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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s336184230
|
p03338
|
u905582793
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 180 |
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()
ls = list("abcdefghijklmnopqrstuvwxyz")
ans = [0]*n
for i in range(1,n):
for j in ls:
if j in s[:i] and j in s[i+1:]:
ans[i]+=1
print(max(ans))
|
s063274901
|
Accepted
| 18 | 3,060 | 178 |
n=int(input())
s=input()
ls = list("abcdefghijklmnopqrstuvwxyz")
ans = [0]*n
for i in range(1,n):
for j in ls:
if j in s[:i] and j in s[i:]:
ans[i]+=1
print(max(ans))
|
s079702888
|
p03151
|
u328131364
| 2,000 | 1,048,576 |
Wrong Answer
| 118 | 19,188 | 666 |
A university student, Takahashi, has to take N examinations and pass all of them. Currently, his _readiness_ for the i-th examination is A_{i}, and according to his investigation, it is known that he needs readiness of at least B_{i} in order to pass the i-th examination. Takahashi thinks that he may not be able to pass all the examinations, and he has decided to ask a magician, Aoki, to change the readiness for as few examinations as possible so that he can pass all of them, while not changing the total readiness. For Takahashi, find the minimum possible number of indices i such that A_i and C_i are different, for a sequence C_1, C_2, ..., C_{N} that satisfies the following conditions: * The sum of the sequence A_1, A_2, ..., A_{N} and the sum of the sequence C_1, C_2, ..., C_{N} are equal. * For every i, B_i \leq C_i holds. If such a sequence C_1, C_2, ..., C_{N} cannot be constructed, print -1.
|
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
diff_m = 0
diff_p = []
mcount = 0
if sum(B) > sum(A):
print("-1")
else:
for i in range(N):
difftmp = A[i] - B[i]
if difftmp < 0:
diff_m += difftmp
mcount += 1
else:
diff_p.append(difftmp)
if diff_m == 0:
print(0)
else:
diff_p.sort(reverse = True)
for j in range(1,N):
print(sum(diff_p[:j]))
if diff_m <= sum(diff_p[:j]):
print(j+1 + mcount)
break
else:
continue
|
s042818120
|
Accepted
| 119 | 18,524 | 656 |
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
diff_m = 0
diff_p = []
mcount = 0
if sum(B) > sum(A):
print(-1)
else:
for i in range(N):
difftmp = A[i] - B[i]
if difftmp < 0:
diff_m += difftmp
mcount += 1
else:
diff_p.append(difftmp)
if diff_m == 0:
print(0)
else:
diff_p.sort(reverse = True)
for j in range(0,N):
diff_m = diff_m + diff_p[j]
if diff_m >= 0:
print(j+1 + mcount)
break
else:
continue
|
s274419564
|
p03860
|
u781025961
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 140 |
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.
|
name1 = input()
name2 = name1.replace(" ","")
name3 = name2[7:]
name4 = name3[:7]
print(name4)
lst1 = list(name4)
print("A" + lst1[0] + "C")
|
s844060409
|
Accepted
| 17 | 2,940 | 128 |
name1 = input()
name2 = name1.replace(" ","")
name3 = name2[7:]
name4 = name3[:7]
lst1 = list(name4)
print("A" + lst1[0] + "C")
|
s808609282
|
p03564
|
u943004959
| 2,000 | 262,144 |
Wrong Answer
| 20 | 2,940 | 50 |
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.
|
R = int(input())
G = int(input())
print(2 * G - R)
|
s143198091
|
Accepted
| 17 | 2,940 | 154 |
def solve():
N = int(input())
K = int(input())
ans = 1
for i in range(N):
ans = min(2 * ans, ans + K)
print(ans)
solve()
|
s267745963
|
p03149
|
u765556930
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 3,064 | 20 |
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".
|
x = input()
print(x)
|
s769178567
|
Accepted
| 17 | 2,940 | 150 |
x = input().split()
test = ["1", "9", "7", "4"]
for _ in x:
if _ in test:
test.remove(_)
if len(test) == 0:
print("YES")
else:
print("NO")
|
s853222802
|
p03162
|
u090068671
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 400 |
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
def resolve():
N = int(input())
ABC = [
map(int, input().split())
for _ in range(N)
]
Va, Vb, Vc = ABC.pop()
for _ in range(N-1):
a, b, c = ABC.pop()
Va, Vb, Vc = max(Vb, Vc) + a, max(Va, Vc) + b, max(Va, Vb) + c
print(max(Va, Vb, Vc))
|
s770196301
|
Accepted
| 478 | 54,424 | 410 |
def resolve():
N = int(input())
ABC = [
map(int, input().split())
for _ in range(N)
]
Va, Vb, Vc = ABC.pop()
for _ in range(N-1):
a, b, c = ABC.pop()
Va, Vb, Vc = max(Vb, Vc) + a, max(Va, Vc) + b, max(Va, Vb) + c
print(max(Va, Vb, Vc))
resolve()
|
s125718947
|
p03607
|
u760961723
| 2,000 | 262,144 |
Wrong Answer
| 207 | 16,660 | 197 |
You are playing the following game with Joisino. * Initially, you have a blank sheet of paper. * Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times. * Then, you are asked a question: How many numbers are written on the sheet now? The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?
|
N = int(input())
s = [int(input()) for i in range(N)]
import collections
c = collections.Counter(s)
print(list(c.values()))
ans_lis = [x for x in list(c.values()) if x %2 != 0]
print(len(ans_lis))
|
s347308390
|
Accepted
| 196 | 16,660 | 174 |
N = int(input())
s = [int(input()) for i in range(N)]
import collections
c = collections.Counter(s)
ans_lis = [x for x in list(c.values()) if x %2 != 0]
print(len(ans_lis))
|
s497478814
|
p03693
|
u630657312
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 106 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r, g, b = map(int, input().split())
if (100*r + 10*g + b) % 4 == 0:
print('Yes')
else:
print('No')
|
s475355798
|
Accepted
| 18 | 2,940 | 106 |
r, g, b = map(int, input().split())
if (100*r + 10*g + b) % 4 == 0:
print('YES')
else:
print('NO')
|
s395248356
|
p02831
|
u006883624
| 2,000 | 1,048,576 |
Wrong Answer
| 39 | 3,060 | 266 |
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
A, B = [int(v) for v in input().split()]
minV = min(A, B)
maxV = max(A, B)
if maxV % minV == 0:
print(maxV)
if maxV == minV + 1:
print(maxV * minV)
v = maxV * 2
for i in range(2, minV + 1):
if v % minV != 0:
v += maxV
continue
print(v)
exit()
|
s176160014
|
Accepted
| 39 | 3,064 | 284 |
A, B = [int(v) for v in input().split()]
minV = min(A, B)
maxV = max(A, B)
if maxV % minV == 0:
print(maxV)
exit()
if maxV == minV + 1:
print(maxV * minV)
exit()
v = maxV * 2
for i in range(2, minV + 1):
if v % minV != 0:
v += maxV
continue
print(v)
exit()
|
s615495820
|
p03847
|
u560301743
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,168 | 469 |
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
|
def solve(n: int) -> int:
mod = 1e9 + 7
n_bin = '0' + bin(n)[2:]
dp = [[0, 0, 0] for _ in range(len(n_bin))]
dp[-1][0] = 1
for i in range(len(n_bin) - 2, -1, -1):
for s in range(3):
for k in range(3):
s2 = min(2, 2 * s + int(n_bin[len(n_bin) - i - 1]) - k)
if s2 >= 0:
dp[i][s2] = (dp[i][s2] + dp[i + 1][s]) % mod
return sum(dp[0]) % mod
n = int(input())
print(solve(n))
|
s099005878
|
Accepted
| 28 | 9,224 | 479 |
def solve(n: int) -> int:
mod = 1e9 + 7
n_bin = '0' + bin(n)[2:]
dp = [[0, 0, 0] for _ in range(len(n_bin))]
dp[-1][0] = 1
for i in range(len(n_bin) - 2, -1, -1):
for s in range(3):
for k in range(3):
s2 = min(2, 2 * s + int(n_bin[len(n_bin) - i - 1]) - k)
if s2 >= 0:
dp[i][s2] = int((dp[i][s2] + dp[i + 1][s]) % mod)
return int(sum(dp[0]) % mod)
n = int(input())
print(solve(n))
|
s434403533
|
p02747
|
u340585574
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 192 |
A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string.
|
n=input()
if(len(n)%2==1):
print("No")
exit()
for i in range(int(len(n)/2)):
if(n[i]=='h'and n[i+1]=='i'):
flag=1
else:
print("No")
exit()
print("Yes")
|
s192885630
|
Accepted
| 17 | 2,940 | 86 |
n=input()
n2=n.replace('hi','')
if(len(n2)==0):
print("Yes")
else:
print("No")
|
s431814405
|
p04043
|
u260068288
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 93 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a =input()
if a.count("5") == 2 and a.count("7") == 1:
print("Yes")
else:
print("No")
|
s012316878
|
Accepted
| 17 | 2,940 | 119 |
abc =input()
five = abc.count("5")
seven = abc.count("7")
if five==2 and seven==1:
print("YES")
else:
print("NO")
|
s993738658
|
p02742
|
u623953567
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 289 |
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())
if(H == 1 or W == 1):
print(1)
else:
if(H%2 == 0):
i = H//2
if(H%2 != 0):
i = H//2+1
j = H-i
if(W%2 == 0):
sum = i*(W//2) + j*(W//2)
if(W%2 != 0):
sum = i*(W/2+1) + j*(W/2)
print(sum)
|
s675611825
|
Accepted
| 17 | 3,064 | 296 |
H,W = map(int,input().split())
if(H == 1 or W == 1):
print(1)
else:
if(H%2 == 0):
i = H//2
if(H%2 != 0):
i = H//2 + 1
j = H-i
if(W%2 == 0):
sum = i*(W//2) + j*(W//2)
if(W%2 != 0):
sum = i*(W//2 + 1) + j*(W//2)
print(sum)
|
s314625695
|
p03730
|
u189397279
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 158 |
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
import sys
A, B, C = map(int, sys.stdin.readline().split())
for i in range(1, B + 1):
if (A * i) % B == C:
print("Yes")
exit()
print("No")
|
s902524214
|
Accepted
| 17 | 2,940 | 158 |
import sys
A, B, C = map(int, sys.stdin.readline().split())
for i in range(1, B + 1):
if (A * i) % B == C:
print("YES")
exit()
print("NO")
|
s286953324
|
p02646
|
u046826851
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,160 | 205 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a,v = map(int, input().split( ))
b,w = map(int, input().split( ))
t = int(input())
dis0 = b - a
dis = v - w
if dis <= 0:
print('No')
exit()
d = dis*t
if d < dis:
print('No')
else:
print('Yes')
|
s315917850
|
Accepted
| 28 | 9,036 | 205 |
a,v = map(int, input().split( ))
b,w = map(int, input().split( ))
t = int(input())
dis0 = abs(b - a)
dis = v - w
if w>=v:
print("NO")
exit()
if abs(b-a)/(v-w)>t:
print("NO")
else:
print("YES")
|
s272282990
|
p03997
|
u507116804
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 62 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a=int(input())
b=int(input())
h=int(input())
print((a+b)*h/2)
|
s666895461
|
Accepted
| 18 | 2,940 | 80 |
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
|
s731204278
|
p02407
|
u316584871
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,584 | 112 |
Write a program which reads a sequence and prints it in the reverse order.
|
n = int(input())
li = list(map(int, input().split()))
li.sort()
for i in li:
print('{}'.format(i), end=" ")
|
s627744953
|
Accepted
| 20 | 5,612 | 215 |
n = int(input())
l = list(map(int, input().split()))
x = []
for i in range(n):
x.append(l[n-1-i])
for s in x:
if(s == x[n-1]):
print('{}'.format(s))
else:
print('{}'.format(s), end=" ")
|
s996274250
|
p03068
|
u346395915
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 150 |
You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`.
|
n = int(input())
s = list(input())
k = int(input())
mozi = s[k-1]
print(mozi)
for i in range(n):
if s[i] != mozi:
s[i] = "*"
print(*s)
|
s273911432
|
Accepted
| 17 | 2,940 | 146 |
n = int(input())
s = list(input())
k = int(input())
mozi = s[k-1]
for i in range(n):
if s[i] != mozi:
s[i] = "*"
print(*s, sep="")
|
s157023879
|
p03303
|
u119460590
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,188 | 101 |
You are given a string S consisting of lowercase English letters. We will write down this string, starting a new line after every w letters. Print the string obtained by concatenating the letters at the beginnings of these lines from top to bottom.
|
s = input()
N = int(input())
while len(s):
print(s[0:min(len(s), N)])
s = s[min(len(s), N):]
|
s821901665
|
Accepted
| 17 | 2,940 | 50 |
s = input()
N = int(input())
print(s[0:len(s):N])
|
s425124330
|
p03160
|
u020091453
| 2,000 | 1,048,576 |
Wrong Answer
| 147 | 13,928 | 355 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
a = list()
for i in input().split(' '):
a.append(int(i))
cost = list()
for i in range(N):
if i == 0:
cost.append(0)
elif i == 1:
cost.append(abs(a[1] - a[0]))
else:
c1 = abs(a[i] - a[i-1]) + cost[i-1]
c2 = abs(a[i] - a[i-2]) + cost[i-2]
cost.append( c1 if c1<c2 else c2)
cost[-1]
|
s948352857
|
Accepted
| 143 | 13,980 | 372 |
N = int(input())
a = list()
for i in input().split(' '):
a.append(int(i))
cost = list()
for i in range(N):
if i == 0:
cost.append(0)
elif i == 1:
cost.append(abs(a[1] - a[0]))
else:
c1 = abs(a[i] - a[i-1]) + cost[i-1]
c2 = abs(a[i] - a[i-2]) + cost[i-2]
cost.append( c1 if c1<c2 else c2)
print(cost[-1])
|
s766770248
|
p03385
|
u046592970
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
s = sorted(input())
if s == ["a","b","c"]:
print("Yes")
else:
print("No")
print(s)
|
s834313718
|
Accepted
| 18 | 2,940 | 82 |
s = sorted(input())
if s == ["a","b","c"]:
print("Yes")
else:
print("No")
|
s438935175
|
p02257
|
u935329231
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,748 | 506 |
A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.
|
# -*- coding: utf-8 -*-
import math
def is_prime(x):
# Primality Test
if x <= 1 or x % 2 == 0:
return False
if x == 2:
return True
i = 3
while i <= math.sqrt(x):
if x % i == 0:
return False
i += 2
return True
def to_int(v):
return int(v)
if __name__ == '__main__':
l = to_int(input())
seq = [to_int(input()) for i in range(l)]
cnt = 0
for v in seq:
if is_prime(v):
cnt += 1
print(cnt)
|
s866499940
|
Accepted
| 820 | 8,020 | 506 |
# -*- coding: utf-8 -*-
import math
def is_prime(x):
# Primality Test
if x == 2:
return True
if x <= 1 or x % 2 == 0:
return False
i = 3
while i <= math.sqrt(x):
if x % i == 0:
return False
i += 2
return True
def to_int(v):
return int(v)
if __name__ == '__main__':
l = to_int(input())
seq = [to_int(input()) for i in range(l)]
cnt = 0
for v in seq:
if is_prime(v):
cnt += 1
print(cnt)
|
s867194513
|
p03679
|
u846150137
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 114 |
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.
|
a,b,c=map(int,input().split())
if b-c>a:
print("dangerous")
elif b>c:
print("safe")
else:
print("delicious")
|
s974733101
|
Accepted
| 17 | 2,940 | 114 |
a,b,c=map(int,input().split())
if c-b>a:
print("dangerous")
elif c>b:
print("safe")
else:
print("delicious")
|
s953655950
|
p03090
|
u198440493
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 3,956 | 153 |
You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem.
|
n=int(input())
a=[]
i=1
while(i<=n):
j=i+1
while(j<=n):
if j!=n-i:
a+=[(i,j)]
j+=1
i+=1
for x in a:
print(*x)
|
s766519490
|
Accepted
| 25 | 3,956 | 187 |
n=int(input())
a=[]
m=n if n&1 else n+1
i=1
while(i<=n):
j=i+1
while(j<=n):
if j!=m-i:
a+=[(i,j)]
j+=1
i+=1
print(len(a))
for x in a:
print(*x)
|
s849972611
|
p03713
|
u244416763
| 2,000 | 262,144 |
Wrong Answer
| 267 | 9,220 | 722 |
There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle. Snuke is trying to divide the bar as evenly as possible. More specifically, he is trying to minimize S_{max} \- S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece. Find the minimum possible value of S_{max} - S_{min}.
|
h,w = map(int,input().split())
ans = h*w
for i in range(1,h-1):
A = w*i
if w%2 == 0 or h-1%2 == 0:
B = (h-i)*w//2
C = (h-i)*w//2
else:
if h-1 < w:
B = (h-i)*((w//2)+1)
C = (h-i)*(w//2)
else:
B = ((h-i)//2 + 1) * w
C = (h-i)//2 * w
ans = min(ans,max(A,max(B,C)) - min(A,min(B,C)))
for i in range(1,w-1):
A = h*i
if w%2 == 0 or h-1%2 == 0:
B = (w-i)*h//2
C = (w-i)*h//2
else:
if w-1 < h:
B = (w-i)*((h//2)+1)
C = (w-i)*(h//2)
else:
B = ((w-i)//2 + 1) * h
C = (w-i)//2 * h
ans = min(ans,max(A,max(B,C)) - min(A,min(B,C)))
print(ans)
|
s405998746
|
Accepted
| 274 | 9,240 | 736 |
h,w = map(int,input().split())
ans = h*w
for i in range(1,h):
A = w*i
if w%2 == 0 or (h-i)%2 == 0:
B = ((h-i)*w)//2
C = ((h-i)*w)//2
else:
if h-i <= w:
B = (h-i)*((w//2)+1)
C = (h-i)*(w//2)
else:
B = ((h-i)//2 + 1) * w
C = ((h-i)//2) * w
ans = min(ans,max(A,max(B,C)) - min(A,min(B,C)))
for i in range(1,w):
A = h*i
if (w-i)%2 == 0 or h%2 == 0:
B = ((w-i)*h)//2
C = ((w-i)*h)//2
else:
if w-i <= h:
B = (w-i)*((h//2)+1)
C = (w-i)*(h//2)
else:
B = ((w-i)//2 + 1) * h
C = ((w-i)//2) * h
ans = min(ans,max(A,max(B,C)) - min(A,min(B,C)))
print(ans)
|
s248818332
|
p03433
|
u750275181
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 107 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
n = int(input())
a = list(map(int,input().split()))
a.sort(reverse=True)
print(sum(a[0::2])-sum(a[1::2]))
|
s702808182
|
Accepted
| 18 | 2,940 | 91 |
n = int(input())
a = int(input())
if n % 500 <= a:
print("Yes")
else:
print("No")
|
s394535334
|
p03433
|
u442152792
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 122 |
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
input = list(map(str,input().split()))
num = ''.join(input)
if(int(num) % 4 == 0):
print('YES')
else:
print('NO')
|
s157250322
|
Accepted
| 17 | 2,940 | 111 |
pay = int(input())
coin = int(input())
mod = pay % 500
if mod <= coin:
print('Yes')
else:
print('No')
|
s819789229
|
p03371
|
u437215432
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 172 |
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
# ABC095C
a, b, c, x, y = map(int, input().split())
if a > c or b > c:
t = min(x, y)
print("(1)", (x - t)*a + (y - t)*b + t*2*c)
else:
print("(2)", a*x + b*y)
|
s862871461
|
Accepted
| 280 | 12,504 | 308 |
import numpy as np
def f(a, b, c, x, y):
Min = np.inf
for use_c in range(max(x, y) * 2 + 1):
price = use_c * c + max(0, x - use_c // 2) * a + max(0, y - use_c // 2) * b
if price < Min:
Min = price
print(Min)
a, b, c, x, y = map(int, input().split())
f(a, b, c, x, y)
|
s737202436
|
p03737
|
u305732215
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 67 |
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
s1, s2, s3 = input().split()
ans = s1[0] + s2[0] + s3[0]
print(ans)
|
s296607187
|
Accepted
| 19 | 2,940 | 79 |
s1, s2, s3 = input().split()
ans = str.upper(s1[0] + s2[0] + s3[0])
print(ans)
|
s927468118
|
p03380
|
u365364616
| 2,000 | 262,144 |
Wrong Answer
| 108 | 14,060 | 184 |
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
N = int(input())
A = list(map(int, input().split()))
A.sort()
X = A[-1]
X2 = X // 2
dd = A[-1]
for i in range(N - 1):
d = abs(X2 - A[i])
if d < dd:
Y = A[i]
print(X, Y)
|
s540080464
|
Accepted
| 107 | 14,428 | 205 |
N = int(input())
A = list(map(int, input().split()))
A.sort()
X = A[-1]
X2 = (X + 1) // 2
dd = A[-1]
for i in range(N - 1):
d = abs(X2 - A[i])
if d < dd:
dd = d
Y = A[i]
print(X, Y)
|
s564329279
|
p03448
|
u048945791
| 2,000 | 262,144 |
Wrong Answer
| 48 | 3,060 | 256 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
ans = 0
for num_a in range(1, a + 1):
for num_b in range(1, b + 1):
for num_c in range(1, c + 1):
if 500 * num_a + 100 * num_b + 50 * num_c == x:
ans += 1
print(ans)
|
s882400309
|
Accepted
| 50 | 3,060 | 256 |
a = int(input())
b = int(input())
c = int(input())
x = int(input())
ans = 0
for num_a in range(0, a + 1):
for num_b in range(0, b + 1):
for num_c in range(0, c + 1):
if 500 * num_a + 100 * num_b + 50 * num_c == x:
ans += 1
print(ans)
|
s294748723
|
p03854
|
u494058663
| 2,000 | 262,144 |
Wrong Answer
| 59 | 3,956 | 503 |
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 = list(input())
tmp = ''
L = len(S)
S.append('0')
S.append('0')
ans = 0
for i in range(L):
tmp += str(S[i])
if tmp=='dream':
if tmp+str(S[i+1])+str(S[i+2])=='dreamer':
ans += len(tmp)+2
tmp = ''
i+=2
else:
ans += len(tmp)
tmp = ''
if tmp == 'erase':
if tmp+str(S[i+1])=='eraser':
ans += len(tmp)+1
tmp = ''
i+=1
if ans == len(S):
print('YES')
else:
print('NO')
|
s393224230
|
Accepted
| 63 | 4,652 | 267 |
S = list(input())
tmp = ''
L = len(S)
S = S[-1:-L-1:-1]
Kouho = ['maerd','remaerd','esare','resare']
ans = 0
for i in range(L):
tmp += str(S[i])
if tmp in Kouho:
ans += len(tmp)
tmp = ''
if ans == L:
print('YES')
else:
print('NO')
|
s298378212
|
p03149
|
u201660334
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,040 | 104 |
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".
|
n1, n2, n3, n4 = input().split()
if {n1, n2, n3, n4} == {1, 9, 7, 4}:
print("YES")
else:
print("NO")
|
s918163134
|
Accepted
| 29 | 9,160 | 114 |
n1, n2, n3, n4 = map(int, input().split())
if {n1, n2, n3, n4} == {1, 9, 7, 4}:
print("YES")
else:
print("NO")
|
s921771395
|
p03997
|
u779308281
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
result = (a + b) * h / 2
print(result)
|
s489597570
|
Accepted
| 17 | 2,940 | 94 |
a = int(input())
b = int(input())
h = int(input())
result = int((a + b) * h / 2)
print(result)
|
s330673975
|
p04043
|
u980503157
| 2,000 | 262,144 |
Wrong Answer
| 22 | 8,920 | 102 |
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.
|
form = input()
form = form.split()
form.sort()
if form == [5,5,7]:
print("YES")
else:
print("NO")
|
s313275510
|
Accepted
| 23 | 9,092 | 93 |
s = input()
s = s.split()
s.sort()
if s == ["5","5","7"]:
print("YES")
else:
print("NO")
|
s016651903
|
p02694
|
u931118906
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 9,156 | 93 |
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())
a = 100
ans = 0
while x >= a:
ans += 1
a = int(a * 1.01)
print(ans)
|
s487630944
|
Accepted
| 25 | 9,152 | 92 |
x = int(input())
a = 100
ans = 0
while x > a:
ans += 1
a = int(a * 1.01)
print(ans)
|
s013205754
|
p02678
|
u614181788
| 2,000 | 1,048,576 |
Wrong Answer
| 1,062 | 48,720 | 987 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
n,m = map(int,input().split())
C = [[0 for j in range(2)] for i in range(m)]
for i in range(m):
C[i] = list(map(int,input().split()))
if C[i][0] > C[i][1]:
C[i][0], C[i][1] = C[i][1],C[i][0]
sortF = lambda val : val[0]
C.sort(key=sortF)
a = [0]*n
b = [0]*n
for i in range(m):
if C[i][0] == 1:
a[C[i][1]-1] = C[i][0]
b[C[i][1]-1] += 1
elif a[C[i][1]-1] != 0:
a[C[i][0]-1] = C[i][1]
b[C[i][0]-1] = b[C[i][1]-1] + 1
elif a[C[i][0]-1] != 0:
a[C[i][1]-1] = C[i][0]
b[C[i][1]-1] = b[C[i][0]-1] + 1
for i in range(m):
if a[C[i][1]-1] != 0:
if b[C[i][1]-1] != 0 and b[C[i][1]-1] + 1 < b[C[i][0]-1]:
a[C[i][0]-1] = C[i][1]
b[C[i][0]-1] = b[C[i][1]-1] + 1
elif a[C[i][0]-1] != 0:
if b[C[i][0]-1] != 0 and b[C[i][1]-1] + 1 < b[C[i][1]-1]:
a[C[i][1]-1] = C[i][0]
b[C[i][1]-1] = b[C[i][0]-1] + 1
print("Yes")
for i in range(n-1):
print(a[i+1])
|
s776279192
|
Accepted
| 715 | 35,596 | 555 |
n,m = map(int,input().split())
V = [None] + [[] for _ in range(n)]
for i in range(m):
a, b= map(int,input().split())
V[a].append(b)
V[b].append(a)
from collections import deque
q = deque()
reach = [None] + [0] + [-1]*(n-1)
goal = [None] + [0]*n
for L in V[1]:
q.append(L)
reach[L] = reach[1] + 1
goal[L] = 1
while q:
p = q.popleft()
for L in V[p]:
if reach[L] == -1:
q.append(L)
reach[L] = reach[p] + 1
goal[L] = p
print("Yes")
for i in range(len(goal)-2):
print(goal[i+2])
|
s945420542
|
p02795
|
u711539583
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 75 |
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.
|
h = int(input())
w = int(input())
n = int(input())
print(int(n / max(h,w)))
|
s884935754
|
Accepted
| 18 | 3,316 | 116 |
h = int(input())
w = int(input())
n = int(input())
m = max(h, w)
if n % m:
print(n // m + 1)
else:
print(n // m)
|
s919309885
|
p02255
|
u599879983
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 290 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
N = int(input())
A = list(map(int,input().split()))
print(A)
def insertionSort(A,N):
for i in range(1,N):
v = A[i]
j = i -1
while j >= 0 and A[j] > v:
A[j+1] = A[j]
j -= 1
A[j+1] = v
print(A)
insertionSort(A,N)
|
s030942077
|
Accepted
| 20 | 5,604 | 372 |
N = int(input())
A = list(map(int,input().split()))
def insertionSort(A,N):
for i in range(1,N):
v = A[i]
j = i -1
while j >= 0 and A[j] > v:
A[j+1] = A[j]
j -= 1
A[j+1] = v
outputNum(A)
def outputNum(A):
tmp = map(str,A)
text = " ".join(tmp)
print(text)
outputNum(A)
insertionSort(A,N)
|
s018813928
|
p03493
|
u582614471
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 42 |
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.
|
n = map(int,input().split())
print(sum(n))
|
s502909234
|
Accepted
| 17 | 2,940 | 40 |
n = map(int,list(input()))
print(sum(n))
|
s527883879
|
p04045
|
u070423038
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,100 | 323 |
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.
|
price, hate = input().split()
no_num = set([x for x in input().split()])
ok_num = sorted(list({'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'} - no_num))
print(ok_num)
result = []
for i in price:
print(i)
for j in ok_num:
if i <= j:
result.append(j)
break
print(''.join(result))
|
s907165552
|
Accepted
| 27 | 9,044 | 534 |
kingaku, n = input().split()
no_num = input().split()
ok_num = [str(i) for i in range(10) if not str(i) in no_num]
result = []
for i in kingaku:
for j in ok_num:
if i == j:
result.append(j)
break
elif i < j:
result.append(j)
print("".join(result).ljust(len(kingaku), ok_num[0]))
exit()
break
else:
n = 1 if ok_num[0] == '0' else 0
print(ok_num[n].ljust(len(kingaku)+1, ok_num[0]))
exit()
print(''.join(result))
|
s799442973
|
p03798
|
u427344224
| 2,000 | 262,144 |
Wrong Answer
| 241 | 3,572 | 1,485 |
Snuke, who loves animals, built a zoo. There are N animals in this zoo. They are conveniently numbered 1 through N, and arranged in a circle. The animal numbered i (2≤i≤N-1) is adjacent to the animals numbered i-1 and i+1. Also, the animal numbered 1 is adjacent to the animals numbered 2 and N, and the animal numbered N is adjacent to the animals numbered N-1 and 1. There are two kinds of animals in this zoo: honest sheep that only speak the truth, and lying wolves that only tell lies. Snuke cannot tell the difference between these two species, and asked each animal the following question: "Are your neighbors of the same species?" The animal numbered i answered s_i. Here, if s_i is `o`, the animal said that the two neighboring animals are of the same species, and if s_i is `x`, the animal said that the two neighboring animals are of different species. More formally, a sheep answered `o` if the two neighboring animals are both sheep or both wolves, and answered `x` otherwise. Similarly, a wolf answered `x` if the two neighboring animals are both sheep or both wolves, and answered `o` otherwise. Snuke is wondering whether there is a valid assignment of species to the animals that is consistent with these responses. If there is such an assignment, show one such assignment. Otherwise, print `-1`.
|
n = int(input())
s = input()
S1 = "SS"
S2 = "WW"
S3 = "SW"
S4 = "WW"
S_list = [S1, S2, S3, S4]
for S in S_list:
last = ""
for i in range(n):
if i == 0:
if S[0] == "S":
if s[i] == "o":
last = S[1]
else:
last = "S" if S[1] == "W" else "W"
else:
if s[i] == "o":
last = "S" if S[1] == "W" else "W"
else:
last = "S" if S[1] == "S" else "W"
elif 1 <= i < n - 1:
if S[i] == "S":
if s[i] == "o":
S += S[i-1]
else:
S += "W" if S[i-1] == "S" else "S"
else:
if s[i] == "o":
S += "S" if S[i-1] == "W" else "S"
else:
S += "S" if S[i-1] == "S" else "W"
elif i == n - 1:
if S[i] == "S":
if s[i] == "o" and S[0] == S[i - 1] and S[i] == last:
print(S)
exit()
if s[i] == "x" and S[0] != S[i - 1] and S[i] == last:
print(S)
exit()
else:
if s[i] == "o" and S[0] != S[i - 1] and S[i] == last:
print(S)
exit()
if s[i] == "x" and S[0] == S[i - 1] and S[i] == last:
print(S)
exit()
print(-1)
|
s501403843
|
Accepted
| 257 | 3,572 | 1,486 |
n = int(input())
s = input()
S1 = "SS"
S2 = "WW"
S3 = "SW"
S4 = "WS"
S_list = [S1, S2, S3, S4]
for S in S_list:
last = ""
for i in range(n):
if i == 0:
if S[0] == "S":
if s[i] == "o":
last = S[1]
else:
last = "S" if S[1] == "W" else "W"
else:
if s[i] == "o":
last = "S" if S[1] == "W" else "W"
else:
last = "S" if S[1] == "S" else "W"
elif 1 <= i < n - 1:
if S[i] == "S":
if s[i] == "o":
S += S[i-1]
else:
S += "W" if S[i-1] == "S" else "S"
else:
if s[i] == "o":
S += "S" if S[i-1] == "W" else "W"
else:
S += "S" if S[i-1] == "S" else "W"
elif i == n - 1:
if S[i] == "S":
if s[i] == "o" and S[0] == S[i - 1] and S[i] == last:
print(S)
exit()
if s[i] == "x" and S[0] != S[i - 1] and S[i] == last:
print(S)
exit()
else:
if s[i] == "o" and S[0] != S[i - 1] and S[i] == last:
print(S)
exit()
if s[i] == "x" and S[0] == S[i - 1] and S[i] == last:
print(S)
exit()
print(-1)
|
s773357384
|
p03855
|
u074220993
| 2,000 | 262,144 |
Wrong Answer
| 2,212 | 238,372 | 606 |
There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities. We will say city A and B are _connected by roads_ if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define _connectivity by railways_ similarly. For each city, find the number of the cities connected to that city by both roads and railways.
|
N, K, L = map(int, input().split())
city = [{"path":{i},"train":{i}} for i in range(N)]
for i in range(K):
p, q = map(int, input().split())
city[p-1]["path"].add(q-1)
city[q-1]["path"].add(p-1)
for i in range(L):
r, s = map(int, input().split())
city[r-1]["train"].add(s-1)
city[s-1]["train"].add(r-1)
for i in range(N):
for x in city[i]["path"]:
city[i]["path"] = city[i]["path"] | city[x]["path"]
for y in city[i]["train"]:
city[i]["train"] = city[i]["train"] | city[y]["train"]
for i in range(N):
print(len(city[i]["path"]&city[i]["train"]), end=" ")
|
s423661919
|
Accepted
| 882 | 55,756 | 599 |
N, K, L = map(int, input().split())
#Parent List(PL)
road = [i for i in range(N)]
rail = [i for i in range(N)]
def fp(x,P): #findParent
if x == P[x]:
return x
else:
P[x] = fp(P[x],P)
return P[x]
for i in range(K+L):
PL = (lambda x:road if x < K else rail)(i)
p, q = map(lambda x:int(x)-1, input().split())
p, q = fp(p,PL), fp(q,PL)
PL[p] = PL[q] = min(p,q)
from collections import defaultdict as dd
Wconnected = dd(lambda:0)
for i in range(N):
Wconnected[fp(i,road),fp(i,rail)] += 1
print(*[Wconnected[fp(i,road),fp(i,rail)] for i in range(N)])
|
s227408170
|
p03067
|
u187606204
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 79 |
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
A,B,C = map(int,input().split())
if A<C & C<B:
print("yes")
else:
print("No")
|
s827723524
|
Accepted
| 17 | 3,060 | 202 |
A,B,C = map(int,input().split())
if A>C & C>B:
print("Yes")
if B>C & C>A:
print("Yes")
if C>A & A>B:
print("No")
if B>A & A>C:
print("No")
if A>B & B>C:
print("No")
if C>B & B>A:
print("No")
|
s617908044
|
p03862
|
u838644735
| 2,000 | 262,144 |
Wrong Answer
| 188 | 20,660 | 1,012 |
There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies. Snuke can perform the following operation any number of times: * Choose a box containing at least one candy, and eat one of the candies in the chosen box. His objective is as follows: * Any two neighboring boxes contain at most x candies in total. Find the minimum number of operations required to achieve the objective.
|
def main():
N, x = map(int, input().split())
A = list(map(int, input().split()))
ans = 0
if A[0] > x:
diff = A[0] - x
A[0] = x
ans += diff
if A[-1] > x:
diff = A[-1] - x
A[-1] = x
ans += diff
B = [a for a in A]
ans1 = 0
for i in range(N-1):
s = A[i] + A[i + 1]
if s <= x:
continue
diff = s - x
if diff > A[i + 1]:
A[i] -= diff - A[i + 1]
A[i + 1] = 0
else:
A[i + 1] -= diff
ans1 += diff
ans2 = 0
for i in range(N - 1):
index = N - 1 - i
s = B[index] + B[index - 1]
if s <= x:
continue
diff = s - x
print(index, s, diff)
if diff > B[index - 1]:
B[index] -= diff - B[index - 1]
B[index - 1] = 0
else:
B[index - 1] -= diff
ans2 += diff
ans += min(ans1, ans2)
print(ans)
if __name__ == '__main__':
main()
|
s731691854
|
Accepted
| 116 | 20,940 | 982 |
def main():
N, x = map(int, input().split())
A = list(map(int, input().split()))
ans = 0
if A[0] > x:
diff = A[0] - x
A[0] = x
ans += diff
if A[-1] > x:
diff = A[-1] - x
A[-1] = x
ans += diff
B = [a for a in A]
ans1 = 0
for i in range(N-1):
s = A[i] + A[i + 1]
if s <= x:
continue
diff = s - x
if diff > A[i + 1]:
A[i] -= diff - A[i + 1]
A[i + 1] = 0
else:
A[i + 1] -= diff
ans1 += diff
ans2 = 0
for i in range(N - 1):
index = N - 1 - i
s = B[index] + B[index - 1]
if s <= x:
continue
diff = s - x
if diff > B[index - 1]:
B[index] -= diff - B[index - 1]
B[index - 1] = 0
else:
B[index - 1] -= diff
ans2 += diff
ans += min(ans1, ans2)
print(ans)
if __name__ == '__main__':
main()
|
s426246160
|
p03487
|
u209594105
| 2,000 | 262,144 |
Wrong Answer
| 87 | 21,532 | 304 |
You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N). Your objective is to remove some of the elements in a so that a will be a **good sequence**. Here, an sequence b is a **good sequence** when the following condition holds true: * For each element x in b, the value x occurs exactly x times in b. For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not. Find the minimum number of elements that needs to be removed so that a will be a good sequence.
|
n = input()
str = input()
nums = str.split()
num_dict = {}
for num in nums:
if num in num_dict:
num_dict[num] += 1
else:
num_dict[num] = 1
result = 0
for k, v in num_dict.items():
if int(k) > v:
result += v
elif int(k) < v:
result += int(k)
print(result)
|
s668081174
|
Accepted
| 96 | 21,528 | 310 |
n = input()
str = input()
nums = str.split()
num_dict = {}
for num in nums:
if num in num_dict:
num_dict[num] += 1
else:
num_dict[num] = 1
result = 0
for k, v in num_dict.items():
if int(k) > v:
result += v
elif int(k) < v:
result += (v - int(k))
print(result)
|
s093050699
|
p03494
|
u361826811
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 487 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
import sys
# import itertools
# import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
def num(n):
cnt = 0
while n:
if n % 2 == 0:
cnt += 1
n /= 2
else:
break
return cnt
N, *A = map(int, read().decode('utf8').split())
ans = 0
for i in A:
ans += num(i)
print(ans)
|
s043085930
|
Accepted
| 153 | 12,396 | 366 |
import sys
# import itertools
import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
N = int(readline())
A = np.array(readline().split(), dtype=np.int64)
ans=0
while np.all(A%2==0):
A//=2
ans+=1
print(ans)
|
s791564566
|
p02256
|
u741801763
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 132 |
Write a program which finds the greatest common divisor of two natural numbers _a_ and _b_
|
x,y = list(map(int,input().split()))
###??????x >= y
if y>x:x,y = y,x
while x % y !=0:
y = x%y
if y > x:x,y = y,x
print(y)
|
s948374321
|
Accepted
| 20 | 5,600 | 130 |
x,y = list(map(int,input().split()))
###??????x >= y
if y>x:x,y = y,x
while x % y !=0:
x = x%y
if y>x:x,y = y,x
print(y)
|
s801564060
|
p02401
|
u685534465
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,576 | 234 |
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.
|
a = 1
b = 1
op = ''
i = 0
while op!='?':
a,op,b = [i for i in input().split()]
a = int(a)
b = int(b)
if op!='?':
if op=='+':
print(a+b)
elif op=='-':
print(a-b)
elif op=='*':
print(a*b)
elif op=='/':
print(a/b)
|
s027399995
|
Accepted
| 20 | 7,556 | 235 |
a = 1
b = 1
op = ''
i = 0
while op!='?':
a,op,b = [i for i in input().split()]
a = int(a)
b = int(b)
if op!='?':
if op=='+':
print(a+b)
elif op=='-':
print(a-b)
elif op=='*':
print(a*b)
elif op=='/':
print(a//b)
|
s050908742
|
p03095
|
u309141201
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 3,444 | 196 |
You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order.
|
from collections import Counter
n = int(input())
s = input()
mod = 10**9 + 7
ans = 1
scnt = Counter(s)
# print(scnt)
for cnt in scnt.values():
print(cnt)
ans *= (cnt+1) % mod
print(ans-1)
|
s215528213
|
Accepted
| 25 | 3,444 | 170 |
from collections import Counter
n = int(input())
s = input()
mod = 10**9 + 7
ans = 1
scnt = Counter(s)
for cnt in scnt.values():
ans = ans*(cnt+1) % mod
print(ans-1)
|
s865176203
|
p03456
|
u043236471
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 132 |
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.
|
import math
a, b = [int(n) for n in input().split()]
res = math.sqrt(a * b)
if res % 1 == 0:
print('YES')
else:
print('NO')
|
s952838362
|
Accepted
| 17 | 2,940 | 145 |
import math
a, b = [int(n) for n in input().split()]
res = math.sqrt(int(str(a)+str(b)))
if res % 1 == 0:
print('Yes')
else:
print('No')
|
s845770738
|
p03543
|
u318427318
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,336 | 253 |
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
#-*-coding:utf-8-*-
import collections
def main():
s = input()
count=collections.Counter(s)
for i in count.values():
if i >=3:
print("Yes")
else:
print("No")
if __name__=="__main__":
main()
|
s950005289
|
Accepted
| 26 | 9,112 | 304 |
#-*-coding:utf-8-*-
def main():
s = input()
string_list=list(str(s))
if string_list[0]==string_list[1]==string_list[2]:
print("Yes")
elif string_list[1]==string_list[2]==string_list[3]:
print("Yes")
else:
print("No")
if __name__=="__main__":
main()
|
s205190140
|
p03416
|
u636822224
| 2,000 | 262,144 |
Wrong Answer
| 127 | 3,060 | 172 |
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
a,b=map(int,input().split())
sum=0
print(b-a+1)
for i in range(b-a+1):
if int(str(a+i)[0])==int(str(a+i)[4]) and int(str(a+i)[1])==int(str(a+i)[3]):
sum+=1
print(sum)
|
s211970666
|
Accepted
| 122 | 3,060 | 159 |
a,b=map(int,input().split())
sum=0
for i in range(b-a+1):
if int(str(a+i)[0])==int(str(a+i)[4]) and int(str(a+i)[1])==int(str(a+i)[3]):
sum+=1
print(sum)
|
s303469547
|
p02748
|
u993310962
| 2,000 | 1,048,576 |
Wrong Answer
| 411 | 18,736 | 238 |
You are visiting a large electronics store to buy a refrigerator and a microwave. The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen. You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time. You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required.
|
a,b,m=map(int, input().split())
A=list(map(int, input().split()))
B=list(map(int, input().split()))
buf=min(A)+min(B)
for i in range(m):
x,y,c=map(int, input().split())
cp=A[x-1]+B[y-1]-c
if buf>=cp:
buf==cp
print(buf)
|
s044387575
|
Accepted
| 400 | 18,736 | 237 |
a,b,m=map(int, input().split())
A=list(map(int, input().split()))
B=list(map(int, input().split()))
buf=min(A)+min(B)
for i in range(m):
x,y,c=map(int, input().split())
cp=A[x-1]+B[y-1]-c
if buf>=cp:
buf=cp
print(buf)
|
s810855988
|
p03997
|
u423665486
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 104 |
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.
|
def resolve():
t = []
for _ in range(3):
t.append(int(input()))
a, b, h = t
print(int((a+b)/2*h))
|
s106384549
|
Accepted
| 17 | 2,940 | 113 |
def resolve():
t = []
for _ in range(3):
t.append(int(input()))
a, b, h = t
print(int((a+b)/2*h))
resolve()
|
s915515039
|
p03997
|
u536177854
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a=int(input())
b=int(input())
h=int(input())
print((a+b)*h/2)
|
s162428226
|
Accepted
| 17 | 2,940 | 62 |
a=int(input())
b=int(input())
h=int(input())
print((a+b)*h//2)
|
s864634447
|
p03079
|
u195912432
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 115 |
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
# coding: utf-8
A, B, C = map(int, input().split())
if A == B and B == C:
print("YES")
else:
print("NO")
|
s712820365
|
Accepted
| 18 | 2,940 | 115 |
# coding: utf-8
A, B, C = map(int, input().split())
if A == B and B == C:
print("Yes")
else:
print("No")
|
s972132988
|
p03699
|
u811000506
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 240 |
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
N = int(input())
S = [int(input()) for i in range(N)]
S.sort()
sum = sum(S)
if sum%10==0:
for i in range(N):
if (sum-S[i])%10!=0:
sum -= S[i]
print(sum)
break
print(0)
else:
print(sum)
|
s444579635
|
Accepted
| 17 | 3,060 | 241 |
N = int(input())
S = [int(input()) for i in range(N)]
S.sort()
sum = sum(S)
if sum%10==0:
for i in range(N):
if (sum-S[i])%10!=0:
sum -= S[i]
print(sum)
exit()
print(0)
else:
print(sum)
|
s365516315
|
p03997
|
u680851063
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 77 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a=int(input())
b=int(input())
c=int(input())
#d=int(input())
print((a+b)*c/2)
|
s447085908
|
Accepted
| 17 | 2,940 | 82 |
a=int(input())
b=int(input())
c=int(input())
#d=int(input())
print(int((a+b)*c/2))
|
s774879150
|
p03845
|
u513081876
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 214 |
Joisino is about to compete in the final round of a certain programming competition. In this contest, there are N problems, numbered 1 through N. Joisino knows that it takes her T_i seconds to solve problem i(1≦i≦N). Also, there are M kinds of drinks offered to the contestants, numbered 1 through M. If Joisino takes drink i(1≦i≦M), her brain will be stimulated and the time it takes for her to solve problem P_i will become X_i seconds. It does not affect the time to solve the other problems. A contestant is allowed to take exactly one of the drinks before the start of the contest. For each drink, Joisino wants to know how many seconds it takes her to solve all the problems if she takes that drink. Here, assume that the time it takes her to solve all the problems is equal to the sum of the time it takes for her to solve individual problems. Your task is to write a program to calculate it instead of her.
|
N = int(input())
T = list(map(int, input().split()))
M = int(input())
p = []
x = []
ans = []
for i in range(M):
p, x = map(int, input().split())
T[p-1] = x
ans.append(sum(T))
for _ in ans:
print(_)
|
s442693587
|
Accepted
| 17 | 3,060 | 212 |
N = int(input())
T = [int(i) for i in input().split()]
M = int(input())
PX = [[int(i) for i in input().split()] for i in range(M)]
for i in range(M):
ans = sum(T) - T[PX[i][0] - 1] + PX[i][1]
print(ans)
|
s707744733
|
p03386
|
u842388336
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 153 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,x = map(int,input().split())
a_set = set(range(a,a+x))
b_set = set(range(b-x,b+1))
answer = a_set.intersection(b_set)
for ans in answer:
print(ans)
|
s809236582
|
Accepted
| 17 | 3,060 | 232 |
a,b,x = map(int,input().split())
a_set = set(range(a,min(a+x,b)))
b_set = set(range(max(a,b-x+1),b+1))
#print(a_set)
#print(b_set)
a_set = a_set.union(b_set)
ans_list = list(a_set)
ans_list.sort()
for ans in ans_list:
print(ans)
|
s182117902
|
p03598
|
u844005364
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 110 |
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
n = int(input())
k = int(input())
arr = list(map(int, input().split()))
print(sum(min(x, k-x) for x in arr))
|
s256503504
|
Accepted
| 19 | 2,940 | 83 |
_,k=input(),int(input())
print(2*sum(min(x,k-x) for x in map(int,input().split())))
|
s748584472
|
p03523
|
u201660334
| 2,000 | 262,144 |
Wrong Answer
| 35 | 9,872 | 117 |
You are given a string S. Takahashi can insert the character `A` at any position in this string any number of times. Can he change S into `AKIHABARA`?
|
import re
s = input()
regex = re.compile(r"A?KIHA?BA?RA?")
if regex.match(s):
print("Yes")
else:
print("No")
|
s329060952
|
Accepted
| 35 | 9,820 | 121 |
import re
s = input()
regex = re.compile(r"A?KIHA?BA?RA?")
if regex.fullmatch(s):
print("YES")
else:
print("NO")
|
s368074065
|
p03478
|
u039623862
| 2,000 | 262,144 |
Wrong Answer
| 26 | 3,064 | 200 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n,a,b=map(int, input().split())
def dsum(n):
res = 0
while n>0:
res *= 10
res += n%10
n //= 10
return res
cnt = 0
for i in range(1,n+1):
if a<=dsum(i)<=b:
cnt += 1
print(cnt)
|
s357937461
|
Accepted
| 25 | 3,060 | 188 |
n,a,b=map(int, input().split())
def dsum(n):
res = 0
while n>0:
res += n%10
n //= 10
return res
cnt = 0
for i in range(1,n+1):
if a<=dsum(i)<=b:
cnt += i
print(cnt)
|
s193068410
|
p03457
|
u156896646
| 2,000 | 262,144 |
Wrong Answer
| 382 | 3,064 | 403 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
t0, x0, y0 = 0, 0, 0
for _ in range(N):
t, x, y = map(int, input().split())
dt = t - t0
dx = x - x0
dy = y - y0
dxy = abs(dx) + abs(dy) #; print(dt, dx, dy, dxy)
isEvenOdd = (dt%2 == 0 and dxy%2 == 0) or (dt%2 != 0 and dxy%2 != 0)
if isEvenOdd and dxy <= dt :
t0, x0, y0 = t, x, y
else:
print('NO')
break
else:
print('YES')
|
s784406969
|
Accepted
| 378 | 3,064 | 334 |
N = int(input())
t0, x0, y0 = 0, 0, 0
for _ in range(N):
t, x, y = map(int, input().split())
dt = t - t0
dx = x - x0
dy = y - y0
dxy = abs(dx) + abs(dy) #; print(dt, dx, dy, dxy)
if dt%2 == dxy%2 and dxy <= dt :
t0, x0, y0 = t, x, y
else:
print('No')
break
else:
print('Yes')
|
s137892677
|
p03434
|
u835283343
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 150 |
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.
|
import math
input()
data = sorted(map(int, input().split(' ')))
s = 0
for i, n in enumerate(data):
s = s + int(math.pow((-1),(i%2)))*n
print(s)
|
s167309323
|
Accepted
| 17 | 2,940 | 164 |
import math
input()
data = sorted(map(int, input().split(' ')), reverse=True)
s = 0
for i, n in enumerate(data):
s = s + int(math.pow((-1),(i%2)))*n
print(s)
|
s157279537
|
p04043
|
u822179469
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 122 |
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.
|
d = list(map(int,input().split()))
print(d)
if d.count(5)==2 and d.count(7) == 1 :
print("YES")
else :
print("NO")
|
s409982184
|
Accepted
| 17 | 2,940 | 113 |
d = list(map(int,input().split()))
if d.count(5)==2 and d.count(7) == 1 :
print("YES")
else :
print("NO")
|
s734709341
|
p03729
|
u422272120
| 2,000 | 262,144 |
Wrong Answer
| 26 | 8,968 | 91 |
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a,b,c = input().split()
print ("Yes") if a[-1] == b[0] and b[-1] == c[0] else print ("No")
|
s443988519
|
Accepted
| 29 | 9,032 | 91 |
a,b,c = input().split()
print ("YES") if a[-1] == b[0] and b[-1] == c[0] else print ("NO")
|
s561105106
|
p03762
|
u814986259
| 2,000 | 262,144 |
Wrong Answer
| 124 | 24,460 | 503 |
On a two-dimensional plane, there are m lines drawn parallel to the x axis, and n lines drawn parallel to the y axis. Among the lines parallel to the x axis, the i-th from the bottom is represented by y = y_i. Similarly, among the lines parallel to the y axis, the i-th from the left is represented by x = x_i. For every rectangle that is formed by these lines, find its area, and print the total area modulo 10^9+7. That is, for every quadruple (i,j,k,l) satisfying 1\leq i < j\leq n and 1\leq k < l\leq m, find the area of the rectangle formed by the lines x=x_i, x=x_j, y=y_k and y=y_l, and print the sum of these areas modulo 10^9+7.
|
n, m = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
mod = 10**9 + 7
H = 0
W = 0
for i in range(n//2):
w = x[n-1-i] - x[i]
if i == n//2 - 1:
if n % 2 == 1:
w *= 2
else:
w *= n-2*i-1
W += w
W %= mod
for j in range(m // 2):
h = y[m-1-j] - y[j]
if j == m//2 - 1:
if m % 2 == 1:
h *= 2
else:
h *= m-1-2*j
H += h
H %= mod
print(H, W)
S = H*W
print(S % mod)
|
s573102080
|
Accepted
| 123 | 24,356 | 840 |
n, m = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
mod = 10**9 + 7
# H = x[-1] - x[0]
# W = y[-1] - y[0]
# S = H*W % mod
# S *= (n-1)*(m-1)
# if n >= 4:
# h = x[-2] - x[1]
# S -= h*W * 2
# if m >= 4:
# w = y[-2] - y[1]
# S -= H*w * 2
H = 0
W = 0
for i in range(n//2):
w = x[n-1-i] - x[i]
if i == n//2 - 1:
if n % 2 == 1:
w *= 2
else:
w *= n-2*i-1
W += w
W %= mod
for j in range(m // 2):
h = y[m-1-j] - y[j]
if j == m//2 - 1:
if m % 2 == 1:
h *= 2
else:
h *= m-1-2*j
H += h
H %= mod
S = H*W
print(S % mod)
|
s781038607
|
p02853
|
u597455618
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 270 |
We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.
|
x, y = list(map(int, input().split()))
ans = 0
if x == 1 and y == 1:
ans += 400000
elif x == 1:
ans += 300000
elif x == 2:
ans += 200000
elif x == 3:
ans += 100000
elif y == 1:
ans += 300000
elif y == 2:
ans += 200000
elif y == 3:
ans += 100000
print(ans)
|
s363862315
|
Accepted
| 17 | 3,064 | 266 |
x, y = list(map(int, input().split()))
ans = 0
if x == 1 and y == 1:
ans += 400000
if x == 1:
ans += 300000
elif x == 2:
ans += 200000
elif x == 3:
ans += 100000
if y == 1:
ans += 300000
elif y == 2:
ans += 200000
elif y == 3:
ans += 100000
print(ans)
|
s476891508
|
p03044
|
u047668580
| 2,000 | 1,048,576 |
Wrong Answer
| 1,556 | 63,068 | 676 |
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.
|
from collections import defaultdict
import queue
N = int(input())
d = defaultdict(lambda: defaultdict(int))
for i in range(N-1):
u, v, w = list(map(int, input().split()))
d[u-1][v-1] = w
d[v-1][u-1] = w
for i in range(N):
if len(d[i]) == 1:
start = i
break
visited = [False for _ in range(N)]
visited[start] = True
all_length = [0 for _ in range(N)]
q = queue.Queue()
q.put(start)
while not q.empty():
i = q.get()
for v, w in d[i].items():
if not visited[v]:
q.put(v)
visited[v] = True
all_length[v] = all_length[i] + w
print(all_length)
for length in all_length:
print(length % 2)
|
s703266254
|
Accepted
| 1,657 | 59,876 | 658 |
from collections import defaultdict
import queue
N = int(input())
d = defaultdict(lambda: defaultdict(int))
for i in range(N-1):
u, v, w = list(map(int, input().split()))
d[u-1][v-1] = w
d[v-1][u-1] = w
for i in range(N):
if len(d[i]) == 1:
start = i
break
visited = [False for _ in range(N)]
visited[start] = True
all_length = [0 for _ in range(N)]
q = queue.Queue()
q.put(start)
while not q.empty():
i = q.get()
for v, w in d[i].items():
if not visited[v]:
q.put(v)
visited[v] = True
all_length[v] = all_length[i] + w
for length in all_length:
print(length % 2)
|
s710121181
|
p03351
|
u639592190
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 103 |
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d=map(int,input().split())
print("YES" if abs(a-c)<=d or (abs(a-b)<=d and abs(b-c)<=d) else "NO")
|
s288701799
|
Accepted
| 17 | 2,940 | 103 |
a,b,c,d=map(int,input().split())
print("Yes" if abs(a-c)<=d or (abs(a-b)<=d and abs(b-c)<=d) else "No")
|
s073792598
|
p03352
|
u814986259
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 64 |
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.
|
import math
X=int(input())
ans=(math.ceil(X**0.5)**2)
print(ans)
|
s583558095
|
Accepted
| 18 | 3,060 | 208 |
X = int(input())
ans = 1
for j in range(10, 1, -1):
if int(pow(X, (1/j)) + 1)**j <= X:
ans = max(int(pow(X, (1/j))+1)**j, ans)
else:
ans = max(int(pow(X, (1/j)))**j, ans)
print(ans)
|
s525190511
|
p04029
|
u690037900
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 31 |
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)
|
s092665911
|
Accepted
| 17 | 2,940 | 36 |
N=int(input())
print(int(N*(N+1)/2))
|
s344234770
|
p03400
|
u023229441
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 114 |
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
n=int(input())
d,x=map(int,input().split())
ans=x
for i in range(n):
k=int(input())
ans+=(n-1)//k+1
print(ans)
|
s407063783
|
Accepted
| 18 | 2,940 | 118 |
n=int(input())
d,x=map(int,input().split())
ans=x
for i in range(n):
k=int(input())
ans+=(d-1)//k+1
print(ans)
|
s114284469
|
p03385
|
u067267880
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
S = input()
if "a" in S and "b" in S and "c" in S:
print("YES")
else:
print("NO")
|
s518342080
|
Accepted
| 17 | 2,940 | 90 |
S = input()
if "a" in S and "b" in S and "c" in S:
print("Yes")
else:
print("No")
|
s389072056
|
p02614
|
u306060982
| 1,000 | 1,048,576 |
Wrong Answer
| 46 | 9,216 | 618 |
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
h,w,k = map(int,input().split())
c = []
for _ in range(h):
cc = input()
c.append(cc)
d = 0
kk = 0
g= []
for i in range(2 ** h):
gg=[]
for j in range(h):
if ((i>>j) &1):
gg.append(j)
g.append(gg)
import copy
gb = copy.deepcopy(g)
for lis in g:
ii = lis
for lisj in gb:
jj = lisj
for i in range(h):
if i in ii:
continue
for j in range(w):
if j in jj:
continue
if c[i][j]=='#':
kk+=1
if kk==k:
d+=1
kk = 0
print(d)
|
s712693308
|
Accepted
| 45 | 9,056 | 715 |
h,w,k = map(int,input().split())
c = []
for _ in range(h):
cc = input()
c.append(cc)
d = 0
kk = 0
g= []
gb=[]
for i in range(2 ** h):
gg=[]
for j in range(h):
if ((i>>j) &1):
gg.append(j)
g.append(gg)
for i in range(2 ** w):
gg=[]
for j in range(w):
if ((i>>j) &1):
gg.append(j)
gb.append(gg)
for lis in g:
ii = lis
for lisj in gb:
jj = lisj
for i in range(h):
if i in ii:
continue
for j in range(w):
if j in jj:
continue
if c[i][j]=='#':
kk+=1
if kk==k:
d+=1
kk = 0
print(d)
|
s291995813
|
p03415
|
u181215519
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 93 |
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}. Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.
|
s1 = str( input() )
s2 = str( input() )
s3 = str( input() )
print( s1[0] + s1[1] + s1[2] )
|
s898570511
|
Accepted
| 17 | 2,940 | 93 |
s1 = str( input() )
s2 = str( input() )
s3 = str( input() )
print( s1[0] + s2[1] + s3[2] )
|
s597924222
|
p02608
|
u595375942
| 2,000 | 1,048,576 |
Wrong Answer
| 56 | 10,576 | 384 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
from collections import Counter
from itertools import combinations_with_replacement
N = int(input())
L = [i for i in range(1, 50)]
ans = []
for c in combinations_with_replacement(L, 3):
x = c[0]
y = c[1]
z = c[2]
tmp = x**2+y**2+z**2+x*y+y*z+z*x
ans.append(tmp)
C = Counter(ans)
for i in range(1, N+1):
if C[i]:
print(C[i]*3)
else:
print(0)
|
s615225467
|
Accepted
| 225 | 10,016 | 510 |
from collections import Counter
from itertools import combinations_with_replacement
N = int(input())
L = [i for i in range(1, 100)]
C = Counter()
for i in range(1, N+1):
C[i] = 0
for c in combinations_with_replacement(L, 3):
x = c[0]
y = c[1]
z = c[2]
tmp = x**2+y**2+z**2+x*y+y*z+z*x
if tmp <= N:
if len(set(c)) == 2:
C[tmp] += 3
elif len(set(c)) == 3:
C[tmp] += 6
else:
C[tmp] += 1
for i in range(1, N+1):
print(C[i])
|
s574759798
|
p02865
|
u189487046
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 26 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n=int(input())
print(n//2)
|
s552336271
|
Accepted
| 17 | 2,940 | 33 |
n = int(input())
print((n-1)//2)
|
s532832424
|
p01540
|
u509278866
| 5,000 | 199,680 |
Wrong Answer
| 50 | 9,052 | 2,095 |
太郎君はある広場にお宝を探しにやってきました。この広場にはたくさんのお宝が埋められていますが、太郎君は最新の機械を持っているので、どこにお宝が埋まっているかをすべて知っています。広場はとても広いので太郎君は領域を決めてお宝を探すことにしましたが、お宝はたくさんあるためどのお宝がその領域の中にあるかすぐにわかりません。そこで太郎君はその領域の中にあるお宝の数を数えることにしました。
|
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
class Ruiwa():
def __init__(self, a):
self.H = h = len(a)
self.W = w = len(a[0])
self.R = r = a
for i in range(h):
for j in range(1,w):
r[i][j] += r[i][j-1]
for i in range(1,h):
for j in range(w):
r[i][j] += r[i-1][j]
def search(self, x1, y1, x2, y2):
print(x1, y1, x2, y2)
if x1 > x2 or y1 > y2:
return 0
r = self.R
rr = r[y2][x2]
if x1 > 0 and y1 > 0:
return rr - r[y1-1][x2] - r[y2][x1-1] + r[y1-1][x1-1]
if x1 > 0:
rr -= r[y2][x1-1]
if y1 > 0:
rr -= r[y1-1][x2]
return rr
def main():
n,m = LI()
na = [LI() for _ in range(n)]
ma = [LI() for _ in range(m)]
xd = set()
yd = set()
for x,y in na:
xd.add(x)
yd.add(y)
for x1,y1,x2,y2 in ma:
xd.add(x1)
yd.add(y1)
xd.add(x2)
yd.add(y2)
xl = sorted(list(xd))
yl = sorted(list(yd))
xx = {}
yy = {}
for i in range(len(xl)):
xx[xl[i]] = i
for i in range(len(yl)):
yy[yl[i]] = i
a = [[0]*(len(yl)+1) for _ in range(len(xl)+1)]
for x,y in na:
a[xx[x]][yy[y]] += 1
rui = Ruiwa(a)
r = []
for x1,y1,x2,y2 in ma:
r.append(rui.search(yy[y1],xx[x1],yy[y2],xx[x2]))
return '\n'.join(map(str,r))
print(main())
|
s512326261
|
Accepted
| 13,370 | 959,732 | 2,102 |
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
class Ruiwa():
def __init__(self, a):
self.H = h = len(a)
self.W = w = len(a[0])
self.R = r = a
for i in range(h):
for j in range(1,w):
r[i][j] += r[i][j-1]
for i in range(1,h):
for j in range(w):
r[i][j] += r[i-1][j]
def search(self, x1, y1, x2, y2):
if x1 > x2 or y1 > y2:
return 0
r = self.R
rr = r[y2][x2]
if x1 > 0 and y1 > 0:
return rr - r[y1-1][x2] - r[y2][x1-1] + r[y1-1][x1-1]
if x1 > 0:
rr -= r[y2][x1-1]
if y1 > 0:
rr -= r[y1-1][x2]
return rr
def main():
n,m = LI()
na = [LI() for _ in range(n)]
xd = set()
yd = set()
for x,y in na:
xd.add(x)
yd.add(y)
xl = sorted(list(xd))
yl = sorted(list(yd))
xx = {}
yy = {}
for i in range(len(xl)):
xx[xl[i]] = i
for i in range(len(yl)):
yy[yl[i]] = i
a = [[0]*(len(yl)+1) for _ in range(len(xl)+1)]
for x,y in na:
a[xx[x]][yy[y]] += 1
rui = Ruiwa(a)
r = []
for _ in range(m):
x1,y1,x2,y2 = LI()
xx1 = bisect.bisect_left(xl, x1)
yy1 = bisect.bisect_left(yl, y1)
xx2 = bisect.bisect(xl, x2) - 1
yy2 = bisect.bisect(yl, y2) - 1
r.append(rui.search(yy1,xx1,yy2,xx2))
return '\n'.join(map(str,r))
print(main())
|
s102166329
|
p03659
|
u584459098
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 24,832 | 189 |
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
N = int(input())
a = list(map(int, input().split()))
diff = []
for i in range(N-1):
print(sum(a[:i+1]), sum(a[i+1:]))
diff.append(abs(sum(a[:i+1]) - sum(a[i+1:])))
print(min(diff))
|
s380237809
|
Accepted
| 197 | 29,440 | 282 |
N = int(input())
a = list(map(int, input().split()))
sunuke = []
diff = []
total = sum(a)
for i, x in enumerate(a[0:-1]):
if(i==0):
sunuke.append(x)
else:
sunuke.append(x + sunuke[i-1])
for i in sunuke:
diff.append(abs((i * 2) - total))
print(min(diff))
|
s992575710
|
p03434
|
u366886346
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 182 |
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
n=int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
alice=0
bob=0
for i in range(n):
if n%2==0:
alice+=a[i]
else:
bob+=a[i]
print(alice-bob)
|
s718425384
|
Accepted
| 17 | 3,060 | 182 |
n=int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
alice=0
bob=0
for i in range(n):
if i%2==0:
alice+=a[i]
else:
bob+=a[i]
print(alice-bob)
|
s754962804
|
p03545
|
u522205105
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,064 | 484 |
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
n = list(map(int,list(input())))
limit = 7
d = [ [] for i in range(len(n))]
op = [ [] for i in range(len(n))]
for i,val in enumerate(n):
if i == 0:
d[0].append(n[0])
op[0].append("")
else:
for j in d[i-1]:
d[i].append(j+val)
d[i].append(j-val)
for j in op[i-1]:
op[i].append(j + "+")
op[i].append(j + "-")
_op = op[len(n)-1][d[len(n)-1].index(limit)]
s = ""
for i,val in enumerate(n):
s += str(val)
if i == len(n)-1:
break
else:
s += _op[i]
print(s)
|
s089908089
|
Accepted
| 18 | 3,064 | 502 |
n = list(map(int,list(input())))
limit = 7
d = [ [] for i in range(len(n))]
op = [ [] for i in range(len(n))]
for i,val in enumerate(n):
if i == 0:
d[0].append(n[0])
op[0].append("")
else:
for j in d[i-1]:
d[i].append(j+val)
d[i].append(j-val)
for j in op[i-1]:
op[i].append(j + "+")
op[i].append(j + "-")
_op = op[len(n)-1][d[len(n)-1].index(limit)]
s = ""
for i,val in enumerate(n):
s += str(val)
if i == len(n)-1:
break
else:
s += _op[i]
s += "="+str(limit)
print(s)
|
s527497726
|
p03024
|
u143051858
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 8,916 | 90 |
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()
n = len(s)
if 8 <= 15-n+s.count('o'):
print('Yes')
else:
print('No')
|
s765199922
|
Accepted
| 30 | 8,996 | 90 |
s = input()
n = len(s)
if 8 <= 15-n+s.count('o'):
print('YES')
else:
print('NO')
|
s098770651
|
p03943
|
u863723142
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,164 | 170 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
str_line = input().split(" ")
num_line = [int(n) for n in str_line]
num_line.sort()
if (num_line[0] == num_line[1] + num_line[2]):
print("Yes")
else:
print("No")
|
s759699600
|
Accepted
| 25 | 9,164 | 187 |
str_line = input().split(" ")
num_line = [int(n) for n in str_line]
num_line.sort()
#print(num_line)
if (num_line[2] == num_line[1] + num_line[0]):
print("Yes")
else:
print("No")
|
s437532075
|
p03778
|
u945418216
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 108 |
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+w<b:
print(a+w-b)
elif b+w<a:
print(a-b-w)
else:
print(0)
|
s155090396
|
Accepted
| 17 | 2,940 | 118 |
w,a,b = map(int, input().split())
if a+w<b:
print(abs(a+w-b))
elif b+w<a:
print(abs(a-b-w))
else:
print(0)
|
s480001739
|
p04045
|
u941438707
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,160 | 126 |
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,*a=map(int,open(0).read().split())
for i in range(n,10**6):
if not set(str(i))&set(a):
print(i)
exit()
|
s348164474
|
Accepted
| 107 | 9,052 | 140 |
n,k,*a=map(int,open(0).read().split())
for i in range(n,10**6):
if all(str(j) not in str(i) for j in a):
print(i)
exit()
|
s707789025
|
p03377
|
u089032001
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 96 |
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 and A+B >= X):
print("Yes")
else:
print("No")
|
s228363150
|
Accepted
| 20 | 2,940 | 97 |
A, B, X = map(int, input().split())
if(A <= X and A+B >= X):
print("YES")
else:
print("NO")
|
s628080828
|
p02845
|
u814781830
| 2,000 | 1,048,576 |
Wrong Answer
| 162 | 28,252 | 317 |
N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green. The person numbered i says: * "In front of me, exactly A_i people are wearing hats with the same color as mine." Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats. Since the count can be enormous, compute it modulo 1000000007.
|
from collections import Counter
N = int(input())
A = list(map(int, input().split()))
print(Counter(A))
MOD = 10 ** 9 + 7
#A.reverse()
ans = 1
cnt = [0] * N
for i, a in enumerate(A):
cnt[a] += 1
if a == 0:
ans *= 4 - cnt[a]
else:
ans *= max(0, cnt[a-1]-cnt[a]+1)
ans %= MOD
print(ans)
|
s829544378
|
Accepted
| 118 | 14,780 | 286 |
from collections import Counter
N = int(input())
A = list(map(int, input().split()))
MOD = 10 ** 9 + 7
ans = 1
cnt = [0] * N
for i, a in enumerate(A):
cnt[a] += 1
if a == 0:
ans *= 4 - cnt[a]
else:
ans *= max(0, cnt[a-1]-cnt[a]+1)
ans %= MOD
print(ans)
|
s941126310
|
p02613
|
u441246928
| 2,000 | 1,048,576 |
Wrong Answer
| 163 | 16,320 | 375 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
S = [input() for i in range(N)]
AC = 0
WA = 0
TLE = 0
RE = 0
for i in range(N) :
if S[i] == 'AC' :
AC += 1
elif S[i] == 'WA' :
WA += 1
elif S[i] == 'TLE' :
TLE += 1
elif S[i] == 'RE' :
RE += 1
print('AC' + 'x' + str(AC) )
print('WA' + 'x' + str(WA) )
print('TLE' + 'x' + str(TLE) )
print('RE' + 'x' + str(RE) )
|
s344621500
|
Accepted
| 160 | 16,320 | 359 |
N = int(input())
S = [input() for i in range(N)]
AC = 0
WA = 0
TLE = 0
RE = 0
for i in range(N) :
if S[i] == 'AC' :
AC += 1
elif S[i] == 'WA' :
WA += 1
elif S[i] == 'TLE' :
TLE += 1
elif S[i] == 'RE' :
RE += 1
print('AC','x',str(AC) )
print('WA','x',str(WA) )
print('TLE','x',str(TLE) )
print('RE','x',str(RE) )
|
s078038294
|
p03828
|
u075595666
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,064 | 605 |
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.
|
def is_prime(q):
if q < 2 or q&1 == 0: return False
return pow(5, q-1, q)*pow(7, q-1, q) == 1
def main():
mod = 10**9+7
n = int(input())
ans = 1
if n == 1:
print(1)
exit()
b = [2]
for i in range(2,min(n,998)):
if i == 5 or i == 7:
b.append(i)
if i == 561:
continue
if is_prime(i):
b.append(i)
for i in b:
p = i
cnt = 1
chk = 1
for j in range(10):
chk += n//p
if p > n:
ans *= chk
ans %= mod
chk = 1
break
p *= i
print(ans)
if __name__ == '__main__':
main()
|
s353106130
|
Accepted
| 19 | 3,064 | 611 |
def is_prime(q):
if q < 2 or q&1 == 0: return False
return pow(5, q-1, q)*pow(7, q-1, q) == 1
def main():
mod = 10**9+7
n = int(input())
ans = 1
if n == 1:
print(1)
exit()
b = [2]
for i in range(2,min(n+1,998)):
if i == 5 or i == 7:
b.append(i)
if i == 561:
continue
if is_prime(i):
b.append(i)
for i in b:
p = i
cnt = 1
chk = 1
for j in range(10):
chk += n//p
if p > n:
ans *= chk
ans %= mod
chk = 1
break
p *= i
print(ans)
if __name__ == '__main__':
main()
|
s743108798
|
p03549
|
u521271655
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 77 |
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
|
n,m = map(int,input().split())
x = 1900*m + 100*(n-m)
p = 1/2**m
print(x//p)
|
s123018365
|
Accepted
| 18 | 2,940 | 82 |
n,m = map(int,input().split())
x = 1900*m + 100*(n-m)
p = 1/2**m
print(int(x//p))
|
s054483549
|
p03563
|
u565204025
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 640 |
Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it.
|
# -*- coding: utf-8 -*-
s = list(input())
t = list(input())
slen = len(s)
tlen = len(t)
cand = []
for i in range(slen - tlen + 1):
match = True
for j in range(tlen):
if s[i + j] == "?" or s[i + j] == t[j]:
pass
else:
match = False
if match:
cand.append(i)
if len(cand) != 0:
cand.reverse()
ans = ""
for i in range(slen):
if not (cand[0] <= i < cand[0] + tlen):
if s[i] == "?":
ans += "a"
else:
ans += s[i]
else:
ans += t[i - cand[0]]
print(ans)
else:
print("UNRESTORABLE")
|
s549127055
|
Accepted
| 18 | 2,940 | 77 |
# -*- coding: utf-8 -*-
r = int(input())
g = int(input())
print(2 * g - r)
|
s551766203
|
p04045
|
u396211450
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,004 | 262 |
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.
|
import math
n,k=map(int,input().split())
l=set(map(int,input().split()))
s=n
p=set()
for i in range(10):
if i not in l:
p.add(i)
if k==1 and 0 in l:
print(n)
else:
d=math.floor(math.log(n, 10)+1)
for i in p:
n=n+i*(10**d)
d=d+1
print(n)
|
s691850913
|
Accepted
| 51 | 9,068 | 327 |
def check(n,s):
v=str(n)
for i in v:
if i in s:
return False
return True
def solve():
n,k=map(int,input().split())
s=input().split()
s=set(s)
while True:
if check(n,s):
print(n)
break
else:
n+=1
for _ in range(1):
solve()
|
s980541709
|
p03712
|
u234631479
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 138 |
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
h, w = map(int, input().split())
S = [input() for i in range(h)]
print(S)
print("#"*(w+2))
for s in S:
print("#"+s+"#")
print("#"*(w+2))
|
s865859693
|
Accepted
| 17 | 3,060 | 129 |
h, w = map(int, input().split())
S = [input() for i in range(h)]
print("#"*(w+2))
for s in S:
print("#"+s+"#")
print("#"*(w+2))
|
s622437641
|
p03160
|
u163501259
| 2,000 | 1,048,576 |
Wrong Answer
| 162 | 13,980 | 275 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
h = list(map(int, input().split()))
dp = [0]*N
dp[0] = 0
for i in range(2,N):
dp[i] = dp[i-1] + abs(h[i]-h[i-1])
if i > 1:
dp[i] = min(dp[i-2] + abs(h[i]-h[i-2]),dp[i-1] + abs(h[i]-h[i-1]))
print(dp[N-1])
|
s036914336
|
Accepted
| 174 | 13,980 | 275 |
N = int(input())
h = list(map(int, input().split()))
dp = [0]*N
dp[0] = 0
for i in range(1,N):
dp[i] = dp[i-1] + abs(h[i]-h[i-1])
if i > 1:
dp[i] = min(dp[i-2] + abs(h[i]-h[i-2]),dp[i-1] + abs(h[i]-h[i-1]))
print(dp[N-1])
|
s916618605
|
p03400
|
u806403461
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 144 |
Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp.
|
n = int(input())
d, x = map(int, input().split())
ans = x
for a in range(n):
A = int(input())
ans += int((d - 1)/A + 1)
print(ans+x)
|
s990545335
|
Accepted
| 18 | 2,940 | 145 |
n = int(input())
d, x = map(int, input().split())
ans = x
for a in range(0, n):
A = int(input())
ans += int((d - 1)/A) + 1
print(ans)
|
s626365161
|
p04031
|
u794521826
| 2,000 | 262,144 |
Wrong Answer
| 40 | 10,756 | 248 |
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.
|
import math
from statistics import mean
n = int(input())
a = list(map(int, input().split()))
m = mean(a)
if m - math.floor(m) < 0.5:
m = math.floor(m)
else:
m = math.floor(m) + 1
diff = 0
for v in a:
diff = (m - v) ** 2
print(diff)
|
s805587829
|
Accepted
| 40 | 10,740 | 249 |
import math
from statistics import mean
n = int(input())
a = list(map(int, input().split()))
m = mean(a)
if m - math.floor(m) < 0.5:
m = math.floor(m)
else:
m = math.floor(m) + 1
diff = 0
for v in a:
diff += (m - v) ** 2
print(diff)
|
s518200550
|
p02270
|
u024715419
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,700 | 402 |
In the first line, two integers $n$ and $k$ are given separated by a space character. In the following $n$ lines, $w_i$ are given respectively.
|
n,k = map(int,input().split())
w = [int(input()) for i in range(n)]
p_min = max(w)-1
p_max = sum(w)+1
while p_max - p_min > 1:
p_mid = int(p_min + (p_max - p_min)/2)
k_tmp = 1
w_tmp = 0
for w_i in w:
w_tmp += w_i
if w_tmp > p_mid:
w_tmp = w_i
k_tmp += 1
if k_tmp <= k:
p_max = p_mid
else:
p_min = p_mid
print(p_mid)
|
s849966770
|
Accepted
| 400 | 11,588 | 548 |
n,k = map(int,input().split())
w = [int(input()) for i in range(n)]
def binarySearch(left,right):
mid = 0
while right > left:
mid = int(left + (right - left)//2)
if check(mid,w,k):
right = mid
else:
left = mid + 1
return left
def check(p,w,k):
k_tmp = 1
w_tmp = 0
for w_i in w:
w_tmp += w_i
if w_tmp > p:
w_tmp = w_i
k_tmp += 1
if k_tmp > k:
return False
return True
print(binarySearch(max(w),sum(w)))
|
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