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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s644170469
|
p03024
|
u879077265
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 85 |
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
S = input()
t = S.count("x")
if t <= 7:
print("Yes")
else:
print("No")
|
s032646098
|
Accepted
| 17 | 2,940 | 86 |
S = input()
t = S.count("x")
if t <= 7:
print("YES")
else:
print("NO")
|
s184630832
|
p03636
|
u853900545
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 36 |
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
s=input()
print(s[0]+'len(s)'+s[-1])
|
s047065288
|
Accepted
| 17 | 2,940 | 46 |
s=input()
a=len(s)-2
print(s[0]+str(a)+s[-1])
|
s963042394
|
p03434
|
u557437077
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 203 |
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 = input()
a = [int(i) for i in input().split()]
a = sorted(a)
alice = 0
bob = 0
for i in reversed(range(len(a))):
if i % 2 == 0:
alice += a[i]
else:
bob += a[i]
print(alice-bob)
|
s040012186
|
Accepted
| 17 | 3,060 | 200 |
n = input()
a = [int(i) for i in input().split()]
a = sorted(a)
alice = 0
bob = 0
for i in range(len(a)):
if i % 2 == 0:
alice += a[-i-1]
else:
bob += a[-i- 1]
print(alice-bob)
|
s100959612
|
p03473
|
u786020649
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,080 | 25 |
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
print(24+int(input())%24)
|
s653112244
|
Accepted
| 26 | 9,152 | 26 |
print(48-int(input())%24)
|
s398308816
|
p02412
|
u546968095
| 1,000 | 131,072 |
Wrong Answer
| 30 | 5,652 | 312 |
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
import math
while True:
n,x = map(int,input().split())
if n == x == 0:
break
c = 0
for i in range(math.ceil(x/3), x-3):
for j in range(math.ceil(i/2), i):
k = i - j
if x == i + j + k:
#print(x, i, j, k)
c += 1
print(c)
|
s180606376
|
Accepted
| 490 | 5,596 | 272 |
while True:
n,x = map(int,input().split())
if n == x == 0:
break
n = n + 1
c = 0
for i in range(1,n):
for j in range(i+1,n):
for k in range(j+1,n):
if x == i + j + k:
c += 1
print(c)
|
s494841186
|
p03569
|
u754022296
| 2,000 | 262,144 |
Wrong Answer
| 301 | 3,316 | 228 |
We have a string s consisting of lowercase English letters. Snuke can perform the following operation repeatedly: * Insert a letter `x` to any position in s of his choice, including the beginning and end of s. Snuke's objective is to turn s into a palindrome. Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required.
|
s = input()
ans = 0
while s:
if s[0] == s[-1]:
s = s[1:-2]
else:
if s[0] == "x":
ans += 1
s = s[1:]
elif s[-1] == "x":
ans += 1
s = s[:-1]
else:
print(-1)
exit()
print(ans)
|
s243625806
|
Accepted
| 309 | 3,316 | 228 |
s = input()
ans = 0
while s:
if s[0] == s[-1]:
s = s[1:-1]
else:
if s[0] == "x":
ans += 1
s = s[1:]
elif s[-1] == "x":
ans += 1
s = s[:-1]
else:
print(-1)
exit()
print(ans)
|
s721124918
|
p03711
|
u975676823
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 213 |
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
a, b = map(int, input().split())
if(a in [4, 6, 9, 11]):
print("YES" if b in [4, 6, 9, 11] else "NO")
elif(a in [1, 3, 5, 7, 8, 10, 12]):
print("YES" if b in [1, 3, 5, 7, 8, 10, 12] else "NO")
else:
print("NO")
|
s523910885
|
Accepted
| 17 | 3,064 | 217 |
a, b = map(int, input().split())
if(a in [4, 6, 9, 11]):
print("Yes" if (b in [4, 6, 9, 11]) else "No")
elif(a in [1, 3, 5, 7, 8, 10, 12]):
print("Yes" if (b in [1, 3, 5, 7, 8, 10, 12]) else "No")
else:
print("No")
|
s282669725
|
p03697
|
u319805917
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 203 |
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=input()
LIs=list(a)
count=0
for i in range(len(LIs)):
for g in range (len(LIs)):
if LIs[i]==LIs[g]:
count+=1
if count==0:
print("yes")
else:
print("no")
|
s440486559
|
Accepted
| 18 | 2,940 | 82 |
a,b=map(int,input().split())
if a+b>=10:
print("error")
else:
print(a+b)
|
s304069826
|
p03139
|
u286955577
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 147 |
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.
|
def solve(N, A, B):
maxNum = max(A, B)
minNum = max(0, N - (A + B))
print(maxNum, minNum)
N, A, B = map(int, input().split())
solve(N, A, B)
|
s846606817
|
Accepted
| 17 | 3,064 | 148 |
def solve(N, A, B):
maxNum = min(A, B)
minNum = max(0, (A + B) - N)
print(maxNum, minNum)
N, A, B = map(int, input().split())
solve(N, A, B)
|
s686095770
|
p02795
|
u815763296
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,188 | 321 |
We have a grid with H rows and W columns, where all the squares are initially white. You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions: * Choose one row, then paint all the squares in that row black. * Choose one column, then paint all the squares in that column black. At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations.
|
import sys
H = int(input())
W = int(input())
N = int(input())
X = max(H, W)
Y = min(H, W)
count = 0
total = 0
for i in range(Y):
total += X
count += 1
if count >= N:
print(count)
sys.exit()
for i in range(X):
total += Y
count += 1
if count >= N:
print(count)
break
|
s677472079
|
Accepted
| 28 | 9,092 | 322 |
import sys
H = int(input())
W = int(input())
N = int(input())
X = max(H, W)
Y = min(H, W)
count = 0
total = 0
for _ in range(Y):
total += X
count += 1
if total >= N:
print(count)
sys.exit()
for _ in range(X):
total += Y
count += 1
if ctotal >= N:
print(count)
break
|
s506243424
|
p00052
|
u990228206
| 1,000 | 131,072 |
Wrong Answer
| 260 | 5,596 | 322 |
n! = n × (n − 1) × (n − 2) × ... × 3 × 2 × 1 を n の階乗といいます。例えば、12 の階乗は 12! = 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 479001600 となり、末尾に 0 が 2 つ連続して並んでいます。 整数 n を入力して、n! の末尾に連続して並んでいる 0 の数を出力するプログラムを作成してください。ただし、n は 20000 以下の正の整数とします。
|
while 1:
try:
zero=0
number=1
n=int(input())
for i in range(1,n+1):
number*=i
while len(str(number))>7:number=int(str(number)[1:])
while number%10==0:
zero+=1
number=int(number/10)
print(zero)
except:break
|
s526695858
|
Accepted
| 260 | 5,588 | 274 |
while 1:
zero=0
number=1
n=int(input())
if n==0:break
for i in range(1,n+1):
number*=i
while len(str(number))>7:number=int(str(number)[1:])
while number%10==0:
zero+=1
number=int(number/10)
print(zero)
|
s469877092
|
p02396
|
u427088273
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,552 | 109 |
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.
|
num = list(map(int,input().split()))
for i,s in enumerate(num,1):
print("Case {0}: {1}".format(i,s))
|
s672760380
|
Accepted
| 150 | 7,652 | 160 |
count = 0
while(True):
num = int(input())
if num == 0:
break
count += 1
print("Case {0}: {1}".format(count,num))
|
s911961696
|
p03456
|
u743164083
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 91 |
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.
|
n = input().split()
x = int("".join(n))
print("Yes") if x%x**0.5 == x**0.5 else print("No")
|
s688248469
|
Accepted
| 17 | 2,940 | 92 |
n = input().split()
x = int("".join(n))**0.5
print("Yes") if x.is_integer() else print("No")
|
s740494678
|
p03546
|
u293992530
| 2,000 | 262,144 |
Wrong Answer
| 40 | 9,504 | 704 |
Joisino the magical girl has decided to turn every single digit that exists on this world into 1. Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points). She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive). You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows: * If A_{i,j}≠-1, the square contains a digit A_{i,j}. * If A_{i,j}=-1, the square does not contain a digit. Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end.
|
import sys
sys.setrecursionlimit(2147483647)
input=sys.stdin.readline
import math
from heapq import heappush, heappop
def solve(h,w,tables,A):
for i in range(10):
for j in range(10):
for k in range(10):
tables[j][k] = min(tables[j][k], tables[j][i] + tables[i][k])
cost = 0
for i in range(h):
for j in range(w):
p = A[i][j]
cost += tables[p][1]
return cost
def main():
h, w = map(int, input().split(' '))
tables = []
for i in range(10):
tables.append(list(map(int, input().split(' '))))
A = []
for _ in range(h):
A.append(list(map(int, input().split(' '))))
ans = solve(h, w, tables, A)
print(ans)
if __name__=='__main__':
main()
|
s525074344
|
Accepted
| 38 | 9,616 | 739 |
import sys
sys.setrecursionlimit(2147483647)
input=sys.stdin.readline
import math
from heapq import heappush, heappop
def solve(h,w,tables,A):
for i in range(10):
for j in range(10):
for k in range(10):
tables[j][k] = min(tables[j][k], tables[j][i] + tables[i][k])
cost = 0
for i in range(h):
for j in range(w):
p = A[i][j]
if p == -1:
continue
cost += tables[p][1]
return cost
def main():
h, w = map(int, input().split(' '))
tables = []
for i in range(10):
tables.append(list(map(int, input().split(' '))))
A = []
for _ in range(h):
A.append(list(map(int, input().split(' '))))
ans = solve(h, w, tables, A)
print(ans)
if __name__=='__main__':
main()
|
s467045647
|
p03455
|
u209616713
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 183 |
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()
c = a+b
d = int(c)
if d % 2 == 0:
if d % 4 == 0:
print('Yes')
else:
print('No')
else:
if d % 4 == 1:
print('Yes')
else:
print('No')
|
s081647867
|
Accepted
| 17 | 2,940 | 91 |
a,b = map(int,input().split())
c = a*b
if c % 2 == 0:
print('Even')
else:
print('Odd')
|
s886885570
|
p03860
|
u633450100
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 31 |
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.
|
s = input()
print('A'+s[0]+'C')
|
s556242143
|
Accepted
| 17 | 2,940 | 55 |
s = [s for s in input().split()]
print('A'+s[1][0]+'C')
|
s344980789
|
p03854
|
u548303713
| 2,000 | 262,144 |
Wrong Answer
| 2,107 | 3,316 | 950 |
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`.
|
#ABC49-C
S=input()
divide=["dream","dreamer","erase","eraser"]
for i in range(4):
divide[i]=divide[i][::-1]
len_s=len(S)
S=S[::-1]
check=0
i=0
a=[]
while check==0:
if i==len_s:
a.append(1)
break
if len_s-i <5:
check=1
a.append(2)
break
if S[i:i+5] == divide[0]:
i=i+5
a.append(3)
continue
if S[i:i+5] == divide[2]:
i=i+5
a.append(4)
continue
if len_s-i <7:
check=1
a.append(5)
break
if S[i:i+7] == divide[1]:
i=i+7
a.append(6)
continue
if S[i:i+7] == divide[3]:
i=i+7
a.append(7)
continue
if check==0:
print("YES")
else:
print("NO")
print(a)
|
s679652609
|
Accepted
| 34 | 3,444 | 1,164 |
#ABC49-C
S=input()
divide=["dream","dreamer","erase","eraser"]
for i in range(4):
divide[i]=divide[i][::-1]
len_s=len(S)
S=S[::-1]
check=0
i=0
a=[]
while check==0:
if i==len_s:
a.append(1)
break
if len_s-i <5:
check=1
a.append(2)
break
if S[i:i+5] == divide[0]:
i=i+5
a.append(3)
continue
if S[i:i+5] == divide[2]:
i=i+5
a.append(4)
continue
if len_s-i <6:
check=1
a.append(5)
break
if S[i:i+6] == divide[3]:
i=i+6
a.append(6)
continue
if len_s-i <7:
check=1
a.append(7)
break
if S[i:i+7] == divide[1]:
i=i+7
a.append(8)
continue
check=1
if check==0:
print("YES")
else:
print("NO")
#print(a)
|
s068071899
|
p02396
|
u587193722
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,632 | 93 |
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.
|
a = [int(i) for i in input().split()]
for i in range(1, len(a)):
print('Case',i,':',a[i])
|
s833410132
|
Accepted
| 90 | 8,408 | 129 |
a =[]
x = 1
while x > 0:
x = int(input())
a.append(x)
for i in range(0,len(a)-1):
print('Case ',i+1,': ',a[i],sep='')
|
s893435251
|
p03556
|
u088974156
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 28 |
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
n=int(input())
print(n**0.5)
|
s297868218
|
Accepted
| 17 | 3,060 | 36 |
n=int(input())
print(int(n**0.5)**2)
|
s123834903
|
p02256
|
u963402991
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,624 | 209 |
Write a program which finds the greatest common divisor of two natural numbers _a_ and _b_
|
def gcd(a,b):
r = a % b
while r != 0:
print (a, b)
a, b = b, r
r = a % b
if r == 0:
break
return b
a,b = list(map(int,input().split()))
print (gcd(a,b))
|
s116475366
|
Accepted
| 20 | 7,636 | 151 |
def gcd(a,b):
r = a % b
while r != 0:
a, b = b, r
r = a % b
return b
a,b = list(map(int,input().split()))
print (gcd(a,b))
|
s736531050
|
p02742
|
u267983787
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 85 |
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:
|
b, c = map(int, input().split())
a = b*c
if a%2 == 0: print(a/2)
else: print(a/2 + 1)
|
s845264663
|
Accepted
| 17 | 2,940 | 140 |
b, c = map(int, input().split())
a = b*c
if b == 1:print(1)
elif c == 1:print(1)
elif a%2 == 0: print(int(a/2))
else: print(int((a + 1)/2))
|
s331394271
|
p03044
|
u630281418
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 29 |
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.
|
for i in range(5):
print(1)
|
s582175925
|
Accepted
| 677 | 50,556 | 500 |
import sys
input = sys.stdin.readline
N = int(input())
tree = [[] for _ in range(N)]
for i in range(N-1):
u, v, w = map(int, input().split())
tree[u-1].append([v-1,w])
tree[v-1].append([u-1,w])
dist = [None]*N
dist[0] = 0
comp = [0]*N
next_ptr = [0]
for ptr in next_ptr:
comp[ptr] = 1
edges = tree[ptr]
for edge in edges:
if comp[edge[0]] == 0:
dist[edge[0]] = dist[ptr] + edge[1]
next_ptr.append(edge[0])
for item in dist:
print(item%2)
|
s396912861
|
p02645
|
u483331319
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,056 | 24 |
When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.
|
S = input()
print(S[3:])
|
s364536801
|
Accepted
| 26 | 9,024 | 24 |
S = input()
print(S[:3])
|
s277906695
|
p02694
|
u021387650
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,020 | 103 |
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?
|
X = int(input())
Deposit = 100
ans = 0
while Deposit < X:
Deposit *= 1.001
ans += 1
print(ans)
|
s378252281
|
Accepted
| 25 | 9,168 | 118 |
X = int(input())
Deposit = 100
ans = 0
while Deposit < X:
Deposit = int(Deposit * 1.01)
ans += 1
print(ans)
|
s399005469
|
p03623
|
u327248573
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 103 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x, a, b = map(int, input().split(' '))
if abs(a - x) >= abs(b - x):
print('A')
else:
print('B')
|
s117034662
|
Accepted
| 18 | 2,940 | 103 |
x, a, b = map(int, input().split(' '))
if abs(a - x) <= abs(b - x):
print('A')
else:
print('B')
|
s645901164
|
p03338
|
u729938879
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 203 |
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
n = int(input())
s = input()
num = []
for i in range(1,n):
former = set(s[:i])
latter = set(s[i:])
comm = len(former and latter)
#print(s[:i], s[i:])
num.append(comm)
print(max(num))
|
s507625228
|
Accepted
| 18 | 3,060 | 213 |
n = int(input())
s = input()
num = []
for i in range(1,n):
former = set(s[:i])
latter = set(s[i:])
comm = len(former.intersection(latter))
#print(s[:i], s[i:])
num.append(comm)
print(max(num))
|
s840987147
|
p04046
|
u102960641
| 2,000 | 262,144 |
Wrong Answer
| 276 | 18,804 | 572 |
We have a large square grid with H rows and W columns. Iroha is now standing in the top-left cell. She will repeat going right or down to the adjacent cell, until she reaches the bottom-right cell. However, she cannot enter the cells in the intersection of the bottom A rows and the leftmost B columns. (That is, there are A×B forbidden cells.) There is no restriction on entering the other cells. Find the number of ways she can travel to the bottom-right cell. Since this number can be extremely large, print the number modulo 10^9+7.
|
h,w,a,b = map(int, input().split())
mod = 10**9 + 7
n = 10**5 * 2 + 1
fact = [1]*(n+1)
rfact = [1]*(n+1)
r = 1
for i in range(1, n+1):
fact[i] = r = r * i % mod
rfact[n] = r = pow(fact[n], mod-2, mod)
for i in range(n, 0, -1):
rfact[i-1] = r = r * i % mod
def perm(n, k):
return fact[n] * rfact[n-k] % mod
def comb(n, k):
if n == 0 and k == 0:
return 0
return fact[n] * rfact[k] * rfact[n-k] % mod
ans = 0
for i in range(b,w):
ans = (ans + comb(h-a-1+i,i) * comb(w-i+a-2,a-1)) % mod
print(ans)
|
s454560131
|
Accepted
| 265 | 18,804 | 535 |
h,w,a,b = map(int, input().split())
mod = 10**9 + 7
n = 10**5 * 2 + 1
fact = [1]*(n+1)
rfact = [1]*(n+1)
r = 1
for i in range(1, n+1):
fact[i] = r = r * i % mod
rfact[n] = r = pow(fact[n], mod-2, mod)
for i in range(n, 0, -1):
rfact[i-1] = r = r * i % mod
def perm(n, k):
return fact[n] * rfact[n-k] % mod
def comb(n, k):
return fact[n] * rfact[k] * rfact[n-k] % mod
ans = 0
for i in range(b,w):
ans = (ans + comb(h-a-1+i,i) * comb(w-i+a-2,a-1)) % mod
print(ans)
|
s215969402
|
p03544
|
u066455063
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 116 |
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
N = int(input())
luca = [2, 1]
for i in range(101):
luca.append(luca[i]+luca[i+1])
print(luca)
print(luca[N])
|
s922569920
|
Accepted
| 17 | 2,940 | 104 |
N = int(input())
luca = [2, 1]
for i in range(101):
luca.append(luca[i]+luca[i+1])
print(luca[N])
|
s398463139
|
p03379
|
u243689896
| 2,000 | 262,144 |
Wrong Answer
| 2,108 | 26,180 | 366 |
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.
|
#ABC 094 Many Medians
numbers=int(input())
list_num=[int(i) for i in input().split()]
sorted_list=sorted(list_num)
median1=sorted_list[int(numbers/2)-1]
median2=sorted_list[int(numbers/2)]
print('median1:', median1)
print('median2:', median2)
for i in list_num:
if sorted_list.index(i) <= numbers/2-1:
print(median2)
else:
print(median1)
|
s876015087
|
Accepted
| 300 | 25,472 | 264 |
numbers=int(input())
list_num=[int(i) for i in input().split()]
sorted_list=sorted(list_num)
median1=sorted_list[int(numbers/2)-1]
median2=sorted_list[int(numbers/2)]
for i in list_num:
if i <= median1:
print(median2)
else:
print(median1)
|
s161907103
|
p03673
|
u746419473
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 26,020 | 192 |
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
n = int(input())
*a, = map(int, input().split())
b = []
for i in range(n):
if i%2 == 0:
b.append(a[i])
else:
b.insert(0, a[i])
if n%2 == 1:
b.reverse()
print(b)
|
s976564819
|
Accepted
| 172 | 26,020 | 178 |
n = int(input())
*a, = map(int, input().split())
right = a[1::2]
right.reverse()
left = a[0::2]
b = []
b.extend(right)
b.extend(left)
if n%2 == 1:
b.reverse()
print(*b)
|
s848413590
|
p02678
|
u131638468
| 2,000 | 1,048,576 |
Wrong Answer
| 374 | 63,944 | 543 |
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,*ab=map(int,open(0).read().split())
path={}
for i in range(n):
path[i]=[]
for i in range(m):
a,b=ab[i*2]-1,ab[i*2+1]-1
path[a].append(b)
path[b].append(a)
direction=[i for i in range(n)]
fin=[False for i in range(n)]
fin[0]=True
i=0
queue=[0]
while i<n:
i+=1
queue2=queue.copy()
queue.clear()
if len(queue)==0:
break
for j in queue2:
fin[j]=True
for k in path[j]:
if not fin[k]:
direction[k]=j
fin[k]=True
queue.append(k)
print('Yes')
for i in range(1,n):
print(direction[i]+1)
|
s748262601
|
Accepted
| 542 | 63,520 | 544 |
n,m,*ab=map(int,open(0).read().split())
path={}
for i in range(n):
path[i]=[]
for i in range(m):
a,b=ab[i*2]-1,ab[i*2+1]-1
path[a].append(b)
path[b].append(a)
direction=[i for i in range(n)]
fin=[False for i in range(n)]
fin[0]=True
i=0
queue=[0]
while i<n:
i+=1
queue2=queue.copy()
queue.clear()
if len(queue2)==0:
break
for j in queue2:
fin[j]=True
for k in path[j]:
if not fin[k]:
direction[k]=j
fin[k]=True
queue.append(k)
print('Yes')
for i in range(1,n):
print(direction[i]+1)
|
s136630155
|
p03377
|
u488178971
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
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.
|
#094
A,B,X= map(int,input().split())
if A+B>=X >=A:
print('Yes')
else:
print('No')
|
s753066535
|
Accepted
| 17 | 2,940 | 90 |
#094
A,B,X= map(int,input().split())
if A+B>=X >=A:
print('YES')
else:
print('NO')
|
s855066046
|
p03352
|
u645568816
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,096 | 132 |
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.
|
N = int(input())
ans = 0
for i in range(1,35):
for k in range(1,10):
if i**k <= N and i**k >= ans:
ans = i**k
print(ans)
|
s209836598
|
Accepted
| 29 | 9,160 | 132 |
N = int(input())
ans = 0
for i in range(1,35):
for k in range(2,10):
if i**k <= N and i**k >= ans:
ans = i**k
print(ans)
|
s722724248
|
p03998
|
u537976628
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 637 |
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
A, B, C = (input() for i in range(3))
cnt_a = 0; cnt_b = -1; cnt_c = -1
current_card = A[cnt_a]
while True:
print(current_card)
if current_card == "a":
cnt_a += 1
if cnt_a == len(A):
print("A")
exit()
else:
current_card = A[cnt_a]
elif current_card == "b":
cnt_b += 1
if cnt_b == len(B):
print("B")
exit()
else:
current_card = B[cnt_b]
elif current_card == "c":
cnt_c += 1
if cnt_c == len(C):
print("C")
exit()
else:
current_card = C[cnt_c]
|
s501580406
|
Accepted
| 17 | 3,064 | 613 |
A, B, C = (input() for i in range(3))
cnt_a = 0; cnt_b = -1; cnt_c = -1
current_card = A[cnt_a]
while True:
if current_card == "a":
cnt_a += 1
if cnt_a == len(A):
print("A")
exit()
else:
current_card = A[cnt_a]
elif current_card == "b":
cnt_b += 1
if cnt_b == len(B):
print("B")
exit()
else:
current_card = B[cnt_b]
elif current_card == "c":
cnt_c += 1
if cnt_c == len(C):
print("C")
exit()
else:
current_card = C[cnt_c]
|
s987128878
|
p02613
|
u657173608
| 2,000 | 1,048,576 |
Wrong Answer
| 154 | 9,680 | 555 |
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.
|
from sys import stdin, stdout
import heapq
import cProfile
from collections import Counter, defaultdict, deque
from functools import reduce
import math
from bisect import bisect,bisect_right,bisect_left
def get_int():
return int(stdin.readline().strip())
def get_tuple():
return map(int, stdin.readline().split())
def get_list():
return list(map(int, stdin.readline().split()))
n = get_int()
dic = defaultdict(int)
for _ in range(n):
st = input()
dic[st] += 1
for val in ['AC','WA','TLE','RE']:
print(val+" * "+str(dic[val]))
|
s831767837
|
Accepted
| 144 | 9,656 | 555 |
from sys import stdin, stdout
import heapq
import cProfile
from collections import Counter, defaultdict, deque
from functools import reduce
import math
from bisect import bisect,bisect_right,bisect_left
def get_int():
return int(stdin.readline().strip())
def get_tuple():
return map(int, stdin.readline().split())
def get_list():
return list(map(int, stdin.readline().split()))
n = get_int()
dic = defaultdict(int)
for _ in range(n):
st = input()
dic[st] += 1
for val in ['AC','WA','TLE','RE']:
print(val+" x "+str(dic[val]))
|
s007042969
|
p03407
|
u992910889
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 96 |
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 or B==C or A+B==C:
print('Yes')
else:
print('No')
|
s142196280
|
Accepted
| 17 | 2,940 | 80 |
A,B,C=map(int,input().split())
if A+B>=C:
print('Yes')
else:
print('No')
|
s186028980
|
p03476
|
u027929618
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,105 | 30,196 | 349 |
We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.
|
def is_prime(t):
i=2
while i**2<=t:
if t%i==0:
return 0
i+=1
return 1
q=int(input())
lr=[list(map(int,input().split())) for _ in range(q)]
n=10**5
p=[0]*(n+1)
d=[]
for i in range(2,n):
if is_prime(i):
d.append(i)
if i in d and (i+1)//2 in d:
p[i]=p[i-1]+1
else:
p[i]=p[i-1]
for l,r in lr:
print(p[r]-p[l-1])
|
s196956500
|
Accepted
| 1,379 | 29,860 | 349 |
def is_prime(t):
i=2
while i**2<=t:
if t%i==0:
return 0
i+=1
return 1
q=int(input())
lr=[list(map(int,input().split())) for _ in range(q)]
n=10**5
p=[0]*(n+1)
d=set()
for i in range(2,n):
if is_prime(i):
d.add(i)
if i in d and (i+1)//2 in d:
p[i]=p[i-1]+1
else:
p[i]=p[i-1]
for l,r in lr:
print(p[r]-p[l-1])
|
s466910506
|
p03377
|
u564906058
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 86 |
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 = map(int,input().split())
if a <= x <= a+b:
print("Yes")
else:
print("No")
|
s651326701
|
Accepted
| 17 | 2,940 | 86 |
a,b,x = map(int,input().split())
if a <= x <= a+b:
print("YES")
else:
print("NO")
|
s880701408
|
p03998
|
u782654209
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 386 |
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
S = [input() for x in range(3)]
winner = ''
turn = 0
while True:
if S[turn][0] == 'a':
S[0] = S[0][1:]
turn = 0
elif S[turn][0] == 'b':
S[1] = S[1][1:]
turn = 1
elif S[turn][0] == 'c':
S[2] = S[2][1:]
turn = 2
for i in range(3):
if S[i]=='': winner='abc'[i]
if winner!='':
break
print(winner)
|
s468262118
|
Accepted
| 17 | 3,060 | 201 |
S = [str(input()) for x in range(3)]
turn = 0
nextturn = 0
while True:
nextturn = 'abc'.find(S[turn][0])
S[turn] = S[turn][1:]
turn = nextturn
if S[turn] == '':
print('ABC'[turn])
break
|
s761627683
|
p03854
|
u246253778
| 2,000 | 262,144 |
Wrong Answer
| 65 | 9,092 | 392 |
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`.
|
S = input()
while len(S) != 0:
if S[-2:] == 'er':
S = S[:-2]
if S[-5:] == 'dream':
S = S[:-5]
elif S[-4:] == 'eras':
S = S[:-4]
else:
print('No')
exit()
else:
if S[-5:] == 'dream' or S[-5:] == 'erase':
S = S[:-5]
else:
print('No')
exit()
print('Yes')
|
s281895398
|
Accepted
| 69 | 9,136 | 392 |
S = input()
while len(S) != 0:
if S[-2:] == 'er':
S = S[:-2]
if S[-5:] == 'dream':
S = S[:-5]
elif S[-4:] == 'eras':
S = S[:-4]
else:
print('NO')
exit()
else:
if S[-5:] == 'dream' or S[-5:] == 'erase':
S = S[:-5]
else:
print('NO')
exit()
print('YES')
|
s470017224
|
p02401
|
u628732336
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,584 | 265 |
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.
|
while True:
a, op, b = input().split()
a = int(a)
b = int(b)
if op == '?':
break
if op == '+':
print(a + b)
if op == '-':
print(a - b)
if op == '*':
print(a * b)
if op == '/':
print(a / b)
|
s639191512
|
Accepted
| 20 | 7,616 | 266 |
while True:
a, op, b = input().split()
a = int(a)
b = int(b)
if op == '?':
break
if op == '+':
print(a + b)
if op == '-':
print(a - b)
if op == '*':
print(a * b)
if op == '/':
print(a // b)
|
s168335150
|
p04029
|
u980492406
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 33 |
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*(n+1)/2)
|
s988640111
|
Accepted
| 17 | 2,940 | 38 |
n = int(input())
print(int(n*(n+1)/2))
|
s135116654
|
p02262
|
u599130514
| 6,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 563 |
Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$
|
def insertionSort(A, n, g):
cnt = 0
for i in range(g, n):
v = A[i]
j = i - g
while j >= 0 and A[j] > v:
A[j+g] = A[j]
j = j - g
cnt += 1
A[j+g] = v
return cnt
def shellSort(A, n):
cnt = 0
m = 2
G = [n - m + 1, m - 1]
for i in range(m):
cnt += insertionSort(A, n, G[i])
print(m)
print(*G)
print(cnt)
print(*A)
arr_length = int(input())
arr_num = []
for i in range(arr_length):
arr_num.append(input())
shellSort(arr_num, arr_length)
|
s746345748
|
Accepted
| 18,480 | 125,972 | 686 |
def insertionSort(A, n, g):
cnt = 0
for i in range(g, n):
v = A[i]
j = i - g
while j >= 0 and A[j] > v:
A[j+g] = A[j]
j = j - g
cnt += 1
A[j+g] = v
return cnt
def shellSort(A, n):
G = []
h = 1
while True:
if h > n:
break
G.append(h)
h = 3 * h + 1
cnt = 0
print(len(G))
G = list(reversed(G))
for g in G:
cnt += insertionSort(A, n, g)
print(*G)
print(cnt)
print('\n'.join(map(str, A)))
arr_length = int(input())
arr_num = []
for i in range(arr_length):
arr_num.append(int(input()))
shellSort(arr_num, arr_length)
|
s309812723
|
p02399
|
u286589639
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,624 | 109 |
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 = map(int, input().split())
d = a // b
r = a % b
f = a / b
print(str(d) + " " + str(r) + " " + str(f))
|
s713320524
|
Accepted
| 20 | 7,652 | 127 |
a, b = map(int, input().split())
d = a // b
r = a % b
f = "{0:.5f}".format(a / b)
print(str(d) + " " + str(r) + " " + str(f))
|
s169196208
|
p03379
|
u669322232
| 2,000 | 262,144 |
Wrong Answer
| 2,105 | 27,156 | 231 |
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.
|
import copy
N = int(input())
numList = list(map(int, input().split()))
numList.sort()
for i in range(0,N):
numListCopy = copy.deepcopy(numList)
numListCopy.pop(i)
medIndex = int(len(numListCopy)/2)
print(numListCopy[medIndex])
|
s884433794
|
Accepted
| 482 | 26,528 | 316 |
import copy
N = int(input())
numList = list(map(int, input().split()))
numListCopy = copy.deepcopy(numList)
numList = sorted(numList)
medIndex = int((N)/2)
med1 = numList[medIndex-1]
med2 = numList[medIndex]
for i in range(0,N):
if numListCopy[i] <= med1:
print(med2)
elif med2 <= numListCopy[i]:
print(med1)
|
s284408044
|
p03671
|
u367393577
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 55 |
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.
|
a,b,c=map(int,input().split())
print(max(a+b,b+c,c+a))
|
s716761514
|
Accepted
| 17 | 2,940 | 55 |
a,b,c=map(int,input().split())
print(min(a+b,b+c,c+a))
|
s478150720
|
p02902
|
u803611972
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 3,956 | 343 |
Given is a directed graph G with N vertices and M edges. The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i. It is guaranteed that the graph contains no self-loops or multiple edges. Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph. Here the null graph is not considered as a subgraph.
|
from collections import*
n,m,*t=map(int,open(0).read().split())
i,o,r=[0]*n,[[]for _ in'_'*n],[]
for a,b in zip(*[iter(t)]*2):
o[a-1]+=b-1,
i[b-1]+=1
q=deque(v for v in range(n)if i[v]<1)
while q:
v=q.popleft()
r+=v,
for w in o[v]:
i[w]-=1
if i[w]==0:q+=w,
print(-(len(r)==n))
for j,k in enumerate(i):
if i[j]>0: print(j+1)
|
s217644980
|
Accepted
| 30 | 3,956 | 708 |
import sys
from collections import deque
def serp(s):
prev = [-1]*n
q=deque([s])
while q:
v = q.pop()
for nv in e[v]:
if nv==s: prev[nv]=v; return(s,prev)
if prev[nv]<0: q+=nv,; prev[nv] =v
return(-1,prev)
n,m,*t=map(int,open(0).read().split())
e,ab = [[] for i in range(n)],[]
for a,b in zip(*[iter(t)]*2):
e[a-1]+= b-1,
ab += [(a-1,b-1)]
for v in range(n):
v0,prev = serp(v)
if v0>=0: break
if v0<0: print(-1); sys.exit()
c=set()
c.add(v0)
pv = prev[v0]
while pv!=v0:
c.add(pv)
pv=prev[pv]
for a,b in ab:
if a in c and b in c and prev[b]!=a:
pv = prev[b]
while pv !=a:
c.remove(pv)
pv=prev[pv]
prev[b]=a
print(len(c))
for i in c: print(i+1)
|
s062001923
|
p03679
|
u746428948
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 97 |
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())
d = b - a
if d <= x: print('delicious')
else: print('dangerous')
|
s341132049
|
Accepted
| 17 | 2,940 | 124 |
x,a,b = map(int,input().split())
d = b - a
if d <= 0: print('delicious')
elif d <= x: print('safe')
else: print('dangerous')
|
s334483422
|
p03433
|
u951601135
| 2,000 | 262,144 |
Wrong Answer
| 157 | 12,504 | 182 |
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 numpy as np
N = int(input())
value = list(map(int,input().split()))
value=np.sort(value)[::-1]
print(N,value)
a=0
b=0
a = np.sum(value[::2])
b = np.sum(value[1::2])
print(a,b)
|
s059463798
|
Accepted
| 18 | 2,940 | 66 |
n=int(input())
a=int(input())
print('Yes' if (n%500)<=a else 'No')
|
s677395293
|
p04043
|
u882209234
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 118 |
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.
|
abc = list(map(int, input().split()))
if abc[0] == abc[2] == 5 and abc[1] == 7:
print('YES')
else:
print('NO')
|
s028583532
|
Accepted
| 17 | 3,060 | 209 |
abc = list(map(int, input().split()))
lengths = [0,0]
for i in abc:
if i == 5: lengths[0] += 1
if i == 7: lengths[1] += 1
if lengths[0] == 2 and lengths[1] == 1:
print('YES')
else:
print('NO')
|
s905119373
|
p03131
|
u852719059
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 3,064 | 341 |
Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations.
|
# C When I hit my pocket
k,a,b = map(int,input().split())
bis = 1
yen = 0
if b <= a + 2 or k <= a + 1:
print(1 + k)
else:
for i in range(k):
if not yen == 0:
bis += b
yen -= 1
elif bis >= a:
bis -= a
yen += 1
else:
bis += 1
print(bis)
|
s631666006
|
Accepted
| 17 | 3,060 | 247 |
# C When I hit my pocket
k,a,b = map(int,input().split())
bis = 1
if k <= 1 or b <= a + 2 or k < a + 1:
print(1 + k)
exit()
else:
bis = b
k = k - ( a + 1)
if not k == 0:
bis += int(k / 2) * (b - a) + k % 2
print(bis)
|
s626133179
|
p02399
|
u248424983
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,720 | 134 |
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)
|
import sys
(a, b) = [int(i) for i in input().split(' ')]
d = a // b
r = a % b
f = a / b
print(repr(d) +" "+ repr(r) +" "+ repr(f))
|
s267059511
|
Accepted
| 20 | 7,724 | 112 |
(a, b) = [int(i) for i in input().split(' ')]
print( repr(a // b) +" "+ repr(a % b) +" "+ ('%0.5f' % (a / b)))
|
s345172934
|
p03485
|
u148551245
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 64 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a, b = (int(i) for i in input().split())
mean = (a + b) // 2 + 1
|
s982982186
|
Accepted
| 17 | 2,940 | 114 |
a, b = map(int, input().split())
m = (a + b) // 2
x = (a + b) / 2
if x % 1 == 0:
print(m)
else:
print(m+1)
|
s269674866
|
p02742
|
u948522631
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 88 |
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:
|
h,w=map(int,input().split())
print(int(h*w/2)+1 if type(h*w/2)==float else int(h*w/2) )
|
s599547066
|
Accepted
| 17 | 3,060 | 122 |
h,w=map(int,input().split())
if h<2 or w<2:
print(1)
elif h%2==0 or w%2==0:
print(int(h*w/2))
else:
print(int(h*w/2)+1)
|
s207670852
|
p02390
|
u114315703
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,568 | 51 |
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
N = int(input())
print(N // 3600, N // 60, N % 60)
|
s195767345
|
Accepted
| 20 | 5,576 | 80 |
N = int(input())
print('{0}:{1}:{2}'.format(N // 3600, N % 3600// 60, N % 60))
|
s529226275
|
p03440
|
u047816928
| 2,000 | 262,144 |
Wrong Answer
| 464 | 33,904 | 1,002 |
You are given a forest with N vertices and M edges. The vertices are numbered 0 through N-1. The edges are given in the format (x_i,y_i), which means that Vertex x_i and y_i are connected by an edge. Each vertex i has a value a_i. You want to add edges in the given forest so that the forest becomes connected. To add an edge, you choose two different vertices i and j, then span an edge between i and j. This operation costs a_i + a_j dollars, and afterward neither Vertex i nor j can be selected again. Find the minimum total cost required to make the forest connected, or print `Impossible` if it is impossible.
|
class UF:
def __init__(self, N):
self.uf = [-1]*N
self.n = N
def find(self, x):
if self.uf[x]<0: return x
self.uf[x] = self.find(self.uf[x])
return self.uf[x]
def size(self, x):
return -self.uf[self.find(x)]
def union(self, x, y):
x, y = self.find(x), self.find(y)
if x==y: return
if self.size(x) > self.size(y): x, y = y, x
self.uf[y] += self.uf[x]
self.uf[x] = y
self.n -= 1
N, M = map(int, input().split())
A = list(map(int, input().split()))
uf = UF(N)
for _ in range(M):
x, y = map(int, input().split())
uf.union(x,y)
D = {}
for x, a in enumerate(A):
c = uf.find(x)
if c not in D: D[c]=[]
D[c].append(a)
ans = 0
L = []
for d in D.values():
d.sort()
ans += d[0]
L += d[1:]
print(d[0])
print(L, ans)
L.sort()
P = len(D)-2
if len(D)==1:
ans = 0
elif len(L)>=P:
ans += sum(L[:len(D)-2])
else:
ans = 'Impossible'
print(ans)
|
s594720585
|
Accepted
| 449 | 33,612 | 971 |
class UF:
def __init__(self, N):
self.uf = [-1]*N
self.n = N
def find(self, x):
if self.uf[x]<0: return x
self.uf[x] = self.find(self.uf[x])
return self.uf[x]
def size(self, x):
return -self.uf[self.find(x)]
def union(self, x, y):
x, y = self.find(x), self.find(y)
if x==y: return
if self.size(x) > self.size(y): x, y = y, x
self.uf[y] += self.uf[x]
self.uf[x] = y
self.n -= 1
N, M = map(int, input().split())
A = list(map(int, input().split()))
uf = UF(N)
for _ in range(M):
x, y = map(int, input().split())
uf.union(x,y)
D = {}
for x, a in enumerate(A):
c = uf.find(x)
if c not in D: D[c]=[]
D[c].append(a)
ans = 0
L = []
for d in D.values():
d.sort()
ans += d[0]
L += d[1:]
L.sort()
P = len(D)-2
if len(D)==1:
ans = 0
elif len(L)>=P:
ans += sum(L[:len(D)-2])
else:
ans = 'Impossible'
print(ans)
|
s213689852
|
p03657
|
u814265211
| 2,000 | 262,144 |
Wrong Answer
| 31 | 9,148 | 86 |
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())
print('Possible' if not A + B % 3 else 'Impossible')
|
s554440721
|
Accepted
| 26 | 9,160 | 120 |
A, B = map(int, input().split())
if (A+B) % 3 and A % 3 and B % 3:
print('Impossible')
else:
print('Possible')
|
s277213706
|
p03455
|
u453117183
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 96 |
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%2 == 0 or b%2 == 0:
print("even")
else:
print("odd")
|
s683948478
|
Accepted
| 17 | 2,940 | 96 |
a,b = map(int,input().split())
if a%2 == 0 or b%2 == 0:
print("Even")
else:
print("Odd")
|
s132330179
|
p03599
|
u050428930
| 3,000 | 262,144 |
Wrong Answer
| 3,156 | 3,064 | 665 |
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())
ans=100
for i in range(f//(100*a)+1):
for j in range(f//(100*b)+1):
s=100*a*i+100*b*j
if s==0:
break
for ii in range((f-100*a)//c+1):
for jj in range((f-100*a)//d+1):
t=c*ii+d*jj
if t==0:
break
u=(100*t)/(s+t)
if s+t<=f and u==f:
print(str(s+t)+" "+str(t))
exit()
if s+t<=f and u<f:
if ans>u:
ans=u
x,y=s,t
print([ans,x,y])
|
s308890189
|
Accepted
| 222 | 3,064 | 474 |
a,b,c,d,e,f=map(int,input().split())
t=100
for i in range(0,f+1,a*100):
for j in range(0,f-i+1,b*100):
k=i+j
if k==0:
continue
s=min(f-i-j,(i+j)*e//100)
for ii in range(0,s+1,c):
for jj in range(0,s-ii+1,d):
if t>e-(ii+jj)*100/(i+j):
t=e-(ii+jj)*100/(i+j)
x,y=ii+jj+i+j,ii+jj
print(str(x)+" "+str(y))
|
s669845689
|
p03597
|
u865413330
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 50 |
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())
print(n * 2 - a)
|
s070258306
|
Accepted
| 17 | 2,940 | 50 |
n = int(input())
a = int(input())
print(n * n - a)
|
s782022358
|
p03997
|
u485319545
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 454 |
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.
|
s_a = list(input())
s_b = list(input())
s_c = list(input())
S=[s_a,s_b,s_c]
order=0
while True:
if S[order][0]=='a':
del S[0][0]
order=0
elif S[order][0]=='b':
del S[1][0]
order=1
else:
del S[2][0]
order=2
if len(S[order])==0:
print('A')
exit(0)
elif len(S[order])==0:
print('B')
exit(0)
elif len(S[order])==0:
print('C')
exit(0)
|
s885187797
|
Accepted
| 17 | 3,064 | 67 |
a=int(input())
b=int(input())
h=int(input())
print(int((a+b)*h/2))
|
s589989666
|
p03475
|
u245870380
| 3,000 | 262,144 |
Wrong Answer
| 107 | 3,064 | 364 |
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())
C, S, F = [], [], []
for i in range(N-1):
c, s, f = map(int, input().split())
C.append(c)
S.append(s)
F.append(f)
print(C)
print(S)
print(F)
for i in range(N):
t = 0
for j in range(i,N-1):
if t < S[j]:
t = S[j]
elif t % F[j] != 0:
t += F[j] - t % F[j]
t += C[j]
print(t)
|
s033977349
|
Accepted
| 98 | 3,064 | 336 |
N = int(input())
C, S, F = [], [], []
for i in range(N-1):
c, s, f = map(int, input().split())
C.append(c)
S.append(s)
F.append(f)
for i in range(N):
t = 0
for j in range(i,N-1):
if t < S[j]:
t = S[j]
elif t % F[j] != 0:
t += F[j] - t % F[j]
t += C[j]
print(t)
|
s467263008
|
p02418
|
u971748390
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,352 | 81 |
Write a program which finds a pattern $p$ in a ring shaped text $s$.
|
s=input()
p=input()
ring=s+s
if p in ring :
print("True")
else :
print("False")
|
s570824814
|
Accepted
| 40 | 7,496 | 77 |
s=input()
p=input()
ring=s+s
if p in ring :
print("Yes")
else :
print("No")
|
s682277929
|
p03610
|
u247211039
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,316 | 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[0::1])
|
s392263326
|
Accepted
| 37 | 3,316 | 96 |
s = input()
x = s[0]
for i in range(len(s)):
if i % 2 == 0:
x= x + s[i]
print(x[1:])
|
s811141985
|
p02806
|
u152614052
| 2,525 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 208 |
Niwango created a playlist of N songs. The title and the duration of the i-th song are s_i and t_i seconds, respectively. It is guaranteed that s_1,\ldots,s_N are all distinct. Niwango was doing some work while playing this playlist. (That is, all the songs were played once, in the order they appear in the playlist, without any pause in between.) However, he fell asleep during his work, and he woke up after all the songs were played. According to his record, it turned out that he fell asleep at the very end of the song titled X. Find the duration of time when some song was played while Niwango was asleep.
|
n = int(input())
li1 = []
li2 = []
for i in range(n):
a, b = input().split()
li1.append(a)
li2.append(int(b))
x = input()
print(*li1)
print(*li2)
asleep = li1.index(x)
print(sum(li2[asleep+1:]))
|
s121419361
|
Accepted
| 17 | 3,060 | 183 |
n = int(input())
li1 = []
li2 = []
for i in range(n):
a, b = input().split()
li1.append(a)
li2.append(int(b))
x = input()
asleep = li1.index(x)
print(sum(li2[asleep+1:]))
|
s265132615
|
p03852
|
u534308356
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 204 |
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
S = input()
l = ['dream', 'dreamer', 'erase', 'eraser']
s = "".join(S[-1::-1].split(l[0][-1::-1]))
for i in range(1, 4):
s = "".join(s.split(l[i][-1::-1]))
print("YES") if s == "" else print("NO")
|
s876401081
|
Accepted
| 26 | 8,984 | 109 |
data = ["a", "e", "i", "o", "u"]
c = input()
if c in data:
print("vowel")
else:
print("consonant")
|
s359395775
|
p02603
|
u941644149
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,160 | 829 |
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
n = int(input())
given_A = list(map(int, input().split()))
A = []
for kabuka in given_A:
if len(A) > 0 and kabuka == A[-1]:
continue
else:
A.append(kabuka)
print(A)
if len(A) == 1:
print(1000)
exit()
money = 1000
stock = 0
n = len(A)
for i in range(len(A)):
if i == 0:
if A[i] < A[i + 1]:
stock = money // A[i]
money = money % A[i]
elif i == n - 1:
if A[i - 1] < A[i]:
money += stock*A[i]
stock = 0
else:
if (A[i - 1] < A[i]) & (A[i] > A[i + 1]):
money += stock*A[i]
stock = 0
elif (A[i - 1] > A[i]) & (A[i] < A[i + 1]):
stock = money // A[i]
money = money % A[i]
print(money)
|
s926648913
|
Accepted
| 30 | 9,220 | 816 |
n = int(input())
given_A = list(map(int, input().split()))
A = []
for kabuka in given_A:
if len(A) > 0 and kabuka == A[-1]:
continue
else:
A.append(kabuka)
if len(A) == 1:
print(1000)
exit()
money = 1000
stock = 0
n = len(A)
for i in range(len(A)):
if i == 0:
if A[i] < A[i + 1]:
stock = money // A[i]
money = money % A[i]
elif i == n - 1:
if A[i - 1] < A[i]:
money += stock*A[i]
stock = 0
else:
if (A[i - 1] < A[i]) & (A[i] > A[i + 1]):
money += stock*A[i]
stock = 0
elif (A[i - 1] > A[i]) & (A[i] < A[i + 1]):
stock = money // A[i]
money = money % A[i]
print(money)
|
s511888465
|
p03370
|
u344959959
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,056 | 113 |
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.
|
a,b = map(int,input().split())
c = [int(input()) for i in range(a)]
print(min(c))
print(a+((b-(sum(c)))//min(c)))
|
s178089763
|
Accepted
| 26 | 9,168 | 99 |
a,b = map(int,input().split())
c = [int(input()) for i in range(a)]
print(a+((b-(sum(c)))//min(c)))
|
s284511249
|
p03449
|
u625963200
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 180 |
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
n=int(input())
A=[list(map(int,input().split())) for _ in range(2)]
print(A)
ans=A[0][0]
for i in range(1,n+1):
ans2=sum(A[0][:i])+sum(A[1][i-1:])
ans=max(ans,ans2)
print(ans)
|
s339043073
|
Accepted
| 17 | 3,060 | 171 |
n=int(input())
A=[list(map(int,input().split())) for _ in range(2)]
ans=A[0][0]
for i in range(1,n+1):
ans2=sum(A[0][:i])+sum(A[1][i-1:])
ans=max(ans,ans2)
print(ans)
|
s268422940
|
p04025
|
u888092736
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 165 |
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
N = int(input())
A = map(int, input().split())
ans = float("inf")
for i in range(-100, 101):
ans = min(ans, sum(map(lambda x: (x - i) * (x - i), A)))
print(ans)
|
s824353245
|
Accepted
| 25 | 3,060 | 200 |
N = int(input())
A = list(map(int, input().split()))
ans = float("inf")
for k in range(-100, 101):
tmp = 0
for i in range(N):
tmp += (A[i] - k) ** 2
ans = min(ans, tmp)
print(ans)
|
s500203881
|
p03493
|
u931209993
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 164 |
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.
|
a = int(input())
cnt = 0
if a%10 == 1:
cnt += 1
a = int(a/10)
if a%10 == 1:
cnt += 1
a = int(a/10)
if a%10 == 1:
cnt += 1
a = int(a/10)
print(cnt)
|
s683849789
|
Accepted
| 17 | 2,940 | 114 |
a = input()
cnt = 0
if a[0] == '1':
cnt += 1
if a[1] == '1':
cnt += 1
if a[2] == '1':
cnt += 1
print(cnt)
|
s353669709
|
p03369
|
u086503932
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 47 |
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.
|
print(700+list(input().split()).count('o')*100)
|
s516955177
|
Accepted
| 17 | 2,940 | 33 |
print(700+input().count('o')*100)
|
s134453146
|
p02975
|
u814986259
| 2,000 | 1,048,576 |
Wrong Answer
| 112 | 14,468 | 664 |
Snuke has N hats. The i-th hat has an integer a_i written on it. There are N camels standing in a circle. Snuke will put one of his hats on each of these camels. If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No`. * The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself. What is XOR? The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non- negative integers x_1, x_2, \ldots, x_n is defined as follows: - When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even. For example, 3 \oplus 5 = 6.
|
import collections
N = int(input())
a = list(map(int, input().split()))
A = set()
table = collections.defaultdict(int)
for i in range(1, N):
table[a[i]] += 2
a.sort()
prev = a[0]
now = a[0]
last = a[0]
for i in range(1, N):
if (a[0] ^ a[i]) in table and table[a[0] ^ a[i]] > 0:
table[a[0] ^ a[i]] -= 1
last = a[i]
now = a[0] | a[i]
break
print(prev, now, last)
for i in range(2, N-2):
if prev ^ now in table and table[now ^ prev] > 0:
prev = now
now = now ^ prev
table[now ^ prev] -= 1
else:
print("No")
exit(0)
if (last ^ now) == a[0]:
print("Yes")
else:
print("No")
|
s801727715
|
Accepted
| 189 | 14,468 | 777 |
import collections
N = int(input())
a = list(map(int, input().split()))
A = set()
table = collections.defaultdict(int)
for i in range(1, N):
table[a[i]] += 1
ans = [-1]*N
ans[0] = a[0]
for i in range(1, N):
if (a[0] ^ a[i]) in table and table[a[0] ^ a[i]] >= 1:
if a[0] == 0:
if table[a[i]] == 1:
continue
ans[1] = (a[0] ^ a[i])
ans[2] = a[i]
table[ans[1]] -= 1
table[ans[2]] -= 1
break
if ans[1] == -1:
print("No")
exit(0)
for i in range(3, N):
if (ans[i-1] ^ ans[i-2]) in table and table[ans[i-1] ^ ans[i-2]] >= 1 and table[ans[i-1] ^ ans[i-2]] >= 1:
ans[i] = ans[i-1] ^ ans[i-2]
table[ans[i]] -= 1
else:
print("No")
exit(0)
print("Yes")
|
s567737792
|
p03063
|
u639444166
| 2,000 | 1,048,576 |
Wrong Answer
| 452 | 6,140 | 390 |
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
#! -*- coding: utf-8 -*-
n = int(input())
line = input()
w = 0
for i in range(n):
if line[i] == ".":
w += 1
b = n-w
w_coutner = 0
b_counter = 0
ans = n
for i in range(n):
if line[i] == ".":
w_coutner += 1
b_counter = i+1 - w_coutner
wtob = b_counter
btow = w - w_coutner
ans_i = wtob + btow
print(i,ans_i)
ans = min(ans,ans_i)
print(ans)
|
s528677696
|
Accepted
| 195 | 3,500 | 411 |
#! -*- coding: utf-8 -*-
n = int(input())
line = input()
w = 0
for i in range(n):
if line[i] == ".":
w += 1
b = n-w
w_coutner = 0
b_counter = 0
ans = n
for i in range(n):
if line[i] == ".":
w_coutner += 1
b_counter = i+1 - w_coutner
wtob = b_counter
btow = w - w_coutner
ans_i = wtob + btow
#print(i,ans_i)
ans = min(ans,ans_i)
ans = min(w,b,ans)
print(ans)
|
s074087792
|
p03447
|
u730710086
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 113 |
You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?
|
# -*- coding: <encoding name> -*-
x = int(input())
a = int(input())
b = int(input())
n = x - a
print(x - n % b)
|
s272301525
|
Accepted
| 17 | 2,940 | 109 |
# -*- coding: <encoding name> -*-
x = int(input())
a = int(input())
b = int(input())
n = x - a
print(n % b)
|
s538095039
|
p02678
|
u941438707
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,148 | 11 |
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.
|
print("No")
|
s952826878
|
Accepted
| 560 | 60,284 | 304 |
from collections import*
(n,m),*c=[[*map(int,i.split())]for i in open(0)]
g=[[]for _ in range(n+1)]
for a,b in c:
g[a]+=[b]
g[b]+=[a]
q=deque([1])
r=[0]*(n+1)
while q:
v=q.popleft()
for i in g[v]:
if r[i]==0:
r[i]=v
q.append(i)
print("Yes",*r[2:],sep="\n")
|
s669565645
|
p02602
|
u598684283
| 2,000 | 1,048,576 |
Wrong Answer
| 2,206 | 41,024 | 337 |
M-kun is a student in Aoki High School, where a year is divided into N terms. There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows: * For the first through (K-1)-th terms: not given. * For each of the K-th through N-th terms: the multiplication of the scores in the last K exams, including the exam in the graded term. M-kun scored A_i in the exam at the end of the i-th term. For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is **strictly** greater than the grade for the (i-1)-th term.
|
n,k = map(int,input().split())
a = list(input().split())
count = 0
tmp = 1
check = []
for _ in range(n - k + 1):
for i in range(k):
tmp *= int(a[i + count])
check.append(tmp)
tmp = 1
count += 1
print(check)
for j in range(n - k):
if check[j] < check[j + 1]:
print("Yes")
else:
print("No")
|
s971113599
|
Accepted
| 154 | 31,668 | 207 |
n, k = input().split()
n = int(n)
k = int(k)
a = [int(s) for s in input().split()]
sums = []
K = k - 1
for i in range(K,n - 1):
if(a[i - K] < a[i + 1]):
print("Yes")
else:
print("No")
|
s291104453
|
p03474
|
u297045966
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 92 |
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 = input().strip().split(' ')
A, B = [int(A),int(B)]
S = input()
print(S[:A]+'-'+S[A:])
|
s355239636
|
Accepted
| 17 | 3,060 | 174 |
A, B = input().strip().split(' ')
A, B = [int(A),int(B)]
S = input()
T = list(S.strip().split('-'))
if len(T[0])==A and len(T[1])==B:
print('Yes')
else:
print('No')
|
s157429929
|
p02409
|
u139687801
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,624 | 1,027 |
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())
buffer = []
i = 0
while i<n:
b,f,r,v = map(int,input().split())
buffer.append([b,f,r,v])
i = i + 1
room = []
for h in range(15):
if h == 3 or h == 7 or h == 11:
room.append(['**']*10)
else:
room.append([0]*10)
for y in range(n):
if buffer[y][0] == 1:
room[buffer[y][1]-1][buffer[y][2]-1] = buffer[y][3] + room[buffer[y][1]-1][buffer[y][2]-1]
elif buffer[y][0] == 2:
room[buffer[y][1]-1+4][buffer[y][2]-1] = buffer[y][3] + room[buffer[y][1]-1+4][buffer[y][2]-1]
elif buffer[y][0] == 3:
room[buffer[y][1]-1+8][buffer[y][2]-1] = buffer[y][3] + room[buffer[y][1]-1+8][buffer[y][2]-1]
elif buffer[y][0] == 4:
room[buffer[y][1]-1+12][buffer[y][2]-1] = buffer[y][3] + room[buffer[y][1]-1+12][buffer[y][2]-1]
for x in range(15):
if x == 3 or x == 7 or x == 11:
print("********************")
else:
for y in range(10):
print(" "+str(room[x][y]), end = "")
print()
|
s307055121
|
Accepted
| 30 | 5,628 | 1,027 |
n = int(input())
buffer = []
i = 0
while i<n:
b,f,r,v = map(int,input().split())
buffer.append([b,f,r,v])
i = i + 1
room = []
for h in range(15):
if h == 3 or h == 7 or h == 11:
room.append(['**']*10)
else:
room.append([0]*10)
for y in range(n):
if buffer[y][0] == 1:
room[buffer[y][1]-1][buffer[y][2]-1] = buffer[y][3] + room[buffer[y][1]-1][buffer[y][2]-1]
elif buffer[y][0] == 2:
room[buffer[y][1]-1+4][buffer[y][2]-1] = buffer[y][3] + room[buffer[y][1]-1+4][buffer[y][2]-1]
elif buffer[y][0] == 3:
room[buffer[y][1]-1+8][buffer[y][2]-1] = buffer[y][3] + room[buffer[y][1]-1+8][buffer[y][2]-1]
elif buffer[y][0] == 4:
room[buffer[y][1]-1+12][buffer[y][2]-1] = buffer[y][3] + room[buffer[y][1]-1+12][buffer[y][2]-1]
for x in range(15):
if x == 3 or x == 7 or x == 11:
print("####################")
else:
for y in range(10):
print(" "+str(room[x][y]), end = "")
print()
|
s262598910
|
p03722
|
u547167033
| 2,000 | 262,144 |
Wrong Answer
| 319 | 35,476 | 303 |
There is a directed graph with N vertices and M edges. The i-th edge (1≤i≤M) points from vertex a_i to vertex b_i, and has a weight c_i. We will play the following single-player game using this graph and a piece. Initially, the piece is placed at vertex 1, and the score of the player is set to 0. The player can move the piece as follows: * When the piece is placed at vertex a_i, move the piece along the i-th edge to vertex b_i. After this move, the score of the player is increased by c_i. The player can end the game only when the piece is placed at vertex N. The given graph guarantees that it is possible to traverse from vertex 1 to vertex N. When the player acts optimally to maximize the score at the end of the game, what will the score be? If it is possible to increase the score indefinitely, print `inf`.
|
import scipy.sparse.csgraph
n,m=map(int,input().split())
graph=[[0]*n for i in range(n)]
for i in range(m):
u,v,c=map(int,input().split())
graph[u-1][v-1]=-c
try:
l=scipy.sparse.csgraph.johnson(graph)
print(-l[0][-1])
except scipy.sparse.csgraph._shortest_path.NegativeCycleError:
print('inf')
|
s507310603
|
Accepted
| 582 | 3,368 | 406 |
import sys
INF=float('inf')
def Bellmanford(n,edges,r):
d=[INF]*n
d[r]=0
for i in range(n):
for u,v,c in edges:
if d[u]!=INF and d[v]>d[u]+c:
d[v]=d[u]+c
if i==n-1 and v==n-1:
return 'inf'
return (-1)*d[n-1]
n,m=map(int,input().split())
edges=[0]*m
for i in range(m):
a,b,c=map(int,input().split())
edges[i]=(a-1,b-1,-c)
ans=Bellmanford(n,edges,0)
print(ans)
|
s683821848
|
p03447
|
u759412327
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 75 |
You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?
|
a = [int(input()) for i in range(3)]
b = a[0]-a[1]
c = a[2]
print(b-b//c*b)
|
s194174575
|
Accepted
| 29 | 9,156 | 54 |
X,A,B = map(int,open(0).read().split())
print((X-A)%B)
|
s724055011
|
p03525
|
u736729525
| 2,000 | 262,144 |
Wrong Answer
| 126 | 3,064 | 1,233 |
In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi. Takahashi checked and found that the _time gap_ (defined below) between the local times in his city and the i-th person's city was D_i hours. The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours. Here, we are using 24-hour notation. That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example. Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours. Find the maximum possible value of s.
|
def diff(s, t):
return min(abs(s-t), 24 - abs(s - t))
def main():
N = int(input())
D = [int(x) for x in input().split()]
solve(D)
def solve(D):
def check(d, di, dj, i, j):
for k in range(N):
if k == i or k == j:
continue
if diff(D[k], di) < d:
return None
if diff(D[k], dj) < d:
return None
return d
D = [0]+D
D.sort()
N = len(D)
answer = 0
for i in range(N-1):
Di = D[i]
Di_ = 24 - D[i]
for j in range(i+1, N):
Dj = D[j]
Dj_ = 24 - D[j]
m = None
for di, dj in [(Di, Dj), (Di_, Dj), (Di, Dj_), (Di_, Dj_)]:
Dij = diff(di, dj)
dm = check(Dij, di, dj, i, j)
if dm is None:
continue
if m is None:
m = dm
m = max(m, dm)
answer = max(answer, m) if m else answer
#print(answer)
return answer
def test(D):
return
print(D, solve(D))
test([7, 12, 8])
test([11, 11])
test([0])
main()
|
s349764199
|
Accepted
| 17 | 3,060 | 195 |
N=int(input())
D =[0]+[int(x) for x in input().split()]
m=D[1]
D.sort()
for i in range(1,N+1):
D[i]=D[i] if i&1 else 24-D[i]
D.sort()
for i in range(1,N+1):
m= min(m,D[i]-D[i-1])
print(m)
|
s813412685
|
p02300
|
u519227872
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,672 | 1,158 |
Find the convex hull of a given set of points P. In other words, find the smallest convex polygon containing all the points of P. Here, in a convex polygon, all interior angles are less than or equal to 180 degrees. Please note that you should find all the points of P on both corner and boundary of the convex polygon.
|
from math import sqrt
h = {1: 'COUNTER_CLOCKWISE',
2: 'CLOCKWISE',
3: 'ONLINE_BACK',
4: 'ONLINE_FRONT',
5: 'ON_SEGMENT',}
def dot(a, b):
return sum([i * j for i,j in zip(a, b)])
def sub(a, b):
return [a[0] - b[0],a[1] - b[1]]
def cross(a, b):
return a[0] * b[1] - a[1] * b[0]
def _abs(a):
return sqrt(a[0] ** 2 + a[1] ** 2)
def ccw(a, b, c):
x = sub(b, a)
y = sub(c, a)
if cross(x, y) > 0: return 1
if cross(x, y) < 0: return 2
if dot(x, y) < 0: return 3
if _abs(x) < _abs(y): return 4
return 5
def CCW(a,b,c):
return h[ccw(a,b,c)]
n = int(input())
c = [list(map(int, input().split())) for i in range(n)]
c.sort(key=lambda x: (x[0], x[1]))
U = c[:2]
for i in range(2, n):
j = len(U)
while j >= 2 and CCW(U[-1],U[-2],c[i]) == "COUNTER_CLOCKWISE":
U.pop()
j -= 1
U.append(c[i])
c = c[::-1]
L = c[:2]
for i in range(2, n):
j = len(L)
while j >= 2 and CCW(L[-1], L[-2], c[i]) == "COUNTER_CLOCKWISE":
L.pop()
j -= 1
L.append(c[i])
for x,y in U:
print(x,y)
for x, y in L:
if not [x,y] in U:
print(x,y)
|
s319037633
|
Accepted
| 1,000 | 31,168 | 876 |
from math import sqrt
from collections import deque
def sub(a, b):
return [a[0] - b[0],a[1] - b[1]]
def cross(a, b):
return a[0] * b[1] - a[1] * b[0]
def ccw(a, b, c):
x = sub(b, a)
y = sub(c, a)
return cross(x, y) > 0
n = int(input())
c = [list(map(int, input().split())) for i in range(n)]
c.sort(key=lambda x: (x[0], x[1]))
U = deque(c[:2])
for i in range(2, n):
j = len(U)
while j >= 2 and ccw(U[-1],U[-2],c[i]):
U.pop()
j -= 1
U.append(c[i])
c = c[::-1]
L = deque(c[:2])
for i in range(2, n):
j = len(L)
while j >= 2 and ccw(L[-1], L[-2], c[i]):
L.pop()
j -= 1
L.append(c[i])
ans = U
for i in range(1,len(L)-1):
x, y = L[i]
ans.append([x,y])
print(len(ans))
i = ans.index(sorted(ans, key=lambda x: (x[1],x[0]))[0])
ans = list(ans)
for x,y in ans[i:] + ans[:i]:
print(x, y)
|
s241850556
|
p03556
|
u247211039
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 67 |
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
import math
N = int(input())
a = math.sqrt(N)
a = int(a)
print(a)
|
s241976422
|
Accepted
| 17 | 2,940 | 71 |
import math
N = int(input())
a = math.sqrt(N)
a = int(a)
print(a*a)
|
s912896712
|
p03712
|
u928784113
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 155 |
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.
|
N,M = map(int,input().split())
A = []
for i in range(N):
A.append(str(input()))
print("#"*(M+2))
print("
print("#"*(M+2))
|
s033258394
|
Accepted
| 18 | 3,060 | 159 |
N,M = map(int,input().split())
A = []
for i in range(N):
A.append(str(input()))
print("#"*(M+2))
for i in range(N):
print("#"+A[i]+"#")
print("#"*(M+2))
|
s140590611
|
p03352
|
u396391104
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 3,572 | 165 |
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())
ans = 1
for a in range(1,x+1):
for b in range(2,x+1):
print(a,b)
if a**b > x:
break
elif a**b > ans:
ans = a**b
print(ans)
|
s304247854
|
Accepted
| 18 | 2,940 | 150 |
x = int(input())
ans = 1
for a in range(1,x+1):
for b in range(2,x+1):
if a**b > x:
break
elif a**b > ans:
ans = a**b
print(ans)
|
s231908750
|
p03069
|
u660750079
| 2,000 | 1,048,576 |
Wrong Answer
| 364 | 22,272 | 456 |
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
N=int(input())
S=input()
list1=[0]*N
for i in reversed(range(N)):
if i==N-1 and S[i]==".":
list1[i]+=1
elif i==N-1:
list1[i]=0
elif S[i]==".":
list1[i]=list1[i+1]+1
else:
list1[i]=list1[i+1]
print(list1)
list2=[]
for i in range(N+1):
if i==0:
tmp=N-list1[0]
elif i==N:
tmp=list1[0]
else:
tmp=i-list1[0]+2*list1[i]
print(tmp)
list2.append(tmp)
print(min(list2))
|
s022776081
|
Accepted
| 205 | 19,304 | 427 |
N=int(input())
S=input()
list1=[0]*N
for i in reversed(range(N)):
if i==N-1 and S[i]==".":
list1[i]+=1
elif i==N-1:
list1[i]=0
elif S[i]==".":
list1[i]=list1[i+1]+1
else:
list1[i]=list1[i+1]
list2=[]
for i in range(N+1):
if i==0:
tmp=N-list1[0]
elif i==N:
tmp=list1[0]
else:
tmp=i-list1[0]+2*list1[i]
list2.append(tmp)
print(min(list2))
|
s829054367
|
p03149
|
u057964173
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 147 |
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
def resolve():
l=list(map(int, input().split()))
if 1 in l and 9 in l and 7 in l and 4 in l:
print('YES')
else:
print('NO')
|
s376879439
|
Accepted
| 18 | 3,060 | 194 |
import sys
def input(): return sys.stdin.readline().strip()
def resolve():
l=set(map(int,input().split()))
if l=={1,7,9,4}:
print('YES')
else:
print('NO')
resolve()
|
s625386379
|
p00001
|
u823517752
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,468 | 110 |
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.
|
m = []
for i in range(0, 10):
m.append(input())
m.sort(reverse=True)
for i in range(0, 3):
print(m[i])
|
s616315756
|
Accepted
| 30 | 7,604 | 115 |
m = []
for i in range(0, 10):
m.append(int(input()))
m.sort(reverse=True)
for i in range(0, 3):
print(m[i])
|
s730946096
|
p03589
|
u125205981
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 2,940 | 298 |
You are given an integer N. Find a triple of positive integers h, n and w such that 4/N = 1/h + 1/n + 1/w. If there are multiple solutions, any of them will be accepted.
|
def main():
N = int(input())
for h in range(N, 0, -1):
for n in range(N, 0, -1):
w = int((4 - N / h - N / n) * N)
if w <= 0:
continue
elif N / h + N / n + N / w == 4:
print(h, n, w)
return
main()
|
s635974991
|
Accepted
| 1,459 | 2,940 | 300 |
def main():
N = int(input())
for h in range(1, 3501):
for n in range(1, 3501):
x = 4 * h * n - N * n - N * h
if x <= 0: continue
w, mod_ = divmod(N * h * n, x)
if not mod_:
print(h, n, w)
return
main()
|
s919095179
|
p03827
|
u107077660
| 2,000 | 262,144 |
Wrong Answer
| 22 | 3,064 | 120 |
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()
x = 0
ans = 0
for l in S:
if l == "I":
x += 1
elif l == "D":
x -= 1
ans = max(ans,x)
|
s407548255
|
Accepted
| 26 | 3,064 | 131 |
N = int(input())
S = input()
x = 0
ans = 0
for l in S:
if l == "I":
x += 1
elif l == "D":
x -= 1
ans = max(ans,x)
print(ans)
|
s500089526
|
p02854
|
u888092736
| 2,000 | 1,048,576 |
Wrong Answer
| 208 | 27,180 | 271 |
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters. Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut: * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan). * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen. Find the minimum amount of money needed before cutting the bar into two parts with the same length.
|
from itertools import accumulate
N = int(input())
A = [0] + list(accumulate(map(int, input().split())))
print(A)
min_diff = float("inf")
for i in range(N - 1):
if abs(A[N] - 2 * A[i + 1]) < min_diff:
min_diff = abs(A[N] - 2 * A[i + 1])
print(abs(min_diff))
|
s073489197
|
Accepted
| 175 | 31,792 | 273 |
from itertools import accumulate
N, *A = map(int, open(0).read().split())
A_acc = list(accumulate(A, initial=0))
min_diff = float("inf")
for i in range(1, N):
left, right = A_acc[i], A_acc[N] - A_acc[i]
min_diff = min(min_diff, abs(right - left))
print(min_diff)
|
s760155332
|
p03910
|
u987164499
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 198 |
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him. Determine the set of problems that should be solved.
|
from sys import stdin
n = int(stdin.readline().rstrip())
point = 0
for i in range(1,n+1):
point += i
if point >= n:
print(i)
if n-i != 0:
print(n-i)
break
|
s127702697
|
Accepted
| 21 | 3,572 | 221 |
from sys import stdin
n = int(stdin.readline().rstrip())
point = 0
for i in range(1,n+1):
point += i
if point >= n:
sa = point-n
for k in [j for j in range(1,i+1) if j != sa]:print(k)
break
|
s039639105
|
p03623
|
u864900001
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 104 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
#ABC71
x, a, b = map(int, input().split())
if(abs(x-a)>(abs(x-b))):
print("A")
else:
print("B")
|
s258268369
|
Accepted
| 17 | 2,940 | 104 |
#ABC71
x, a, b = map(int, input().split())
if(abs(x-a)>(abs(x-b))):
print("B")
else:
print("A")
|
s735615036
|
p03778
|
u393512980
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 119 |
AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved.
|
w,a,b=map(int,input().split())
if a>b:
t=a
a=b
b=t
if b-a<abs(b-a-w)+1:
print(b-a)
else:
print(abs(b-a-w)+1)
|
s548386021
|
Accepted
| 17 | 2,940 | 103 |
w,a,b=map(int,input().split())
if a>b:
t=a
a=b
b=t
if a<=b<=a+w:
print(0)
else:
print(b-a-w)
|
s846920872
|
p03361
|
u387774811
| 2,000 | 262,144 |
Wrong Answer
| 24 | 3,064 | 568 |
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j). Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`. However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black. Determine if square1001 can achieve his objective.
|
h,w=map(int,input().split())
list1=[[0 for i in range(w)]for j in range(h)]
list2=[[0 for i in range(w)]for j in range(h)]
dx=[-1,-1,0,0]
dy=[0,0,-1,1]
a=0
for j in range(h):
s=input()
for i in range(w):
list1[j][i]=s[i]
for j in range(h):
for i in range(w):
if list1[j][i]=="#":
for k in range(4):
xk=i+dx[k]
yk=j+dy[k]
if xk>=0 and xk<w and yk>=0 and yk<h:
list2[yk][xk]+=1
for j in range(h):
for i in range(w):
if list1[j][i]=="#" and list2[j][i]==0:
None
else:
a+=1
if a==h*w:
print("Yes")
else:
print("No")
|
s958405556
|
Accepted
| 24 | 3,064 | 566 |
h,w=map(int,input().split())
list1=[[0 for i in range(w)]for j in range(h)]
list2=[[0 for i in range(w)]for j in range(h)]
dx=[-1,1,0,0]
dy=[0,0,-1,1]
a=0
for j in range(h):
s=input()
for i in range(w):
list1[j][i]=s[i]
for j in range(h):
for i in range(w):
if list1[j][i]=="#":
for k in range(4):
xk=i+dx[k]
yk=j+dy[k]
if xk>=0 and xk<w and yk>=0 and yk<h:
list2[yk][xk]+=1
for j in range(h):
for i in range(w):
if list1[j][i]=="#" and list2[j][i]==0:
None
else:
a+=1
if a==h*w:
print("Yes")
else:
print("No")
|
s380728863
|
p03150
|
u589716761
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 3,064 | 357 |
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
S=input()
K=list('keyence')
n=0
flg=0
i_cut=[0,0]
for i,c in enumerate(S):
if c==K[n]:
if flg==1:
i_cut[1] = i-1
break
n+=1
if n>=7:
if flg==0:
i_cut = [i+1,len(S)]
break
else:
if flg==0:
i_cut[0] = i
flg +=1
if S[:i_cut[0]]+S[1+i_cut[1]:]=='keyence':
print("Yes")
else:
print('No')
|
s276171522
|
Accepted
| 18 | 3,060 | 175 |
S=input()
N=len(S)
for i in range(0,N+1):
for j in range(i,N+1):
if S[:i] + S[j:] == 'keyence':
print("YES")
exit(0)
else:
print('NO')
|
s701123289
|
p03139
|
u477320129
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 77 |
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.
|
N, A, B = list(map(int, input().split()))
print(min(A, B), abs(A - (N - B)))
|
s786899742
|
Accepted
| 17 | 2,940 | 78 |
N, A, B = list(map(int, input().split()))
print(min(A, B), max(A + B - N, 0))
|
s075744306
|
p02392
|
u410114382
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,644 | 80 |
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
a,b,c = map(int,input().split())
if a<b<c:
print('yes')
else:
print('no')
|
s584353514
|
Accepted
| 20 | 7,640 | 78 |
a,b,c = map(int,input().split())
if a<b<c:
print('Yes')
else:
print('No')
|
s734618616
|
p04045
|
u504662715
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,205 | 9,052 | 272 |
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.
|
n,k = input().split()
d = list(map(int,input().split()))
num = [i for i in range(10)]
poss = list(set(d) ^ set(num))
an = int(n)
while True:
sum = 0
for p in poss:
sum += n.count(str(p))
if sum ==len(n):
break
else:
an +=1
print(an)
|
s972999559
|
Accepted
| 81 | 9,088 | 141 |
n,k = map(int,input().split())
d=set(input().split())
while True:
if len(set(str(n))&d)!=0:
n+=1
else:
break
print(n)
|
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