
|
s802704864
|
Accepted
| 17 | 2,940 | 98 |
n, k = [int(i) for i in input().split()]
if n >= (k-1)*2+1:
print("YES")
else:
print("NO")
|
s079885416
|
p03573
|
u054782559
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,172 | 90 |
You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers.
|
a, b, c=map(int,input().split())
if a<b:
print(a)
elif a==b:
print(c)
else:
print(b)
|
s998945895
|
Accepted
| 28 | 9,108 | 91 |
a, b, c=map(int,input().split())
if a==b:
print(c)
elif c==b:
print(a)
else:
print(b)
|
s218713674
|
p03303
|
u238876549
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,188 | 194 |
You are given a string S consisting of lowercase English letters. We will write down this string, starting a new line after every w letters. Print the string obtained by concatenating the letters at the beginnings of these lines from top to bottom.
|
strings=[input() for string in range(2)]
string=strings[0]
num=int(strings[1])
i=0
check=True
while check:
s=string[i:i+num]
if len(s)<num:
check=False
print(s)
i=i+num
|
s011200754
|
Accepted
| 19 | 3,060 | 192 |
strings=[input() for string in range(2)]
string=strings[0]
num=int(strings[1])
result=""
for i in range(len(string)):
if i% num==0:
result=result+string[i]
i=i+1
print(result)
|
s954177764
|
p03251
|
u785505707
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 193 |
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
head=[int(n) for n in input().split()]
list_x=[int(n) for n in input().split()]
list_y=[int(n) for n in input().split()]
if max(list_x) < min(list_y):
print("War")
else:
print("No War")
|
s664439973
|
Accepted
| 17 | 3,060 | 239 |
head=[int(n) for n in input().split()]
list_x=[int(n) for n in input().split()]
list_x.append(head[2])
list_y=[int(n) for n in input().split()]
list_y.append(head[3])
if max(list_x) < min(list_y):
print("No War")
else:
print("War")
|
s233333102
|
p02853
|
u600588566
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 170 |
We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.
|
x,y=map(int,input().split())
if x<4:
score=40000-x*10000
else:
score=0
if y<4:
score+=40000-y*10000
else:
score+=0
if score==60000:
score=100000
print(score)
|
s799816092
|
Accepted
| 19 | 3,060 | 102 |
x,y=map(int,input().split())
score = max(4-x,0)+max(4-y,0)
if score==6:
score+=4
print(score*100000)
|
s189538286
|
p03470
|
u252828980
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 52 |
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
n,*L = map(int,open(0).read().split())
print(set(L))
|
s662799993
|
Accepted
| 17 | 2,940 | 57 |
n,*L = map(int,open(0).read().split())
print(len(set(L)))
|
s851388535
|
p03408
|
u736479342
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,168 | 221 |
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
n = int(input())
s = [input() for i in range(n)]
m = int(input())
t = [input() for j in range(m)]
S = list(set(s))
T = list(set(t))
for x in range(len(S)):
a = s.count(S[x]) - t.count(S[x])
ans = max(0, a)
print(ans)
|
s804529648
|
Accepted
| 23 | 9,040 | 245 |
n = int(input())
s = [input() for i in range(n)]
m = int(input())
t = [input() for j in range(m)]
S = list(set(s))
T = list(set(t))
b = []
for x in range(len(S)):
a = s.count(S[x]) - t.count(S[x])
b.append(a)
ans = max(0, max(b))
print(ans)
|
s454078027
|
p03370
|
u819135704
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 146 |
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
N, X = map(int, input().split())
m_list = [int(input()) for i in range(N)]
Sum = sum(m_list)
Min = min(m_list)
a = (N - Sum) // Min
print(N + a)
|
s600254827
|
Accepted
| 17 | 2,940 | 146 |
N, X = map(int, input().split())
m_list = [int(input()) for i in range(N)]
Sum = sum(m_list)
Min = min(m_list)
a = (X - Sum) // Min
print(N + a)
|
s090769515
|
p03455
|
u955481356
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 95 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = (int(i) for i in input().split())
if (a*b)%2 == 0:
print('even')
else:
print('odd')
|
s055837124
|
Accepted
| 18 | 2,940 | 96 |
a, b = (int(i) for i in input().split())
if (a*b)%2 == 0:
print('Even')
else:
print('Odd')
|
s112168955
|
p03456
|
u043877190
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 108 |
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
a,b=input().split()
n=a+b
for i in range(1,101):
i+=1
if n==i**2:
print("Yes")
else:
print("No")
|
s049837706
|
Accepted
| 17 | 2,940 | 140 |
a,b=input().split()
n=a+b
for i in range(1,1000):
n=int(n)
if n==i**2:
print("Yes")
break
else:
print("No")
|
s105629202
|
p03860
|
u999893056
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 28 |
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
print(input().split()[1][0])
|
s657342641
|
Accepted
| 17 | 2,940 | 25 |
print("A"+input()[8]+"C")
|
s395645495
|
p02936
|
u017415492
| 2,000 | 1,048,576 |
Wrong Answer
| 1,459 | 49,592 | 353 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
n, q = map(int, input().split())
arr = [[] for _ in range(n)]
for _ in range(n-1):
a, b = map(int, input().split())
arr[a-1].append(b)
ans = [0 for _ in range(n)]
for _ in range(q):
p, x = map(int, input().split())
ans[p-1] += x
for i, c in enumerate(ans):
for j in arr[i]:
ans[j-1] += c
print(ans)
|
s146536979
|
Accepted
| 1,722 | 56,088 | 646 |
from collections import deque
n,q=map(int,input().split())
edge=[[] for i in range(n+1)]
for i in range(n-1):
a,b=map(int,input().split())
edge[a].append(b)
edge[b].append(a)
x_list=[0]*(n+1)
for i in range(q):
p,x1=map(int,input().split())
x_list[p]+=x1
def bfs(x,x_list):
qu=deque([])
qu.append(0)
qu.append(x)
node_number=1
while qu:
pre=qu.popleft()
node_number = qu.popleft()
nodes = edge[node_number]
for i in nodes:
if pre != i:
qu.append(node_number)
qu.append(i)
x_list[i]+=x_list[node_number]
else:continue
return x_list
answer=bfs(1,x_list)[1:]
print(*answer)
|
s643043047
|
p03944
|
u469953228
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 753 |
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
L = list(map(int,input().split()))
w = L[0]
h = L[1]
n = L[2]
X = list(range(w+1))
Y = list(range(h+1))
X_1 = [0]
Y_1 = [0]
X_2 = [w]
Y_2 = [h]
for i in range(n):
li = list(map(int,input().split()))
if li[2] == 1 and li[0] > max(X_1):
X = [n for n in X if n >= li[0]]
X_1.append(li[0])
elif li[2] == 2 and li[0] < min(X_2):
X = [n for n in X if n <= li[0]]
X_2.append(li[0])
elif li[2] == 3 and li[1] > max(Y_1):
Y = [n for n in Y if n >= li[1]]
Y_1.append(li[1])
elif li[2] == 4 and li[1] < min(Y_2):
Y = [n for n in Y if n <= li[1]]
Y_2.append(li[1])
if X == [] or Y == []:
print(0)
else:
if X[0] == 0:
X.remove(0)
if Y[0] == 0:
Y.remove(0)
print((max(X)-min(X))*(max(Y)-min(Y)))
|
s286697462
|
Accepted
| 17 | 3,064 | 293 |
W, H, N = map(int,input().split())
l = 0
r = W
u = H
d = 0
for i in range(N):
x, y, a = map(int,input().split())
if a == 1:
l = max(x,l)
elif a == 2:
r = min(x,r)
elif a == 3:
d = max(y,d)
else:
u = min(y,u)
print(max((r-l),0)*max((u-d),0))
|
s081765657
|
p02396
|
u075836834
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,580 | 100 |
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
l = list(map(int,input().split()))
for i in range(1,len(l)+1):
print("Case"+str(i)+":"+str(l[i-1]))
|
s830339784
|
Accepted
| 130 | 7,616 | 91 |
cnt = 1
while 1:
x=int(input())
if x == 0:
break;
print("Case %d: %d"%(cnt,x))
cnt+=1
|
s518138432
|
p02578
|
u907676137
| 2,000 | 1,048,576 |
Wrong Answer
| 161 | 32,192 | 228 |
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
N = int(input())
numbers = input().split()
numbers = list(map(int,numbers))
m = 0
gross_count = 0
for n in numbers:
if m > n:
gross_count += (m-n)
print(gross_count)
else:
m = n
print(gross_count)
|
s280141049
|
Accepted
| 107 | 32,192 | 201 |
N = int(input())
numbers = input().split()
numbers = list(map(int,numbers))
m = 0
gross_count = 0
for n in numbers:
if m > n:
gross_count += (m-n)
else:
m = n
print(gross_count)
|
s461089094
|
p02394
|
u677096240
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 142 |
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
w, h, x, y, r = map(int, input().split())
if 0 <= (x-r) and (x+r) <= w and 0 <= (y-r) and (y+r) <= h:
print("yes")
else:
print("No")
|
s366374910
|
Accepted
| 20 | 5,596 | 142 |
w, h, x, y, r = map(int, input().split())
if 0 <= (x-r) and (x+r) <= w and 0 <= (y-r) and (y+r) <= h:
print("Yes")
else:
print("No")
|
s387464094
|
p03455
|
u142510648
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 103 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
def even(input):
return 'Even' if int(input.split()[0]) * int(input.split()[1]) % 2 == 0 else 'Odd'
|
s821707236
|
Accepted
| 18 | 2,940 | 111 |
list = list(map(int, input().split()))
if list[0] * list[1] % 2 == 0:
print('Even')
else:
print('Odd')
|
s873005871
|
p00340
|
u623996423
| 1,000 | 262,144 |
Wrong Answer
| 20 | 5,600 | 124 |
アイヅ放送協会の教育番組(ＡＨＫ教育)では、子ども向けの工作番組「あそんでつくろ」を放送しています。今日は棒で長方形を作る回ですが、用意した４本の棒を使って長方形ができるかを確かめたいと思います。ただし、棒は切ったり折ったりしてはいけません。 ４本の棒の長さが与えられるので、それらすべてを辺とする長方形が作れるかどうか判定するプログラムを作成せよ。
|
a = list(map(int, input().split()))
print('yes' if ((a[0] in a) and (a[1] in a) and (a[2] in a) and (a[3] in a)) else 'no')
|
s493720831
|
Accepted
| 20 | 5,544 | 57 |
print('yes' if len(set(input().split())) <= 2 else 'no')
|
s418423820
|
p04029
|
u683956577
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 74 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N = int(input())
print(N)
x = 0
for i in range(1, N+1):
x = x+i
print(x)
|
s634748615
|
Accepted
| 17 | 2,940 | 65 |
N = int(input())
x = 0
for i in range(1, N+1):
x = x+i
print(x)
|
s151869771
|
p02694
|
u727072969
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 9,168 | 148 |
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
X = int(input())
n = 100
i = 0
while n < X:
n *= 1.01
A = math.floor(n)
i += 1
if A > X:
print(i)
break
|
s242851714
|
Accepted
| 27 | 9,064 | 139 |
X = int(input())
n = 100
A = 100
i = 0
while n < X:
n = A*1.01
A = int(n)
i += 1
if A >= X:
print(i)
break
|
s890587053
|
p04045
|
u892487306
| 2,000 | 262,144 |
Wrong Answer
| 581 | 3,316 | 455 |
Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money. Find the amount of money that she will hand to the cashier.
|
from collections import Counter
def ok(n, D):
counter = Counter(str(n))
for d in D:
if counter[str(d)] > 0:
return False
return True
def main():
N, K = list(map(int, input().split(' ')))
D = list(map(int, input().split(' ')))
print(ok(1334, [0, 1, 2, 4]))
while not ok(N, D):
N += 1
print(N)
if __name__ == '__main__':
main()
|
s092697360
|
Accepted
| 561 | 3,316 | 421 |
from collections import Counter
def ok(n, D):
counter = Counter(str(n))
for d in D:
if counter[str(d)] > 0:
return False
return True
def main():
N, K = list(map(int, input().split(' ')))
D = list(map(int, input().split(' ')))
while not ok(N, D):
N += 1
print(N)
if __name__ == '__main__':
main()
|
s628506533
|
p03457
|
u152589615
| 2,000 | 262,144 |
Wrong Answer
| 459 | 12,580 | 363 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
# coding: utf-8
import sys
N = int(input())
t = [0]
x = [0]
y = [0]
for i in range(N):
txy = [int(x) for x in input().split()]
t.append(txy[0])
x.append(txy[1])
y.append(txy[2])
for i in range(N - 1):
d = abs(x[i + 1] - x[i]) + abs(y[i + 1] - y[i])
print(d)
dt = t[i + 1] - t[i]
if dt < d or (dt - d) % 2 == 1:
print("No")
sys.exit()
print("Yes")
|
s907024756
|
Accepted
| 417 | 11,816 | 347 |
# coding: utf-8
import sys
N = int(input())
t = [0]
x = [0]
y = [0]
for i in range(N):
txy = [int(x) for x in input().split()]
t.append(txy[0])
x.append(txy[1])
y.append(txy[2])
for i in range(N):
d = abs(x[i + 1] - x[i]) + abs(y[i + 1] - y[i])
dt = t[i + 1] - t[i]
if dt < d or dt % 2 != d % 2:
print("No")
sys.exit()
print("Yes")
|
s153836893
|
p02678
|
u263660661
| 2,000 | 1,048,576 |
Wrong Answer
| 1,482 | 53,788 | 656 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
n, m = map(int, input().split())
a, b = list(), list()
for i in range(m):
tmp = list(map(int, input().split()))
a.append(tmp[0])
b.append(tmp[1])
graph = [[] for _ in range(n)]
for i in range(m):
graph[a[i] - 1].append(b[i] - 1)
graph[b[i] - 1].append(a[i] - 1)
print(graph)
pathway = [None for _ in range(n)]
queue = [0]
depth = [-1]*n
depth[0] = 0
pre = [-1]*n
while queue:
now = queue.pop(0)
for i in graph[now]:
if depth[i] == -1:
if pre[i] == -1:
pre[i] = now
depth[i] = depth[now] + 1
queue.append(i)
print("Yes")
for i in range(1, n):
print(pre[i] + 1)
|
s575296313
|
Accepted
| 1,446 | 51,088 | 643 |
n, m = map(int, input().split())
a, b = list(), list()
for i in range(m):
tmp = list(map(int, input().split()))
a.append(tmp[0])
b.append(tmp[1])
graph = [[] for _ in range(n)]
for i in range(m):
graph[a[i] - 1].append(b[i] - 1)
graph[b[i] - 1].append(a[i] - 1)
pathway = [None for _ in range(n)]
queue = [0]
depth = [-1]*n
depth[0] = 0
pre = [-1]*n
while queue:
now = queue.pop(0)
for i in graph[now]:
if depth[i] == -1:
if pre[i] == -1:
pre[i] = now
depth[i] = depth[now] + 1
queue.append(i)
print("Yes")
for i in range(1, n):
print(pre[i] + 1)
|
s495014882
|
p03593
|
u751663243
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,064 | 555 |
We have an H-by-W matrix. Let a_{ij} be the element at the i-th row from the top and j-th column from the left. In this matrix, each a_{ij} is a lowercase English letter. Snuke is creating another H-by-W matrix, A', by freely rearranging the elements in A. Here, he wants to satisfy the following condition: * Every row and column in A' can be read as a palindrome. Determine whether he can create a matrix satisfying the condition.
|
H,W = map(int,input().strip().split())
lines = []
for i in range(H):
lines.append(input().strip())
chars = {}
for line in lines:
for c in line:
if c not in chars:
chars[c]=1
chars[c]+=1
def judge(H,W,chars):
ones = 1
twos = H+W-2
for c in chars:
if ones<0 or twos<0:
return False
if chars[c]%4==0:
continue
if chars[c]%2==0:
twos-=1
else:
ones-=1
return True
if judge(H, W, chars):
print("Yes")
else:
print("No")
|
s620660732
|
Accepted
| 19 | 3,064 | 841 |
H,W = map(int,input().strip().split())
lines = []
for i in range(H):
lines.append(input().strip())
chars = {}
for line in lines:
for c in line:
if c not in chars:
chars[c]=0
chars[c]+=1
def judge(H,W,chars):
if H%2 + W%2==2:
ones = 1
twos = (H + W - 2)//2
elif H%2 + W%2==0:
ones = 0
twos = 0
elif H%2 == 1:
ones = 0
twos = W//2
else:
ones = 0
twos = H//2
for c in chars:
if ones<0 or twos<0:
return False
chars[c] %= 4
if chars[c]==0:
continue
if chars[c] == 2:
twos-=1
elif chars[c] == 3:
twos-=1
ones-=1
else:
ones-=1
return True
if judge(H, W, chars):
print("Yes")
else:
print("No")
|
s189139681
|
p02928
|
u578850957
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 693 |
We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.
|
#N,K = map(int,input().split())
N,K = map(int,'10 998244353'.split())
#Alist = list(map(int,input().split()))
Alist = list(map(int,'10 9 8 7 5 6 3 4 2 1'.split()))
dabuleAlist = Alist + Alist
beforeTnumber = [0]*N
afterTnumber = [0]*N
beforesum = 0
aftersum = 0
mod_num = 10**9+7
for now in range(N):
for before in range(now):
if Alist[before] < Alist[now]:
beforeTnumber[now] += 1
for after in range(now,N):
if Alist[after] < Alist[now]:
afterTnumber[now] += 1
for i in range(N):
beforesum += beforeTnumber[i]
aftersum += afterTnumber[i]
Ksum = ((K/2)*(K+1))%mod_num
print((aftersum*(Ksum-1) + beforesum*(Ksum))%mod_num)
|
s061114915
|
Accepted
| 791 | 3,188 | 727 |
N,K = map(int,input().split())
#N,K = map(int,'10 998244353'.split())
#N,K = map(int,'10 2'.split())
Alist = list(map(int,input().split()))
#Alist = list(map(int,'10 9 8 7 5 6 3 4 2 1'.split()))
beforeTnumber = [0]*N
afterTnumber = [0]*N
beforesum = 0
aftersum = 0
mod_num = 10**9+7
for now in range(N):
for before in range(now):
if Alist[before] < Alist[now]:
beforeTnumber[now] += 1
for after in range(now,N):
if Alist[after] < Alist[now]:
afterTnumber[now] += 1
beforesum = sum(beforeTnumber)
aftersum = sum(afterTnumber)
Kaftersum = ((K*(K+1))//2)%mod_num
Kbeforesum = ((K*(K-1))//2)%mod_num
print(int(((aftersum)*(Kaftersum)+(beforesum*Kbeforesum))%mod_num))
|
s026457072
|
p02612
|
u863371419
| 2,000 | 1,048,576 |
Wrong Answer
| 37 | 9,144 | 30 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(n%1000)
|
s312239782
|
Accepted
| 29 | 9,148 | 113 |
n = int(input())
if n%1000 == 0:
print(0)
else:
ans1 = n//1000+1
ans2 = ans1*1000
print(ans2-n)
|
s874574171
|
p03359
|
u934868410
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 47 |
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a,b = map(int,input().split())
print(a - b < a)
|
s398599154
|
Accepted
| 17 | 2,940 | 49 |
a,b = map(int,input().split())
print(a - (b < a))
|
s627742890
|
p03612
|
u390694622
| 2,000 | 262,144 |
Wrong Answer
| 53 | 13,880 | 165 |
You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this.
|
N = int(input())
a = list(map(int,input().split()))
ans = 0
for i in range(N):
if i == a[i]:
ans += 1
if i > 0 and i-1 == a[i-1]:
ans -= 1
print(ans)
|
s059461976
|
Accepted
| 61 | 14,008 | 226 |
N = int(input())
a = list(map(int,input().split()))
ans = 0
left = False
for i in range(N):
if i+1 == a[i]:
if left:
left = False
else:
ans += 1
left = True
else:
left = False
print(ans)
|
s554006915
|
p03643
|
u477696265
| 2,000 | 262,144 |
Wrong Answer
| 30 | 8,984 | 44 |
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
a = input()
b = ('abc')
c = b + a
print( c )
|
s818858388
|
Accepted
| 31 | 9,044 | 44 |
a = input()
b = ('ABC')
c = b + a
print( c )
|
s716523202
|
p02601
|
u327421986
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,180 | 339 |
M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.
|
import sys
input = sys.stdin.readline
def main(a, b, c, k):
for i in range(k):
if b <= a: b*= 2
elif c <= b: c *= 2
print(a, b, c)
if b > a and c > b:
return "Yes"
return "No"
a, b, c = list(map(int, input().split()))
k = int(input())
print(main(a, b, c, k))
|
s814535380
|
Accepted
| 27 | 9,180 | 320 |
import sys
input = sys.stdin.readline
def main(a, b, c, k):
for i in range(k):
if b <= a: b*= 2
elif c <= b: c *= 2
if b > a and c > b:
return "Yes"
return "No"
a, b, c = list(map(int, input().split()))
k = int(input())
print(main(a, b, c, k))
|
s273580831
|
p03139
|
u850266651
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 148 |
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
li = [int(i) for i in input().split()]
n = li[0]
a = li[1]
b = li[2]
res1 = max(a,b)
if a+b-n<0:
res2 =0
else:
res2=a+b-n
print(res1, res2)
|
s812338606
|
Accepted
| 17 | 3,060 | 148 |
li = [int(i) for i in input().split()]
n = li[0]
a = li[1]
b = li[2]
res1 = min(a,b)
if a+b-n<0:
res2 =0
else:
res2=a+b-n
print(res1, res2)
|
s043083234
|
p03387
|
u863076295
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,316 | 466 |
You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
lst = list(map(int,input().split()))
m = 0
for i in range(len(lst)):
m = max(m,lst[i])
for i in range(len(lst)):
if lst[i]==m:
idx = i
idx_1 = (idx + 1 + 3)%3
idx_2 = (idx - 1 + 3)%3
cnt_1 = (lst[idx]-lst[idx_1])//2 + (lst[idx]-lst[idx_1])%2
lst[idx_1] += 2*cnt_1
cnt_2 = (lst[idx]-lst[idx_2])//2 + (lst[idx]-lst[idx_2])%2
lst[idx_2] += 2*cnt_2
if lst[0] == lst[1] and lst[1] == lst[2]:
print(cnt_1 + cnt_2)
else: print(cnt_1 + cnt_2 +1)
|
s522630798
|
Accepted
| 17 | 3,064 | 510 |
lst = list(map(int,input().split()))
m = 0
for i in range(len(lst)):
m = max(m,lst[i])
for i in range(len(lst)):
if lst[i]==m:
idx = i
idx_1 = (idx + 1 + 3)%3
idx_2 = (idx - 1 + 3)%3
cnt_1 = (lst[idx]-lst[idx_1])//2
lst[idx_1] += 2*cnt_1
cnt_2 = (lst[idx]-lst[idx_2])//2
lst[idx_2] += 2*cnt_2
if lst[idx_1] == lst[idx_2] and lst[idx_1] == lst[idx]: print(cnt_1 + cnt_2)
elif lst[idx_1] == lst[idx_2] and lst[idx_1]+1 == lst[idx]: print(cnt_1 + cnt_2 + 1)
else: print(cnt_1 + cnt_2 + 2)
|
s255531306
|
p03997
|
u844895214
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 131 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
from sys import stdin
def I(): return int(stdin.readline().rstrip())
a,b,h = [I() for i in range(3)]
ans = (a+b)*h/2
print(ans)
|
s156012409
|
Accepted
| 16 | 2,940 | 136 |
from sys import stdin
def I(): return int(stdin.readline().rstrip())
a,b,h = [I() for i in range(3)]
ans = int((a+b)*h/2)
print(ans)
|
s260836688
|
p02613
|
u618373524
| 2,000 | 1,048,576 |
Wrong Answer
| 152 | 16,156 | 231 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n =int(input())
list=[]
for i in range(n):
s = input()
list.append(s)
print('AC × '+str(list.count('AC')))
print('WA × '+str(list.count('WA')))
print('TLE × '+str(list.count('TLE')))
print('RE × '+str(list.count('RE')))
|
s394237274
|
Accepted
| 152 | 16,168 | 227 |
n =int(input())
list=[]
for i in range(n):
s = input()
list.append(s)
print('AC x '+str(list.count('AC')))
print('WA x '+str(list.count('WA')))
print('TLE x '+str(list.count('TLE')))
print('RE x '+str(list.count('RE')))
|
s105117881
|
p03475
|
u974935538
| 3,000 | 262,144 |
Wrong Answer
| 35 | 3,064 | 283 |
A railroad running from west to east in Atcoder Kingdom is now complete. There are N stations on the railroad, numbered 1 through N from west to east. Tomorrow, the opening ceremony of the railroad will take place. On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated. The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds. Here, it is guaranteed that F_i divides S_i. That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains. For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains.
|
N=int(input())
csflist=[]
for i in range(N-1):
c,s,f=map(int,input().split())
csflist.append((c,s,f))
for i in range(N):
sec=0
for j in range(i,N-1):
c,s,f=csflist[j]
if sec<s:
sec=s
elif sec%f>0:
sec+=f-(sec%f)
sec+=c
print(sec)
|
s309568418
|
Accepted
| 92 | 3,060 | 306 |
N=int(input())
csflist=[]
for i in range(N-1):
c,s,f=map(int,input().split())
csflist.append((c,s,f))
#print(csflist)
for i in range(N):
sec=0
for j in range(i,N-1):
c,s,f=csflist[j]
if sec<s:
sec=s
elif sec%f>0:
sec+=f-(sec%f)
sec+=c
#print(i,j,sec)
print(sec)
|
s278949925
|
p03719
|
u271176141
| 2,000 | 262,144 |
Wrong Answer
| 29 | 9,152 | 188 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
A,B,C = map(int,input().split())
if A <= B and B <= C:
print("Yes")
else:
print("No")
|
s596143133
|
Accepted
| 27 | 9,092 | 188 |
A,B,C = map(int,input().split())
if A <= C and C <= B:
print("Yes")
else:
print("No")
|
s582076188
|
p03067
|
u881316390
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 216 |
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
def check_between(a,b,c):
if (b < c) and (c < a) :
print('YES')
elif (a < c) and (c <b) :
print('YES')
else:
print('NO')
a, b, c = map(int, input().split())
check_between(a,b, c)
|
s812478393
|
Accepted
| 17 | 3,060 | 215 |
def check_between(a,b,c):
if (b < c) and (c < a) :
print('Yes')
elif (a < c) and (c <b) :
print('Yes')
else:
print('No')
a, b, c = map(int, input().split())
check_between(a,b,c)
|
s380821110
|
p03577
|
u205561862
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 25 |
Rng is going to a festival. The name of the festival is given to you as a string S, which ends with `FESTIVAL`, from input. Answer the question: "Rng is going to a festival of what?" Output the answer. Here, assume that the name of "a festival of s" is a string obtained by appending `FESTIVAL` to the end of s. For example, `CODEFESTIVAL` is a festival of `CODE`.
|
print(input()+"FESTIVAL")
|
s205897822
|
Accepted
| 17 | 2,940 | 20 |
print(input()[:-8])
|
s378677076
|
p02742
|
u810787773
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 160 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
def main():
H,W = map(int,input().split())
N = H*W
if N%2 == 0:
ans = N/2
else:
ans = (N//2) + 1
return ans
print(main())
|
s722629943
|
Accepted
| 17 | 2,940 | 232 |
def main():
H,W = map(int,input().split())
N = H*W
#print(N//2)
if H == 1 or W == 1:
return 1
elif N%2 == 0:
ans = int(N/2)
else:
ans = int((N//2) + 1)
return ans
print(main())
|
s962058336
|
p03997
|
u408620326
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 49 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
print((int(input())+int(input()))*int(input())/2)
|
s935018227
|
Accepted
| 17 | 2,940 | 50 |
print((int(input())+int(input()))*int(input())//2)
|
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