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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s668608974
|
p02842
|
u403331159
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 122 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
import math
n=int(input())
ans=math.ceil(n/1.08)
print(ans*1.08)
if int(ans*1.08)==n:
print(ans)
else:
print(":(")
|
s511403149
|
Accepted
| 17 | 2,940 | 106 |
import math
n=int(input())
ans=math.ceil(n/1.08)
if int(ans*1.08)==n:
print(ans)
else:
print(":(")
|
s482171969
|
p03997
|
u668503853
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
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.
|
a=int(input())
b=int(input())
h=int(input())
print((a+b)/2*2)
|
s541829934
|
Accepted
| 17 | 2,940 | 62 |
a=int(input())
b=int(input())
h=int(input())
print((a+b)*h//2)
|
s009605684
|
p03695
|
u154756110
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 116 |
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
N=int(input())
A=list(map(int,input().split()))
cnt=0
for i in range(N):
if(A[i]>=3200):
cnt+=1
print(N-cnt,N)
|
s926547992
|
Accepted
| 17 | 3,060 | 198 |
N=int(input())
A=list(map(int,input().split()))
cnt=0
X=set()
for i in range(N):
if(A[i]>=3200):
cnt+=1
else:
X.add(A[i]//400)
if(len(X)):
print(len(X),len(X)+cnt)
else:
print(1,cnt)
|
s665477552
|
p02659
|
u973972117
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,084 | 55 |
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
A,B=map(float,input().split())
C=(A*B*10)//10
print(C)
|
s050532102
|
Accepted
| 27 | 9,064 | 103 |
A,B = map(str,input().split())
A = int(A)
B = int(B[0])*100+int(B[2])*10+int(B[3])
print(int(A*B//100))
|
s897697193
|
p00100
|
u040533857
| 1,000 | 131,072 |
Wrong Answer
| 50 | 6,788 | 421 |
There is data on sales of your company. Your task is to write a program which identifies good workers. The program should read a list of data where each item includes the employee ID _i_ , the amount of sales _q_ and the corresponding unit price _p_. Then, the program should print IDs of employees whose total sales proceeds (i.e. sum of p × q) is greater than or equal to 1,000,000 in the order of inputting. If there is no such employees, the program should print "NA". You can suppose that _n_ < 4000, and each employee has an unique ID. The unit price _p_ is less than or equal to 1,000,000 and the amount of sales _q_ is less than or equal to 100,000.
|
while True:
try:
n=int(input())
except:
break
if n==0:
break
data=[]
for i in range(n):
spam=[x for x in map(int,input().split())]
if spam[1]*spam[2]>=1000000:
if (spam[0] in [b[0] for b in data])==False:
data.append(spam)
if len(data)==0:
print('NA')
else:
for i in [b[0] for b in data]:
print(i)
|
s499182252
|
Accepted
| 40 | 6,748 | 505 |
while True:
try:
n=int(input())
except:
break
if n==0:
break
data={}
staff=[]
for i in range(n):
spam=list(map(int,input().split()))
if spam[0] in data.keys():
data[spam[0]]+=spam[1]*spam[2]
else:
data[spam[0]]=spam[1]*spam[2]
staff.append(spam[0])
if max(data.values())<1000000:
print('NA')
else:
for i in staff:
if data[i] >= 1000000:
print(i)
|
s483528794
|
p03377
|
u382431597
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 71 |
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,b,x=list(map(int, input().split()))
print("YES" if (x-a)>b else "NO")
|
s459094454
|
Accepted
| 17 | 2,940 | 75 |
a,b,x=list(map(int, input().split()))
print("YES" if 0<=(x-a)<=b else "NO")
|
s736904857
|
p03827
|
u220345792
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 98 |
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
N = int(input())
S = input()
plus_ = S.count("I")
minus_ = S.count("D")
print(max(plus_-minus_,0))
|
s456867298
|
Accepted
| 18 | 2,940 | 147 |
N = int(input())
S = input()
tmp = 0
ans = 0
for i in S:
if i == "I":
tmp += 1
elif i == "D":
tmp -= 1
ans = max(ans, tmp)
print(ans)
|
s142810194
|
p03079
|
u163449343
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 64 |
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
a, b, c = input().split()
print("YNeos"[(a == b and b == c)::2])
|
s152664185
|
Accepted
| 18 | 2,940 | 67 |
a, b, c = input().split()
print(["No", "Yes"][(a == b and b == c)])
|
s452654153
|
p03352
|
u294385082
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 239 |
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
x = int(input())
l = [1]
for i in range(2,32):
for j in range(1,10):
if i**j <= 1000:
l.append(i**j)
l = sorted(l,reverse = True)
for k in sorted(list(range(1,x+1)),reverse = True):
if k in l:
print(k)
exit()
|
s037877147
|
Accepted
| 18 | 3,060 | 239 |
x = int(input())
l = [1]
for i in range(2,32):
for j in range(2,10):
if i**j <= 1000:
l.append(i**j)
l = sorted(l,reverse = True)
for k in sorted(list(range(1,x+1)),reverse = True):
if k in l:
print(k)
exit()
|
s726057899
|
p03455
|
u708255304
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 86 |
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("Eveb")
else:
print("Odd")
|
s181742844
|
Accepted
| 17 | 2,940 | 86 |
a, b = map(int, input().split())
if a*b % 2 == 0:
print("Even")
else:
print("Odd")
|
s915939551
|
p03434
|
u048945791
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 163 |
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()))
a.sort()
ans = 0
for i in range(len(a)):
if i % 2 == 0:
ans += a[i]
else:
ans -= a[i]
print(ans)
|
s850790990
|
Accepted
| 17 | 3,060 | 175 |
n = int(input())
a = list(map(int, input().split()))
a.sort()
a.reverse()
ans = 0
for i in range(len(a)):
if i % 2 == 0:
ans += a[i]
else:
ans -= a[i]
print(ans)
|
s640906774
|
p04012
|
u771756419
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 180 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
w = list(input())
alp = list("abcdefghijklmnopqrstuvwxyz")
ans = 0
for i in alp:
if w.count(i) % 2 != 0 :
ans += 1
if ans == 0:
print("YES")
else:
print("NO")
|
s167540707
|
Accepted
| 18 | 2,940 | 180 |
w = list(input())
alp = list("abcdefghijklmnopqrstuvwxyz")
ans = 0
for i in alp:
if w.count(i) % 2 != 0 :
ans += 1
if ans == 0:
print("Yes")
else:
print("No")
|
s656533168
|
p03854
|
u479977918
| 2,000 | 262,144 |
Wrong Answer
| 20 | 4,276 | 613 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
# -*- coding: utf-8 -*-
import sys
S = input()
five_sign1 = "dream"
five_sign2 = "erase"
six_sign = "eraser"
sevent_sign = "dreamer"
while(len(S) > 0):
print("S {}".format(S))
if S[0:6] == "eraser":
S = S[6:]
elif S[0:5] == "erase":
S = S[5:]
elif S[0:5] == "dream":
tmp = S[5:8]
if len(tmp) == 0:
S = S[5:]
continue
elif tmp == "era":
S = S[5:]
elif tmp == "erd":
S = S[7:]
else:
print("NO")
sys.exit()
else:
print("NO")
sys.exit()
print("YES")
|
s431715353
|
Accepted
| 76 | 3,316 | 547 |
# -*- coding: utf-8 -*-
import sys
S = input()
while(len(S) > 0):
if S[0:6] == "eraser":
S = S[6:]
elif S[0:5] == "erase":
S = S[5:]
elif S[0:5] == "dream":
tmp = S[5:8]
if len(tmp) == 0:
S = S[5:]
elif tmp == "era":
S = S[5:]
elif tmp == "er":
S = S[7:]
elif tmp == "ere":
S = S[7:]
elif tmp == "erd":
S = S[7:]
else:
S = S[5:]
else:
print("NO")
sys.exit()
print("YES")
|
s218310167
|
p02612
|
u514969825
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,136 | 118 |
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())
for i in range(1, 12):
if (N >= i*1000):
continue
else:
print(N - (i-1)*1000)
break
|
s505912477
|
Accepted
| 28 | 9,120 | 113 |
N = int(input())
for i in range(1, 12):
if (i*1000 < N):
continue
else:
print(i*1000 - N)
break
|
s515448408
|
p04043
|
u434311880
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 180 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
A, B, C = map(int,input().split())
if A == 5 and B == C == 7:
print("YES")
elif A==B==7 and C==5:
print("YES")
elif A==C==7 and B==5:
print("YES")
else:
print("NO")
|
s595752428
|
Accepted
| 17 | 2,940 | 181 |
A, B, C = map(int,input().split())
if A == 7 and B == C == 5:
print("YES")
elif A==B==5 and C==7:
print("YES")
elif A==C==5 and B==7:
print("YES")
else:
print("NO")
|
s564994962
|
p03970
|
u727787724
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 111 |
CODE FESTIVAL 2016 is going to be held. For the occasion, Mr. Takahashi decided to make a signboard. He intended to write `CODEFESTIVAL2016` on it, but he mistakenly wrote a different string S. Fortunately, the string he wrote was the correct length. So Mr. Takahashi decided to perform an operation that replaces a certain character with another in the minimum number of iterations, changing the string to `CODEFESTIVAL2016`. Find the minimum number of iterations for the rewrite operation.
|
s=list(input())
a=list('CODEFESTIVAL2016')
ans=0
for i in range(len(s)):
if s[i]==a[i]:
ans+=1
print(ans)
|
s076849324
|
Accepted
| 17 | 2,940 | 112 |
s=list(input())
a=list('CODEFESTIVAL2016')
ans=0
for i in range(len(s)):
if s[i]!=a[i]:
ans+=1
print(ans)
|
s051476887
|
p02612
|
u969081133
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,148 | 32 |
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())
a=N%1000
print(a)
|
s178035563
|
Accepted
| 28 | 9,164 | 65 |
N=int(input())
a=N%1000
if a==0:
print(0)
else:
print(1000-a)
|
s047231500
|
p03455
|
u622011073
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,188 | 65 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b=input().split();a=int(a+b)**.5
print(['No','Yes'][a==int(a)])
|
s903116366
|
Accepted
| 18 | 2,940 | 57 |
a,b=map(int,input().split())
print(['Even','Odd'][a*b&1])
|
s999386292
|
p02476
|
u861198832
| 1,000 | 262,144 |
Wrong Answer
| 20 | 5,580 | 44 |
Given two integers $A$ and $B$, compute the remainder of $\frac{A}{B}$.
|
a,b = map(int,input().split())
print(a//b)
|
s124399531
|
Accepted
| 20 | 5,584 | 43 |
a,b = map(int,input().split())
print(a%b)
|
s452565543
|
p03379
|
u411858517
| 2,000 | 262,144 |
Wrong Answer
| 396 | 20,884 | 155 |
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.
|
N = int(input())
arr = input().split()
arr.sort()
for i in range(N):
if i < N/2:
print(arr[int(N/2-1)])
else:
print(arr[int(N/2)])
|
s745831822
|
Accepted
| 391 | 25,620 | 199 |
N = int(input())
arr = list(map(int, input().split()))
X = arr.copy()
arr.sort()
for i in range(N):
if X[i] < arr[int(N/2)]:
print(arr[int(N/2)])
else:
print(arr[int(N/2)-1])
|
s308663713
|
p03555
|
u320511454
| 2,000 | 262,144 |
Wrong Answer
| 35 | 9,080 | 65 |
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
s,t=input(),input()
print("Yes") if s == t[::-1] else print("No")
|
s249560696
|
Accepted
| 29 | 9,056 | 65 |
s,t=input(),input()
print("YES") if s == t[::-1] else print("NO")
|
s583243061
|
p02388
|
u830616709
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,724 | 29 |
Write a program which calculates the cube of a given integer x.
|
(lambda x:x**3)(int(input()))
|
s349525481
|
Accepted
| 20 | 7,648 | 41 |
print(str((lambda x:x**3)(int(input()))))
|
s757776373
|
p04012
|
u974935538
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,316 | 184 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
import collections
w = list(input())
tmp = collections.Counter(w)
ans = "YES"
for i in tmp.values():
if i%2 != 0:
ans = "NO"
break
else:
pass
print(ans)
|
s040824499
|
Accepted
| 20 | 3,316 | 184 |
import collections
w = list(input())
tmp = collections.Counter(w)
ans = "Yes"
for i in tmp.values():
if i%2 != 0:
ans = "No"
break
else:
pass
print(ans)
|
s718486331
|
p03377
|
u315838013
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,316 | 227 |
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.
|
import sys
if __name__ == "__main__":
A, B, X = list(map(int, (input().split(' '))))
buf = X - A
if buf < 0:
print('No')
sys.exit()
if buf <= B:
print('Yes')
else:
print('No')
|
s466937545
|
Accepted
| 17 | 2,940 | 227 |
import sys
if __name__ == "__main__":
A, B, X = list(map(int, (input().split(' '))))
buf = X - A
if buf < 0:
print('NO')
sys.exit()
if buf <= B:
print('YES')
else:
print('NO')
|
s417070315
|
p03386
|
u027403702
| 2,000 | 262,144 |
Wrong Answer
| 2,237 | 4,084 | 156 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
A,B,K = map(int, input().split())
List = [i for i in range(A,B+1)]
s = List[0:K]
r = List[::-1]
l = r[0:K]
for i in l:
s.append(i)
print(sorted(set(s)))
|
s422848482
|
Accepted
| 18 | 3,188 | 249 |
A,B,K = map(int, input().split())
if B-A+1 >= K:
s = [i for i in range(A,A+K)]
l = [j for j in range(B-K+1,B+1)]
for k in l:
s.append(k)
w = sorted(set(s))
else:
w = [b for b in range(A,B+1)]
for j in w:
print(j)
|
s876814453
|
p03479
|
u831752983
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 77 |
As a token of his gratitude, Takahashi has decided to give his mother an integer sequence. The sequence A needs to satisfy the conditions below: * A consists of integers between X and Y (inclusive). * For each 1\leq i \leq |A|-1, A_{i+1} is a multiple of A_i and strictly greater than A_i. Find the maximum possible length of the sequence.
|
x,y=map(int,input().split())
n=y//x
ct=0
while n>1:
ct+=1
n//=2
print(ct)
|
s089908774
|
Accepted
| 17 | 3,064 | 77 |
x,y=map(int,input().split())
n=y//x
ct=0
while n>0:
n//=2
ct+=1
print(ct)
|
s106917073
|
p02612
|
u398511319
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 8,956 | 76 |
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= input()
x=int(N)
while True:
x -= 1000
if x <= 1000:
break
print(x)
|
s148433142
|
Accepted
| 28 | 9,084 | 204 |
N= input()
x=int(N)
if x<1000:
print(1000-x)
if x==1000:
print(0)
if x > 1000:
while True:
x -= 1000
if x < 1001:
break
print(1000-x)
|
s014180113
|
p02396
|
u096070545
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 234 |
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.
|
# coding: utf-8
# Your code here!
# TP1_3_B
# N=input()
# print(N)
N=-1
# while N != 0:
# N=int(input())
# print(N)
for i in range(1,100):
N=int(input())
if N == 0:
break
print('Case {0}:{1}'.format(i,N))
|
s841453972
|
Accepted
| 140 | 5,600 | 237 |
# coding: utf-8
# Your code here!
# TP1_3_B
# N=input()
# print(N)
N=-1
# while N != 0:
# N=int(input())
# print(N)
for i in range(1,10001):
N=int(input())
if N == 0:
break
print('Case {0}: {1}'.format(i,N))
|
s048236676
|
p03999
|
u210827208
| 2,000 | 262,144 |
Wrong Answer
| 28 | 3,188 | 240 |
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. Evaluate all possible formulas, and print the sum of the results.
|
import itertools
s=input()
C=['+','']
X=list(itertools.product(C,repeat=len(s)-1))
print(X)
ans=0
for i in range(len(X)):
res=''
for j in range(len(s)-1):
res+=s[j]+X[i][j]
res+=s[len(s)-1]
ans+=eval(res)
print(ans)
|
s587573799
|
Accepted
| 27 | 3,064 | 231 |
import itertools
s=input()
C=['+','']
X=list(itertools.product(C,repeat=len(s)-1))
ans=0
for i in range(len(X)):
res=''
for j in range(len(s)-1):
res+=s[j]+X[i][j]
res+=s[len(s)-1]
ans+=eval(res)
print(ans)
|
s150481217
|
p02271
|
u436634575
| 5,000 | 131,072 |
Wrong Answer
| 30 | 6,716 | 304 |
Write a program which reads a sequence _A_ of _n_ elements and an integer _M_ , and outputs "yes" if you can make _M_ by adding elements in _A_ , otherwise "no". You can use an element only once. You are given the sequence _A_ and _q_ questions where each question contains _M i_.
|
n = int(input())
a = list(map(int, input().strip().split()))
q = int(input())
M = list(map(int, input().strip().split()))
d = []
for x in range(1 << n):
b = '{:0{}b}'.format(x, n)
print(b)
d.append(sum(a[i] for i in range(n) if b[i] == '1'))
for m in M:
print('yes' if m in d else 'no')
|
s767556639
|
Accepted
| 9,940 | 44,760 | 291 |
n = int(input())
a = list(map(int, input().strip().split()))
q = int(input())
M = list(map(int, input().strip().split()))
d = []
for x in range(1 << n):
b = '{:0{}b}'.format(x, n)
d.append(sum(a[i] for i in range(n) if b[i] == '1'))
for m in M:
print('yes' if m in d else 'no')
|
s863959408
|
p03861
|
u175590965
| 2,000 | 262,144 |
Wrong Answer
| 2,107 | 2,940 | 107 |
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?
|
a,b,x = map(int,input().split())
ans = 0
for i in range(a,b):
if i % x ==0:
ans += 1
print(ans)
|
s654620461
|
Accepted
| 17 | 2,940 | 58 |
a,b,x = map(int,input().split())
print((b//x)-((a-1)//x))
|
s773462425
|
p03997
|
u557235596
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,064 | 110 |
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.
|
if __name__ == "__main__":
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2)
|
s485814672
|
Accepted
| 22 | 3,064 | 116 |
if __name__ == "__main__":
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
|
s474471861
|
p02415
|
u313600138
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,548 | 155 |
Write a program which converts uppercase/lowercase letters to lowercase/uppercase for a given string.
|
s=input()
for c in s:
if 'A' <= c and c <= 'Z':
print(c.lower(), end='')
elif 'a' <= c and c <= 'z':
print(c.upper(), end='')
else:
pass
|
s191492663
|
Accepted
| 20 | 5,716 | 174 |
s=input()
for c in s:
if 'A' <= c and c <= 'Z':
print(c.lower(), end='')
elif 'a' <= c and c <= 'z':
print(c.upper(), end='')
else:
print(c,end='')
print()
|
s332167162
|
p03433
|
u624923345
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 146 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
A = int(input())
a = N % 500
if a == 0:
print("yes")
else:
if a <= A:
print("yes")
else:
print("no")
|
s344586063
|
Accepted
| 17 | 2,940 | 146 |
N = int(input())
A = int(input())
a = N % 500
if a == 0:
print("Yes")
else:
if a <= A:
print("Yes")
else:
print("No")
|
s003191077
|
p03575
|
u048004795
| 2,000 | 262,144 |
Wrong Answer
| 35 | 9,432 | 721 |
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.
|
from collections import deque
N, M = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(M)]
c = 0
print(ab)
for i in range(M):
G = [[] for _ in range(N)]
for j in range(M):
if i == j:
continue
a, b = ab[j][0], ab[j][1]
G[a - 1].append(b - 1)
G[b - 1].append(a - 1)
Q = deque()
Q.append(0)
d = [0] * N
d[0] = 1
while Q:
q = Q.popleft()
for g in G[q]:
if d[g] == 0:
d[g] = 1
Q.append(g)
if 0 in d:
c += 1
print(c)
|
s377598804
|
Accepted
| 33 | 9,480 | 711 |
from collections import deque
N, M = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(M)]
c = 0
for i in range(M):
G = [[] for _ in range(N)]
for j in range(M):
if i == j:
continue
a, b = ab[j][0], ab[j][1]
G[a - 1].append(b - 1)
G[b - 1].append(a - 1)
Q = deque()
Q.append(0)
d = [0] * N
d[0] = 1
while Q:
q = Q.popleft()
for g in G[q]:
if d[g] == 0:
d[g] = 1
Q.append(g)
if 0 in d:
c += 1
print(c)
|
s732208793
|
p02255
|
u971076135
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 393 |
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.
|
from sys import stdin
def stdinput():
return stdin.readline().strip()
def main():
n = int(stdinput())
a = list(map(int, stdinput().split()))
for i in range(1, n):
v = a[i]
j = i - 1
while j >= 0 and a[j] > v:
a[j+1] = a[j]
j -= 1
a[j + 1] = v
print(' '.join(map(str, a)))
if __name__ == '__main__':
main()
|
s653572410
|
Accepted
| 20 | 5,608 | 393 |
from sys import stdin
def stdinput():
return stdin.readline().strip()
def main():
n = int(stdinput())
a = list(map(int, stdinput().split()))
for i in range(0, n):
v = a[i]
j = i - 1
while j >= 0 and a[j] > v:
a[j+1] = a[j]
j -= 1
a[j + 1] = v
print(' '.join(map(str, a)))
if __name__ == '__main__':
main()
|
s857786720
|
p03469
|
u947327691
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 24 |
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
s=input()
s="2018"+s[:4]
|
s191495797
|
Accepted
| 17 | 2,940 | 33 |
s=input()
s="2018"+s[4:]
print(s)
|
s592239424
|
p02397
|
u623827446
| 1,000 | 131,072 |
Wrong Answer
| 50 | 7,664 | 116 |
Write a program which reads two integers x and y, and prints them in ascending order.
|
for i in range(3000):
x,y=map(int,input().split())
if x == 0 and y == 0:
break
print('{0} {1}'.format(x,y))
|
s857620497
|
Accepted
| 50 | 7,572 | 131 |
for i in range(3000):
x,y=map(int,input().split())
if x == 0 and y == 0:
break
print('{0} {1}'.format(min(x,y),max(x,y)))
|
s331772608
|
p03644
|
u217627525
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 59 |
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=1
while(ans<=n):
ans*2
print(ans//2)
|
s766301270
|
Accepted
| 17 | 2,940 | 59 |
n=int(input())
ans=1
while ans<=n:
ans*=2
print(ans//2)
|
s595582834
|
p03643
|
u404459933
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 59 |
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.
|
import math
N=int(input())
L=int(math.log(N,2))
print(2**L)
|
s973153943
|
Accepted
| 17 | 2,940 | 20 |
print('ABC'+input())
|
s682706313
|
p03433
|
u353797797
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 224 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
import sys
sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline
int1 = lambda x: int(x) - 1
def main():
n=int(input())
a=int(input())
if a<=n%500:
print("Yes")
else:
print("No")
main()
|
s711845983
|
Accepted
| 18 | 2,940 | 224 |
import sys
sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline
int1 = lambda x: int(x) - 1
def main():
n=int(input())
a=int(input())
if a>=n%500:
print("Yes")
else:
print("No")
main()
|
s263618180
|
p03958
|
u785578220
| 1,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 158 |
There are K pieces of cakes. Mr. Takahashi would like to eat one cake per day, taking K days to eat them all. There are T types of cake, and the number of the cakes of type i (1 ≤ i ≤ T) is a_i. Eating the same type of cake two days in a row would be no fun, so Mr. Takahashi would like to decide the order for eating cakes that minimizes the number of days on which he has to eat the same type of cake as the day before. Compute the minimum number of days on which the same type of cake as the previous day will be eaten.
|
k,t =map(int,input().split())
a = list(map(int,input().split()))
a.append(0)
a.sort()
t = a[-1]
s = sum(a[:-1])
if t == 1:
print(t-1)
else:
print(t-s)
|
s939497099
|
Accepted
| 17 | 3,060 | 185 |
k,f =map(int,input().split())
a = list(map(int,input().split()))
a.append(0)
a.sort()
t = a[-1]
s = sum(a[:-1])
if 1 == f:
print(t-1)
elif t<=s:
print(0)
else:
print(t-s-1)
|
s646036033
|
p03007
|
u186967328
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 14,264 | 175 |
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.
|
N = int(input())
A = list(map(int,input().split()))
A = sorted(A)
for i in range(N-1):
print(A[0],A[-1])
diff = A[0]-A[-1]
A = A[1:len(A)-1]
A.insert(0,diff)
|
s329933759
|
Accepted
| 260 | 14,144 | 528 |
N = int(input())
A = list(map(int,input().split()))
A = sorted(A)
plus = []
minus = []
for i in range(N):
if i == 0:
minus.append(A[i])
elif i == N-1:
plus.append(A[i])
else:
if A[i] > 0:
plus.append(A[i])
else:
minus.append(A[i])
print(sum(plus)-sum(minus))
tmp = minus[0]
for i in range(len(plus)-1):
print(tmp,plus[i])
tmp -= plus[i]
minus[0] = tmp
tmp2 = plus[-1]
for i in range(len(minus)):
print(tmp2,minus[i])
tmp2 -= minus[i]
|
s871614633
|
p03854
|
u832871520
| 2,000 | 262,144 |
Wrong Answer
| 269 | 4,080 | 430 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
div = ['maerd', 'remaerd', 'esare', 'resare']
s = input()
rs = ''.join(list(reversed(s)))
can1 = True
cnt = 0
for i in range(len(s)):
if cnt != 0:
cnt -= 1
continue
print(i)
can2 = False
for d in range(len(div)):
if (rs.find(div[d])) == i:
can2 = True
cnt = len(div[d]) -1
if can2 == False:
can1 = False
if can1:
print('Yes')
else:
print('No')
|
s696685371
|
Accepted
| 23 | 6,516 | 87 |
import re
print("YES" if re.match("^(dream|dreamer|erase|eraser)+$",input()) else "NO")
|
s177549471
|
p04044
|
u953753178
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 92 |
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
A = list(map(str, input().split()))
A.sort()
res = ''
for a in A:
res += a
print(res)
|
s133660042
|
Accepted
| 17 | 3,060 | 183 |
N, L = list(map(int, input().split()))
input_str_list = []
for input_str_index in range(N):
input_str_list.append(input())
input_str_list.sort()
print(''.join(input_str_list))
|
s197827344
|
p03795
|
u953794676
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 64 |
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.
|
e = int(input())
print(int(e/15))
print(e*800 - (int(e/15)*200))
|
s522719249
|
Accepted
| 17 | 2,940 | 47 |
e = int(input())
print(e*800 - (int(e/15)*200))
|
s651512167
|
p03387
|
u235905557
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 296 |
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.
|
a, b, c = map(int, input().split())
M = max(a, max(b, c))
if (a+b+c)%2 == 1:
X = M
else:
X = M + 1
print((3*X - (a+b+c)) / 2)
|
s436088851
|
Accepted
| 17 | 2,940 | 278 |
a, b, c = map(int, input().split())
X = 0
M = max(a, max(b, c))
if (3*M)%2 == (a+b+c)%2:
X = M
else:
X = M + 1
print(int((3*X - (a+b+c)) / 2))
|
s908746328
|
p02612
|
u898109279
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,128 | 32 |
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)
|
s266258121
|
Accepted
| 28 | 9,136 | 88 |
n = int(input())
oturi = n % 1000
if (oturi == 0):
print(0)
else:
print(1000 - oturi)
|
s469741578
|
p02612
|
u248364740
| 2,000 | 1,048,576 |
Wrong Answer
| 32 | 9,144 | 99 |
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.
|
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
n = int(input())
result = n % 1000
print(result)
|
s934877537
|
Accepted
| 29 | 9,144 | 51 |
n = int(input())
print((1000 - (n % 1000)) % 1000)
|
s450152090
|
p02401
|
u316246166
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 150 |
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
for i in range(1000):
a, b, c = input().split()
if b == '?': break
a = int(a)
c = int(c)
print(a, b, c)
|
s724109369
|
Accepted
| 20 | 5,596 | 252 |
for i in range(1000):
a, b, c = input().split()
if b == '?': break
a = int(a)
c = int(c)
if b == '+': print(a + c)
if b == '-': print(a - c)
if b == '*': print(a * c)
if b == '/': print(a // c)
|
s799175798
|
p03712
|
u503111914
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 106 |
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
H,W = map(int,input().split())
print("#"*W)
for i in range(H-2):
print("#" + input() + "#")
print("#"*W)
|
s742762100
|
Accepted
| 18 | 3,060 | 113 |
H,W = map(int,input().split())
print("#"*(W+2))
for i in range(H):
print("#" + input() + "#")
print("#"*(W+2))
|
s689677256
|
p03564
|
u119982147
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.
|
N = int(input())
K = int(input())
A = 1 + ( N * K )
print(A)
|
s872627539
|
Accepted
| 17 | 2,940 | 177 |
N = int(input())
K = int(input())
number = 1
for i in range(N):
if number * 2 < number + K:
number = number * 2
else:
number = number + K
print(number)
|
s193413734
|
p03597
|
u799428010
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 52 |
We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black?
|
N = int(input())
A = int(input())
B=N*N-A*A
print(B)
|
s729727196
|
Accepted
| 17 | 2,940 | 50 |
N = int(input())
A = int(input())
B=N*N-A
print(B)
|
s730550124
|
p03944
|
u853586331
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 230 |
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.
|
W,H,N=map(int, input().split())
b=0
c=W
d=0
e=H
for i in range(N):
x,y,a=map(int, input().split())
if a==1:
b==max(b,x)
elif a==2:
c==min(c,x)
elif a==3:
d==max(d,y)
else:
e==min(e,y)
print(max(0,(c-b))*max(0,(e-d)))
|
s266667890
|
Accepted
| 18 | 3,064 | 225 |
W,H,N=map(int, input().split())
w=0
h=0
for i in range(N):
x,y,a=map(int, input().split())
if a==1:
w=max(w,x)
elif a==2:
W=min(W,x)
elif a==3:
h=max(h,y)
else:
H=min(H,y)
print(max(0,(W-w))*max(0,(H-h)))
|
s125129397
|
p03997
|
u482969053
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 51 |
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)
|
s575157650
|
Accepted
| 16 | 2,940 | 50 |
print((int(input())+int(input()))*int(input())//2)
|
s422891514
|
p03044
|
u367130284
| 2,000 | 1,048,576 |
Wrong Answer
| 647 | 9,604 | 2,084 |
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
class POTENTIAL_UNION_FIND(object):
def __init__(self, n):
self.parent = [-1 for i in range(n)]
self.diff_p = [0 for i in range(n)]
def root(self, x):
if self.parent[x] < 0:
return x
else:
rx = self.root(self.parent[x])
self.diff_p[x] += self.diff_p[self.parent[x]]
self.parent[x] = rx
return rx
def size(self,x):
return -self.parent[self.root(x)]
def merge(self, x, y, dxy):
rx = self.root(x)
ry = self.root(y)
dxy += self.diff_p[x] - self.diff_p[y]
x,y=rx,ry
if x == y:
return False
if self.parent[x] > self.parent[y]:
x,y,dxy=y,x,-dxy
self.parent[x] += self.parent[y]
self.parent[y] = x
self.diff_p[y] = dxy
return True
def issame(self, x, y):
return self.root(x) == self.root(y)
def diff(self,x,y):
if self.root(x) == self.root(y):
return self.diff_p[y] - self.diff_p[x]
else:
return None
n=int(input())
u=POTENTIAL_UNION_FIND(n)
for i in range(n-1):
a,b,c=map(int,input().split())
u.merge(a-1,b-1,c)
for i in u.diff_p:
if i%2==0:
print(1)
else:
print(0)
|
s456475283
|
Accepted
| 876 | 10,252 | 2,050 |
class POTENTIAL_UNION_FIND(object):
def __init__(self, n):
self.parent = [-1 for i in range(n)]
self.diff_p = [0 for i in range(n)]
def root(self, x):
if self.parent[x] < 0:
return x
else:
rx = self.root(self.parent[x])
self.diff_p[x] += self.diff_p[self.parent[x]]
self.parent[x] = rx
return rx
def size(self,x):
return -self.parent[self.root(x)]
def merge(self, x, y, dxy):
rx = self.root(x)
ry = self.root(y)
dxy += self.diff_p[x] - self.diff_p[y]
x,y=rx,ry
if x == y:
return False
if self.parent[x] > self.parent[y]:
x,y,dxy=y,x,-dxy
self.parent[x] += self.parent[y]
self.parent[y] = x
self.diff_p[y] = dxy
return True
def issame(self, x, y):
return self.root(x) == self.root(y)
def diff(self,x,y):
if self.root(x) == self.root(y):
return self.diff_p[y] - self.diff_p[x]
else:
return None
n=int(input())
u=POTENTIAL_UNION_FIND(n)
for i in range(n-1):
a,b,c=map(int,input().split())
u.merge(a-1,b-1,c)
for i in range(n):
print(u.diff(0,i)%2)
|
s860850942
|
p03048
|
u027076164
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 3,060 | 233 |
Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this?
|
R,G,B,N = map(int, input().split())
c = 0
for i in range(int(N/R)):
for j in range(int((N-R*i)/G)):
for k in range(int((N-R*i-G*j)/B)):
if N-R*i-G*j-B*k-int(N-R*i-G*j-B*k) == 0:
c += 1
print(c)
|
s530727503
|
Accepted
| 1,489 | 2,940 | 171 |
R,G,B,N = map(int, input().split())
c = 0
for i in range(0,int(N/R)+1):
for j in range(0,int((N-R*i)/G)+1):
if (N-R*i-G*j)%B == 0:
c += 1
print(c)
|
s006268363
|
p03780
|
u817328209
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 6,896 | 560 |
AtCoDeer the deer has N cards with positive integers written on them. The number on the i-th card (1≤i≤N) is a_i. Because he loves big numbers, he calls a subset of the cards _good_ when the sum of the numbers written on the cards in the subset, is K or greater. Then, for each card i, he judges whether it is _unnecessary_ or not, as follows: * If, for any good subset of the cards containing card i, the set that can be obtained by eliminating card i from the subset is also good, card i is unnecessary. * Otherwise, card i is NOT unnecessary. Find the number of the unnecessary cards. Here, he judges each card independently, and he does not throw away cards that turn out to be unnecessary.
|
import sys
N, K = map(int, input().split())
a = list(map(int, input().split()))
def isNecessary(now) :
psum = 0
for i in range(N) :
if i == now :
continue
psum += a[i]
if psum >= K :
return False
else :
return True
S = sum(a)
ans = 0
a = sorted(a, reverse=True)
l, r = 0, N-1
now = N//2
while True :
if isNecessary(now) :
r = now
else :
l = now
now = (l+r)//2
if r == l :
if N==1 and a[0]<K :
print(1)
else :
print(r)
|
s581091554
|
Accepted
| 21 | 3,572 | 223 |
N, K = map(int, input().split())
a = list(map(int, input().split()))
a = sorted(a)
ans = N
psum = 0
for i in range(N-1, -1, -1) :
if psum+a[i] < K :
psum += a[i]
else :
ans = min(ans, i)
print(ans)
|
s742805115
|
p03433
|
u721316601
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
A = int(input())
if N % 500 == A:
print('Yes')
else:
print('No')
|
s883660559
|
Accepted
| 17 | 2,940 | 91 |
N = int(input())
A = int(input())
if N % 500 <= A:
print('Yes')
else:
print('No')
|
s450069095
|
p02266
|
u874395007
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,560 | 359 |
Your task is to simulate a flood damage. For a given cross-section diagram, reports areas of flooded sections. Assume that rain is falling endlessly in the region and the water overflowing from the region is falling in the sea at the both sides. For example, for the above cross-section diagram, the rain will create floods which have areas of 4, 2, 1, 19 and 9 respectively.
|
lines = input()
stack_1 = []
area = 0
for i, line in enumerate(lines):
if line == '\\':
stack_1.append(i)
elif line == '_':
pass
elif line == '/':
if not stack_1 == []:
this_area = i - stack_1.pop()
area += this_area
else:
raise ValueError('Unexpected line: {i}, {line}')
print(area)
|
s568101207
|
Accepted
| 30 | 6,216 | 1,055 |
lines = input()
stack_1 = []
stack_2 = []
area = 0
for i, line in enumerate(lines):
if line == '\\':
stack_1.append(i)
elif line == '_':
pass
elif line == '/':
if stack_1 == []:
continue
last_index = stack_1.pop()
this_area = i - last_index
area += this_area
if stack_2 == []:
stack_2.append((last_index, this_area))
continue
block_area = this_area
while True:
if stack_2 == []:
break
stack_2_index, stack_2_area = stack_2.pop()
if stack_2_index < last_index:
stack_2.append((stack_2_index, stack_2_area))
break
else:
block_area += stack_2_area
stack_2.append((last_index, block_area))
print(area)
kazu = len(stack_2)
kaku_menseki = [stack_2[i][1] for i in range(len(stack_2))]
kaku_menseki_out = ' '.join(map(str, kaku_menseki))
if kazu > 0:
print(f'{kazu} {kaku_menseki_out}')
else:
print(f'{kazu}')
|
s429228934
|
p04030
|
u810457760
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 94 |
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
S = input()
ans = ""
for i in S:
if i == "B":
S = S[:-1]
else:
ans += i
print(ans)
|
s110562910
|
Accepted
| 17 | 2,940 | 98 |
S = input()
ans = ""
for i in S:
if i == "B":
ans = ans[:-1]
else:
ans += i
print(ans)
|
s204066477
|
p03588
|
u855200003
| 2,000 | 262,144 |
Wrong Answer
| 328 | 17,892 | 197 |
A group of people played a game. All players had distinct scores, which are positive integers. Takahashi knows N facts on the players' scores. The i-th fact is as follows: the A_i-th highest score among the players is B_i. Find the maximum possible number of players in the game.
|
N = int(input())
dic = {}
for i in range(N):
a,b = [int(j) for j in input().split(" ")]
dic[a] = b
ans = len(dic)
max_keys = max(dic.keys())
min_value = dic[max_keys]
print(ans + min_value)
|
s192483940
|
Accepted
| 328 | 17,892 | 185 |
N = int(input())
dic = {}
for i in range(N):
a,b = [int(j) for j in input().split(" ")]
dic[a] = b
max_keys = max(dic.keys())
min_value = dic[max_keys]
print(max_keys+min_value)
|
s249705459
|
p02606
|
u248670151
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,156 | 115 |
How many multiples of d are there among the integers between L and R (inclusive)?
|
a=list(map(int,input().split()))
count=0
for i in range(a[0],a[1]):
if i%a[2]==0:
count+=1
print(count)
|
s124115846
|
Accepted
| 30 | 9,084 | 130 |
a=list(map(int,input().split()))
count=0
for i in range(a[0],a[1]+1):
if i%a[2]==0:
count+=1
print(count)
|
s342188029
|
p03351
|
u598229387
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 104 |
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d = map(int, input().split())
print('YES' if abs(a-c)<=d or abs(a-b)<=d and abs(b-c)<=d else 'No')
|
s591000014
|
Accepted
| 17 | 2,940 | 104 |
a,b,c,d = map(int, input().split())
print('Yes' if abs(a-c)<=d or abs(a-b)<=d and abs(b-c)<=d else 'No')
|
s134255158
|
p03836
|
u276115223
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,100 | 331 |
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
# ABC 051: C – Back and Forth
sx, sy, tx, ty = [int(i) for i in input().split()]
dx = tx - sx
dy = ty - sy
# r1
r1 = 'U' * (dy + 1) + 'R' * (dx + 1)
# r2
r2 = 'D' * (dy + 1) + 'L' * (dx + 1)
# r3
r3 = 'L' + 'U' * (dy + 2) + 'R' * (dx + 2) + 'D'
# r4
r4 = 'R' + 'D' * (dy + 2) + 'L' * (dx + 2) + 'U'
print(r1 + r2 + r3 + r4)
|
s670003353
|
Accepted
| 26 | 9,024 | 307 |
# ABC 051: C – Back and Forth
sx, sy, tx, ty = [int(i) for i in input().split()]
dx = tx - sx
dy = ty - sy
# r1
r1 = 'U' * dy + 'R' * dx
# r2
r2 = 'D' * dy + 'L' * dx
# r3
r3 = 'L' + 'U' * (dy + 1) + 'R' * (dx + 1) + 'D'
# r4
r4 = 'R' + 'D' * (dy + 1) + 'L' * (dx + 1) + 'U'
print(r1 + r2 + r3 + r4)
|
s895424128
|
p03400
|
u225388820
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 102 |
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
n=int(input())
d,x=map(int,input().split())
for i in range(n):
a=int(input())
x+=(d-1)//a
print(x)
|
s527460058
|
Accepted
| 17 | 2,940 | 104 |
n=int(input())
d,x=map(int,input().split())
for i in range(n):
a=int(input())
x+=(d-1)//a+1
print(x)
|
s494490228
|
p02468
|
u567380442
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,724 | 279 |
For given integers m and n, compute mn (mod 1,000,000,007). Here, A (mod M) is the remainder when A is divided by M.
|
def power(m, n):
if n == 1:
return m
div = 1000000007
ret = power(m * m, n // 2) % div
if n % 2:
ret %= div
return ret
from sys import stdin
readline = stdin.readline
m, n = map(int, readline().split())
print(power(m, n))
|
s260959541
|
Accepted
| 30 | 6,728 | 220 |
def power(m, n):
if n:return power(m * m % 1000000007, n // 2) * (m if n % 2 else 1) % 1000000007
return 1
from sys import stdin
readline = stdin.readline
m, n = map(int, readline().split())
print(power(m, n))
|
s070270683
|
p02612
|
u459521923
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 8,836 | 59 |
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 = 111200
n2 = str(n)[-3:]
ans = 1000 - int(n2)
print(ans)
|
s796849987
|
Accepted
| 28 | 9,160 | 135 |
# -*- coding: utf-8 -*-
n = int(input())
n2 = str(n)[-3:]
if int(n2) == 0:
print(0)
else:
ans = 1000 - int(n2)
print(ans)
|
s712564009
|
p03545
|
u750485713
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 262 |
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.
|
S = input()
a = len(str(S))-1
for i in range(2**a):
temp = str(S)[0]
for j in range(a):
if i & (1<<j):
temp += "+"
else:
temp += "-"
temp+=str(S)[j+1]
if eval(temp) == 7:
print(temp)
|
s982381234
|
Accepted
| 18 | 3,060 | 290 |
S = input()
a = len(str(S))-1
for i in range(2**a):
temp = str(S)[0]
for j in range(a):
if i & (1<<j):
temp += "+"
else:
temp += "-"
temp+=str(S)[j+1]
if eval(temp) == 7:
print("{}=7".format(temp))
break
|
s088032377
|
p03599
|
u371763408
| 3,000 | 262,144 |
Wrong Answer
| 74 | 7,412 | 366 |
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
a,b,c,d,e,f=map(int,input().split())
a100=a*100
b100=b*100
ok=[]
for ai in range(31):
for bi in range(31):
for ci in range(31):
for di in range(31):
ab= ai*a100+bi*b100
cd= ci*c+di*d
if ab + cd >f:
break
else:
if (ab/100)*e>=cd:
ok.append([ab+cd,cd])
print(sorted(ok,key=lambda x:x[1])[-1])
|
s583269643
|
Accepted
| 441 | 45,316 | 452 |
a,b,c,d,e,f=map(int,input().split())
a100=a*100
b100=b*100
ok=[]
for ai in range(31):
for bi in range(31):
for ci in range(101):
for di in range(101):
ab= ai*a100+bi*b100
cd= ci*c+di*d
if ab + cd >f:
break
else:
if (ab/100)*e>=cd:
ok.append([ab+cd,cd])
print(*sorted(ok, key=lambda x: (100*x[1] / x[0] if x[0] else 0))[-1])
|
s649468875
|
p03695
|
u667458133
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 507 |
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
N = int(input())
a = list(map(int, input().split()))
array1 = [400, 800, 1200, 1600, 2000, 2400, 2800, 3200]
array2 = [True for _ in range(8)]
result1 = 0
plus = 0
for i in range(N):
print(i)
if a[i] >= 3200:
plus += 1
continue
for j in range(len(array1)):
if a[i] < array1[j]:
if array2[j]:
result1 += 1
array2[j] = False
break
result2 = result1 + plus
if result2 > 8:
result2 = 8
print(result1, result2)
|
s791272072
|
Accepted
| 17 | 3,064 | 460 |
N = int(input())
a = list(map(int, input().split()))
array1 = [400, 800, 1200, 1600, 2000, 2400, 2800, 3200]
array2 = [True for _ in range(8)]
result1 = 0
plus = 0
for i in range(N):
if a[i] >= 3200:
plus += 1
continue
for j in range(8):
if a[i] < array1[j]:
if array2[j]:
result1 += 1
array2[j] = False
break
result2 = result1 + plus
print(max(result1, 1), result2)
|
s695854596
|
p03545
|
u189604332
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 1,235 |
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.
|
class TrainTicket():
def calculate(self, S: list):
for x in range(2):
for y in range(2):
for z in range (2):
if x == 1:
res = S[0] + S[1]
else:
res = S[0] - S[1]
if y ==1:
res += S[2]
else:
res -= S[2]
if z == 1:
res += S[3]
else:
res -= S[3]
if res == 7:
if x == 1:
xr = "+"
else:
xr = "-"
if y == 1:
yr = "+"
else:
yr= "-"
if z == 1:
zr = "+"
else:
zr = "-"
return(str(S[0]) + xr + str(S[1]) + yr + str(S[2]) + zr + str(S[3]))
if __name__ =="__main__":
S = list(map(int, input()))
print(S)
tt = TrainTicket()
res = tt.calculate(S)
print(str(res)+"=7")
|
s266518336
|
Accepted
| 18 | 3,064 | 1,222 |
class TrainTicket():
def calculate(self, S: list):
for x in range(2):
for y in range(2):
for z in range (2):
if x == 1:
res = S[0] + S[1]
else:
res = S[0] - S[1]
if y ==1:
res += S[2]
else:
res -= S[2]
if z == 1:
res += S[3]
else:
res -= S[3]
if res == 7:
if x == 1:
xr = "+"
else:
xr = "-"
if y == 1:
yr = "+"
else:
yr= "-"
if z == 1:
zr = "+"
else:
zr = "-"
return(str(S[0]) + xr + str(S[1]) + yr + str(S[2]) + zr + str(S[3]))
if __name__ =="__main__":
S = list(map(int, input()))
tt = TrainTicket()
res = tt.calculate(S)
print(str(res)+"=7")
|
s604282037
|
p03372
|
u270681687
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 15,832 | 709 |
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter. Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories. Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter. Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
|
n, c = map(int, input().split())
x = []
v = []
vcum = []
cum = 0
for i in range(n):
X, V = map(int, input().split())
x.append(X)
cum += V
v.append(V)
vcum.append(cum)
maxnut = 0
for i in range(n):
maxnut = max(maxnut, vcum[i] - x[i])
for i in range(n):
if i == 0:
maxnut = max(maxnut, vcum[n-1] - (c - x[i]))
else:
maxnut = max(maxnut, vcum[n-1] - vcum[i-1] - (c - x[i]))
for i in range(n):
for j in range(i+1, n):
maxnut = max(maxnut, vcum[i] - 2 * x[i] + vcum[n-1] - vcum[j-1], - (c - x[j]))
for i in range(n):
for j in range(i+1, n):
maxnut = max(maxnut, vcum[i] - x[i] + vcum[n-1] - vcum[j-1], - (c - x[j]) * 2)
print(maxnut)
|
s585115889
|
Accepted
| 538 | 40,948 | 2,125 |
import sys
input = sys.stdin.readline
n, c = map(int, input().split())
sushi = [list(map(int, input().split())) for _ in range(n)]
if n == 1:
print(max(sushi[0][1]-sushi[0][0], sushi[0][1]-(c-sushi[0][0]), 0))
exit()
x = []
for i in range(n):
x.append(sushi[i][0])
vcum = []
v = 0
for i in range(n):
v += sushi[i][1]
vcum.append(v)
ans = 0
for i in range(n):
if ans < vcum[i] - x[i]:
ans = vcum[i] - x[i]
for i in range(n):
if i == 0:
if ans < vcum[n-1] - (c - x[0]):
ans = vcum[n-1] - (c - x[0])
else:
if ans < (vcum[n - 1] - vcum[i - 1]) - (c - x[i]):
ans = (vcum[n - 1] - vcum[i - 1]) - (c - x[i])
right_dp = [0] * n
right_dp[n-1] = vcum[n-1] - vcum[n-2] - (c - x[n-1])
for i in reversed(range(1, n-1)):
right_dp[i] = max(right_dp[i+1], vcum[n-1] - vcum[i-1] - (c - x[i]))
for i in range(n-1):
cal = vcum[i] - 2 * x[i] + right_dp[i+1]
if ans < cal:
ans = cal
left_dp = [0] * n
left_dp[0] = vcum[0] - x[0]
for i in range(1, n):
left_dp[i] = max(left_dp[i-1], vcum[i] - x[i])
for i in range(1, n):
cal = vcum[n-1] - vcum[i-1] - 2 * (c - x[i]) + left_dp[i-1]
if ans < cal:
ans = cal
print(ans)
|
s496994255
|
p03644
|
u037221289
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 69 |
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())
L = [(1,2), (2,4), (3,8), (4,16), (5,32), (6,64)]
|
s464115845
|
Accepted
| 18 | 3,064 | 310 |
N = int(input())
L = [(N,0),(1,0), (2,1), (4,2), (8,3), (16,4), (32,5), (64,6),(128,7)]
sorted_L = sorted(L,key=lambda x:x[0])
for i,j in enumerate(sorted_L):
if j[0] == N:
if sorted_L[i][0] == sorted_L[i+1][0]:
print(sorted_L[i][0])
break
else:
print(sorted_L[i-1][0])
break
|
s183037969
|
p03479
|
u516554284
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,068 | 76 |
As a token of his gratitude, Takahashi has decided to give his mother an integer sequence. The sequence A needs to satisfy the conditions below: * A consists of integers between X and Y (inclusive). * For each 1\leq i \leq |A|-1, A_{i+1} is a multiple of A_i and strictly greater than A_i. Find the maximum possible length of the sequence.
|
x,y=map(int,input().split())
ans=0
while x<=y:
x=x*2
ans+1
print(ans)
|
s052891694
|
Accepted
| 27 | 9,160 | 79 |
x,y=map(int,input().split())
ans=0
while x<=y:
x=x*2
ans+=1
print(ans)
|
s532442032
|
p03737
|
u556371693
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 67 |
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.
|
a,b,c=input().split()
print(a[0].upper(),b[0].upper(),c[0].upper())
|
s726401668
|
Accepted
| 17 | 2,940 | 76 |
a,b,c=input().split()
print(a[0].upper(), b[0].upper(), c[0].upper(),sep="")
|
s076173431
|
p03080
|
u493610446
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 141 |
There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat.
|
name = input()
count = 0
for i in name:
if i == "R":
count += 1
if count * 2 > len(name):
print("Yes")
else:
print("No")
|
s524958756
|
Accepted
| 18 | 2,940 | 149 |
input()
name = input()
count = 0
for i in name:
if i == "R":
count += 1
if count * 2 > len(name):
print("Yes")
else:
print("No")
|
s736876211
|
p02831
|
u699944218
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 127 |
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
def lcm(x, y):
return (x * y) // math.gcd(x, y)
def main(A, B):
A, B = list(map(int,input().split()))
print(lcm(A, B))
|
s915598090
|
Accepted
| 28 | 9,084 | 86 |
from math import gcd
A, B = list(map(int,input().split()))
print(A * B // gcd(A, B))
|
s361610740
|
p03637
|
u941434715
| 2,000 | 262,144 |
Wrong Answer
| 191 | 23,032 | 338 |
We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective.
|
#coding: utf-8
import math
import heapq
import bisect
import numpy as np
from collections import Counter
#from scipy.misc import comb
N = int(input())
A = list(map(int, input().split()))
cnt1 = 0
cnt4 = 0
for a in A:
if a%4 == 0:
cnt4 += 1
else:
cnt1 += 1
if cnt1 <= cnt4:
print("Yes")
else:
print("No")
|
s826118123
|
Accepted
| 196 | 23,096 | 438 |
#coding: utf-8
import math
import heapq
import bisect
import numpy as np
from collections import Counter
#from scipy.misc import comb
N = int(input())
A = list(map(int, input().split()))
cnt1 = 0
cnt2 = 0
cnt4 = 0
for a in A:
if a%4 == 0:
cnt4 += 1
elif a%2 == 0:
cnt2 += 1
else:
cnt1 += 1
if cnt1 <= cnt4:
print("Yes")
elif cnt1 == cnt4 + 1 and cnt2 == 0:
print("Yes")
else:
print("No")
|
s960966141
|
p03448
|
u594956556
| 2,000 | 262,144 |
Wrong Answer
| 40 | 3,060 | 224 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
X = X // 50
ans = 0
for ai in range(A):
for bi in range(B):
for ci in range(C):
if 10*ai + 2*bi + ci == X:
ans += 1
print(ans)
|
s015116250
|
Accepted
| 43 | 3,060 | 231 |
A = int(input())
B = int(input())
C = int(input())
X = int(input())
X = X // 50
ans = 0
for ai in range(A+1):
for bi in range(B+1):
for ci in range(C+1):
if 10*ai + 2*bi + ci == X:
ans += 1
print(ans)
|
s846623970
|
p03494
|
u998733244
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 158 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n = int(input())
A = list(map(int, input().split()))
print(A)
count = 0
while all(a%2==0 for a in A):
A = [a/2 for a in A]
count += 1
print(count)
|
s877085678
|
Accepted
| 19 | 3,060 | 216 |
n = int(input())
A = list(map(int, input().split()))
count = 0
while True:
for i in range(n):
if A[i]%2==1:
print(count)
exit()
A[i] = A[i]/2
count += 1
print(count)
|
s245578983
|
p03994
|
u254871849
| 2,000 | 262,144 |
Wrong Answer
| 62 | 5,968 | 546 |
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times. * Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`. For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`. Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
|
# 2019-11-26 17:11:55(JST)
import sys
from string import ascii_lowercase as alph
ind_alph = dict((char, ind) for ind, char in enumerate(alph))
def main():
s, k = sys.stdin.read().split()
s = list(s)
k = int(k)
n = len(s)
i = 0
while k:
j = ind_alph[s[i]]
diff = (26 - j) % 26
if diff <= k:
s[i] = 'a'
k -= diff
if i == n - 1 and k:
s[i] = alph[(j + k) % 26]
k = 0
i += 1
print(s)
if __name__ == '__main__':
main()
|
s251096259
|
Accepted
| 56 | 4,660 | 554 |
# 2019-11-26 17:11:55(JST)
import sys
from string import ascii_lowercase as alph
ind_alph = dict((char, ind) for ind, char in enumerate(alph))
def main():
s, k = sys.stdin.read().split()
s = list(s)
k = int(k)
n = len(s)
i = 0
while k:
diff = (26 - ind_alph[s[i]]) % 26
if diff <= k:
s[i] = 'a'
k -= diff
if i == n - 1 and k:
s[i] = alph[(ind_alph[s[i]] + k) % 26]
k = 0
i += 1
print(''.join(s))
if __name__ == '__main__':
main()
|
s552036264
|
p03044
|
u068112689
| 2,000 | 1,048,576 |
Wrong Answer
| 807 | 46,632 | 534 |
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
n = int(input())
links = [[] for _ in range(n+1)]
for _ in range(n-1):
u,v,w = map(int, input().split())
links[u].append([v, w])
links[v].append([u, w])
ans = [0] * (n+1)
q = [(1, -1)]
while q:
me, parent = q.pop()
for child, w in links[me]:
if child == parent:
continue
if w % 2 ==0:
ans[me] = 1
ans[child] = 1
q.append((child, me))
print(*ans[1:])
|
s433276334
|
Accepted
| 677 | 38,828 | 469 |
n = int(input())
graph = [[] for i in range(n)]
for i in range(n-1):
u, v, w = map(int, input().split())
graph[u-1].append((v-1, w))
graph[v-1].append((u-1, w))
colors = [-1] * n
colors[0] = 0
Me_Parent = [[0, -1]]
while Me_Parent:
m, p = Me_Parent.pop()
Child = graph[m]
for c, w in Child:
if c != p:
colors[c] = (colors[m] + w) % 2
Me_Parent.append([c, m])
for i in range(n):
print(colors[i])
|
s683187628
|
p02612
|
u465597982
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,148 | 45 |
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())
a1=n//1000
print(n-(a1*1000))
|
s820657904
|
Accepted
| 34 | 9,176 | 151 |
import math
n=int(input())
if(n<1000):
print(1000-n)
elif(n%1000!=0):
a=int(math.floor(n//1000))
print(((a+1)*1000)-n)
else:
print("0")
|
s218531922
|
p03657
|
u691018832
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 172 |
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.
|
# sys.stdin.readline()
import sys
input = sys.stdin.readline
a, b = map(int, input().split())
if (a*b)%3 == 0:
ans = 'Possible'
else:
ans = 'Impossible'
print(ans)
|
s551997407
|
Accepted
| 17 | 2,940 | 196 |
# sys.stdin.readline()
import sys
input = sys.stdin.readline
a, b = map(int, input().split())
if (a+b)%3 == 0 or a%3 == 0 or b%3 == 0:
ans = 'Possible'
else:
ans = 'Impossible'
print(ans)
|
s184741387
|
p03578
|
u598229387
| 2,000 | 262,144 |
Wrong Answer
| 401 | 35,556 | 411 |
Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems.
|
n = int(input())
d = [int(i) for i in input().split()]
m = int(input())
t = [int(i) for i in input().split()]
d = sorted(d)
t = sorted(t)
left = 0
ans = 'Yes'
for i in range(m):
if t[i] < d[left]:
ans='No'
break
elif t[i] == d[left]:
left+=1
if i!= m-1 and left >=n:
ans = 'No'
else:
left+=1
if left>=n:
ans = 'No'
print(ans)
|
s839324442
|
Accepted
| 386 | 35,012 | 571 |
n = int(input())
d = [int(i) for i in input().split()]
m = int(input())
t = [int(i) for i in input().split()]
d = sorted(d)
t = sorted(t)
left = 0
ans = 'YES'
for i in range(m):
if t[i] < d[left]:
ans='NO'
break
elif t[i] == d[left]:
left+=1
if i!= m-1 and left >=n:
ans = 'NO'
break
else:
while t[i] != d[left]:
left+=1
if left >= n:
ans='NO'
break
left+=1
if left>=n:
ans = 'NO'
break
print(ans)
|
s407008611
|
p02389
|
u283452598
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,488 | 40 |
Write a program which calculates the area and perimeter of a given rectangle.
|
a,b=input().split()
print(int(a)*int(b))
|
s910550727
|
Accepted
| 20 | 7,524 | 62 |
x=input().split()
a=int(x[0])
b=int(x[1])
print(a*b,(a*2+b*2))
|
s260109232
|
p03657
|
u932465688
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 165 |
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):
print('Possible')
else:
if ((A+B) % 3 == 0):
print('Posssible')
else:
print('Impossible')
|
s253290751
|
Accepted
| 17 | 2,940 | 164 |
A,B = map(int,input().split())
if (A % 3 == 0 or B % 3 == 0):
print('Possible')
else:
if ((A+B) % 3 == 0):
print('Possible')
else:
print('Impossible')
|
s411603095
|
p03739
|
u136090046
| 2,000 | 262,144 |
Wrong Answer
| 95 | 14,468 | 451 |
You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero. * For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.
|
n = int(input())
array = list(map(int, input().split()))
total = 0
flag = -1
minus = 0
for tmp in range(len(array)):
total += array[tmp]
if total * flag <= 0:
minus += abs(total*flag)+1
total = flag
flag *= -1
total = 0
flag = 1
plus = 0
for tmp in range(len(array)):
total += array[tmp]
if total * flag <= 0:
plus += abs(total*flag)+1
total = flag
flag *= -1
print(min(minus, plus))
|
s847626051
|
Accepted
| 132 | 14,468 | 442 |
n = int(input())
array = list(map(int, input().split()))
total = 0
flag = -1
minus = 0
for tmp in range(len(array)):
total += array[tmp]
if total * flag <= 0:
minus += abs(total*flag)+1
total = flag
flag *= -1
total = 0
flag = 1
plus = 0
for tmp in range(len(array)):
total += array[tmp]
if total * flag <= 0:
plus += abs(total*flag)+1
total = flag
flag *= -1
print(min(minus, plus))
|
s319402219
|
p03433
|
u873973257
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 86 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
pay = input()
oney = input()
print('YES' if (int(pay) % 500) <= int(oney) else 'NO' )
|
s341433835
|
Accepted
| 17 | 2,940 | 86 |
pay = input()
oney = input()
print('Yes' if (int(pay) % 500) <= int(oney) else 'No' )
|
s967946426
|
p04043
|
u536836760
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 190 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
x = list(input().split())
five = 0
seven = 0
for i in x:
if i == 5:
five += five
elif i == 7:
seven += seven
if seven == 1 and five == 2:
print('YES')
else:
print('NO')
|
s017110797
|
Accepted
| 17 | 2,940 | 191 |
x = list(map(int,input().split()))
five = 0
seven = 0
for i in x:
if i == 5:
five += 1
elif i == 7:
seven += 1
if seven == 1 and five == 2:
print('YES')
else:
print('NO')
|
s803525365
|
p02608
|
u688219499
| 2,000 | 1,048,576 |
Wrong Answer
| 489 | 16,772 | 290 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
n = int(input())
for i in range(5):
print(0)
count = [0] * (10**6)
s = 0
for i in range(1, 101):
for j in range(1, 101):
for k in range(1, 101):
s = i * i + j * j + k * k + i * j + j * k + k * i
count[s] += 1
for i in range(n):
print(count[i])
|
s768936039
|
Accepted
| 487 | 16,904 | 262 |
n = int(input())
count = [0] * (10**6)
s = 0
for i in range(1, 101):
for j in range(1, 101):
for k in range(1, 101):
s = i * i + j * j + k * k + i * j + j * k + k * i
count[s - 1] += 1
for i in range(n):
print(count[i])
|
s183854303
|
p02399
|
u279955105
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,664 | 114 |
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
a,b = list(map(int, input().split()))
d = int(a / b)
r = a % b
f = a / b
print(str(d) + " "+ str(r) + " "+ str(f))
|
s620373310
|
Accepted
| 20 | 5,608 | 158 |
a,b = map(int, input().split())
answer1 = a//b
answer2 = a % b
answer3 = a / b
print(str(answer1) + ' ' + str(answer2) + ' ' + str('{:.5f}'.format(answer3)))
|
s436649529
|
p03610
|
u350836088
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,188 | 26 |
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s = input()
print(s[1::2])
|
s298015849
|
Accepted
| 17 | 3,192 | 19 |
print(input()[::2])
|
s875799482
|
p03796
|
u611090896
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 37 |
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
N = int(input())
print(N % (10**9+7))
|
s120857152
|
Accepted
| 154 | 9,908 | 57 |
import math
print(math.factorial(int(input()))%(10**9+7))
|
s414588009
|
p03493
|
u113255362
| 2,000 | 262,144 |
Wrong Answer
| 27 | 8,948 | 106 |
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()
List = [S[0], S[1], S[2]]
res = 0
for i in range(3):
if List[i] == 1:
res +=1
print(res)
|
s120042356
|
Accepted
| 31 | 9,028 | 108 |
S = input()
List = [S[0], S[1], S[2]]
res = 0
for i in range(3):
if List[i] == "1":
res +=1
print(res)
|
s006594847
|
p03679
|
u521323621
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,160 | 95 |
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 x >= b-a:
print("delicious")
else:
print("dangerous")
|
s949670853
|
Accepted
| 28 | 9,004 | 127 |
x,a,b = map(int, input().split())
if 0 >= b-a:
print("delicious")
elif x >= b-a:
print("safe")
else:
print("dangerous")
|
s741804908
|
p03369
|
u775652851
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,044 | 74 |
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()
t=0
for i in range(3):
if a[i]=="o":
t+=1
print(500+t*100)
|
s290442310
|
Accepted
| 29 | 8,980 | 75 |
a=input()
t=0
for i in range(3):
if a[i]=="o":
t+=1
print(700+t*100)
|
s750186630
|
p03455
|
u902361509
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 92 |
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("Odd")
else:
print("Even")
|
s402998564
|
Accepted
| 18 | 2,940 | 91 |
a, b = map(int, input().split())
if (a*b)%2 == 0:
print("Even")
else:
print("Odd")
|
s291650263
|
p03478
|
u210545407
| 2,000 | 262,144 |
Wrong Answer
| 39 | 3,060 | 362 |
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(" "))
wholeSum = 0
def digitSum(n):
basis = 10
transDigit = list()
while True:
transDigit.append(n % basis)
n = n / basis
if n < 1: break
result = sum(transDigit)
return result
for pivot in range(n+1):
target = digitSum(pivot)
if target <= b\
and a <= target:
wholeSum += pivot
print(wholeSum)
|
s489765838
|
Accepted
| 34 | 2,940 | 257 |
n, a, b = map(int, input().split(" "))
wholeSum = 0
def digitSum(n):
result = sum(map(int, list(str(n))))
return result
for pivot in range(n+1):
target = digitSum(pivot)
if target <= b\
and a <= target:
wholeSum += pivot
print(wholeSum)
|
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