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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s241227386
|
p03854
|
u761471989
| 2,000
| 262,144
|
Wrong Answer
| 20
| 5,172
| 890
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S = input()
f = 0
while(1):
print(S)
if len(S) <= 4:
break
if S[:5] == "dream":
if len(S) == 5:
f = 1
break
elif len(S) == 6:
break
if S[:7] == "dreamer":
if len(S) == 7:
f = 1
break
if S[7] != "a":
S = S[7:]
continue
else:
S = S[5:]
else:
S = S[5:]
elif S[:5] == "erase":
if len(S) == 5:
f = 1
break
if S[:6] == "eraser":
if len(S) == 6:
f = 1
break
if S[6] != "a":
S = S[7:]
continue
else:
S = S[:5]
else:
S = S[5:]
else:
break
if f == 0:
print("NO")
else:
print("YES")
|
s446785808
|
Accepted
| 75
| 3,188
| 890
|
S = input()
f = 0
while(1):
#print(S)
if len(S) <= 4:
break
if S[:5] == "dream":
if len(S) == 5:
f = 1
break
elif len(S) == 6:
break
if S[:7] == "dreamer":
if len(S) == 7:
f = 1
break
if S[7] != "a":
S = S[7:]
continue
else:
S = S[5:]
else:
S = S[5:]
elif S[:5] == "erase":
if len(S) == 5:
f = 1
break
if S[:6] == "eraser":
if len(S) == 6:
f = 1
break
if S[6] != "a":
S = S[6:]
continue
else:
S = S[:5]
else:
S = S[5:]
else:
break
if f == 0:
print("NO")
else:
print("YES")
|
s486078638
|
p03712
|
u593567568
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,180
| 284
|
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())
A = [list(input()) for _ in range(H)]
ANS = []
for i in range(H+1):
t = []
for j in range(W+1):
if i == 0 or j == 0 or i == H or j == W:
t.append("#")
else:
t.append(A[i-1][j-1])
ANS.append("".join(t))
print("\n".join(ANS))
|
s250529498
|
Accepted
| 28
| 8,960
| 289
|
H,W = map(int,input().split())
A = [list(input()) for _ in range(H)]
ANS = []
for i in range(H+2):
t = []
for j in range(W+2):
if i == 0 or j == 0 or i == H+1 or j == W+1:
t.append("#")
else:
t.append(A[i-1][j-1])
ANS.append("".join(t))
print("\n".join(ANS))
|
s813470426
|
p03827
|
u105124953
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 175
|
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
n = int(input())
s = input()
tmp_list = [0]
x = 0
for _ in s:
if _ == 'I':
x += 1
else:
x -= 1
tmp_list.append(x)
print(x)
print(max(tmp_list))
|
s599667538
|
Accepted
| 17
| 2,940
| 162
|
n = int(input())
s = input()
tmp_list = [0]
x = 0
for _ in s:
if _ == 'I':
x += 1
else:
x -= 1
tmp_list.append(x)
print(max(tmp_list))
|
s521847013
|
p03957
|
u582243208
| 1,000
| 262,144
|
Wrong Answer
| 22
| 3,064
| 155
|
This contest is `CODEFESTIVAL`, which can be shortened to the string `CF` by deleting some characters. Mr. Takahashi, full of curiosity, wondered if he could obtain `CF` from other strings in the same way. You are given a string s consisting of uppercase English letters. Determine whether the string `CF` can be obtained from the string s by deleting some characters.
|
s=input()
a=""
for i in range(len(s)):
if s[i]=='C':
a+="c"
if s[i]=='F':
a+="f"
if a=="cf":
print("YES")
else:
print("NO")
|
s420412027
|
Accepted
| 23
| 3,064
| 157
|
s=input()
a=""
for i in range(len(s)):
if s[i]=='C':
a+="c"
if s[i]=='F':
a+="f"
if "cf" in a:
print("Yes")
else:
print("No")
|
s988428383
|
p03545
|
u870518235
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 405
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
S = list(input())
ops = []
lst = []
for i in range(8):
ops.append(bin(i))
ops[i] = ops[i][2:].zfill(3)
print(ops[i])
for i in range(8):
res = S[0]
for j in range(3):
if ops[i][j] == "0":
res = res + "+" + S[j+1]
else:
res = res + "-" + S[j+1]
if eval(res) == 7:
res = res + "=7"
lst.append(res)
break
print(lst[0])
|
s601833921
|
Accepted
| 18
| 3,064
| 387
|
S = list(input())
ops = []
lst = []
for i in range(8):
ops.append(bin(i))
ops[i] = ops[i][2:].zfill(3)
for i in range(8):
res = S[0]
for j in range(3):
if ops[i][j] == "0":
res = res + "+" + S[j+1]
else:
res = res + "-" + S[j+1]
if eval(res) == 7:
res = res + "=7"
lst.append(res)
break
print(lst[0])
|
s209373607
|
p03693
|
u219369949
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 99
|
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 g * 10 + b % 4 == 0:
print('YES')
else:
print('NO')
|
s031219517
|
Accepted
| 17
| 2,940
| 101
|
r, g, b = map(int, input().split())
if (g * 10 + b) % 4 == 0:
print('YES')
else:
print('NO')
|
s105620214
|
p03644
|
u375616706
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,188
| 241
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
from math import log2
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
def ceil(n):
return int(-n//1)*(-1)
N = int(input())
ans = ceil(log2(N))
if ans > N:
ans -= 1
print(2**ans)
|
s985189273
|
Accepted
| 18
| 3,060
| 244
|
from math import log2
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
def ceil(n):
return int(-n//1)*(-1)
N = int(input())
ans = ceil(log2(N))
if 2**ans > N:
ans -= 1
print(2**ans)
|
s306079158
|
p03503
|
u379340657
| 2,000
| 262,144
|
Wrong Answer
| 141
| 3,064
| 527
|
Joisino is planning to open a shop in a shopping street. Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods. There are already N stores in the street, numbered 1 through N. You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2. Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}. Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.
|
n = int(input())
f = [list(map(int, input().split())) for _ in range(n)]
p = [list(map(int, input().split())) for _ in range(n)]
benefits = []
for t in range(1,1024):
bits = list(bin(t))
a = [0 for _ in range(n)]
jk = 0
while jk < 10:
b = bits.pop()
if b == "b":
break
elif b == "0":
jk += 1
continue
else:
for i in range(n):
if f[i][jk]:
a[i] += 1
s = 0
for i in range(n):
s += p[i][a[i]]
else:
benefits.append(s)
else:
print(max(benefits))
|
s928100932
|
Accepted
| 139
| 3,064
| 554
|
n = int(input())
f = [list(map(int, input().split())) for _ in range(n)]
p = [list(map(int, input().split())) for _ in range(n)]
benefits = -1000000000
for t in range(1,1024):
bits = list(bin(t))
a = [0 for _ in range(n)]
jk = 0
while jk < 10:
b = bits.pop()
if b == "b":
break
elif b == "0":
jk += 1
continue
else:
for i in range(n):
if f[i][jk]:
a[i] += 1
jk += 1
s = 0
for i in range(n):
s += p[i][a[i]]
else:
benefits = max(s,benefits)
else:
print(benefits)
|
s374883143
|
p03556
|
u618373524
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 35
|
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
n = int(input())
print(int(n**0.5))
|
s774532985
|
Accepted
| 19
| 3,060
| 49
|
n = int(input())
m = int(n**0.5)
print(int(m**2))
|
s673232453
|
p03470
|
u306060982
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 229
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
N = int(input())
l = 0
count = 0
dlist = []
for i in range(N):
dlist.append(int(input()))
for i in range(N):
m = max(dlist)
if(l != m):
count += 1
dlist[dlist.index(m)] = 0
l = m
print(count)
|
s786009456
|
Accepted
| 18
| 2,940
| 122
|
N = int(input())
dlist = []
for i in range(N):
dlist.append(int(input()))
dlistt = list(set(dlist))
print(len(dlistt))
|
s690858001
|
p03149
|
u334712262
| 2,000
| 1,048,576
|
Wrong Answer
| 362
| 7,996
| 1,275
|
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".
|
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub
sys.setrecursionlimit(10000)
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input().strip()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
@mt
def slv(N):
if 1 not in N:
return "No"
if 9 not in N:
return "No"
if 7 not in N:
return "No"
if 4 not in N:
return "No"
return 'Yes'
def main():
N = read_int_n()
print(slv(N))
if __name__ == '__main__':
main()
|
s875535667
|
Accepted
| 258
| 7,868
| 1,275
|
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub
sys.setrecursionlimit(10000)
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input().strip()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
@mt
def slv(N):
if 1 not in N:
return "NO"
if 9 not in N:
return "NO"
if 7 not in N:
return "NO"
if 4 not in N:
return "NO"
return 'YES'
def main():
N = read_int_n()
print(slv(N))
if __name__ == '__main__':
main()
|
s828270975
|
p03672
|
u316386814
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 134
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
s = input()
for n in range(len(s) // 2 - 1, -1, -1):
if s[:n] == s[n: 2 * n]:
ans = s[: 2 * n]
break
print(ans)
|
s436162812
|
Accepted
| 18
| 2,940
| 129
|
s = input()
for n in range(len(s) // 2 - 1, -1, -1):
if s[:n] == s[n: 2 * n]:
ans = 2 * n
break
print(ans)
|
s601084601
|
p02409
|
u554198876
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,708
| 414
|
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
N = int(input())
A = [[[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10]]
print(len(A))
for i in range(N):
b, f, r, v = [int(j) for j in input().split()]
A[b-1][f-1][r-1] += v
for a, b in enumerate(A):
for f in b:
print(' '.join([str(i) for i in f]))
else:
if a != len(A) - 1:
print('#' * 10)
|
s036267522
|
Accepted
| 40
| 7,756
| 406
|
N = int(input())
A = [[[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10], [[0] * 10, [0] * 10, [0] * 10]]
for i in range(N):
b, f, r, v = [int(j) for j in input().split()]
A[b-1][f-1][r-1] += v
for a, b in enumerate(A):
for f in b:
print(' ' + ' '.join([str(i) for i in f]))
else:
if a != len(A) - 1:
print('#' * 20)
|
s429810190
|
p03378
|
u125505541
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 171
|
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
n, m, x = input().split()
n = int(n)
m = int(m)
x = int(x)
c = [0 for _ in range(n+1)]
a = input().split()
for i in range(m):
c[int(a[i])]=1
min(sum(c[:x]),sum(c[x:]))
|
s091020967
|
Accepted
| 17
| 3,064
| 178
|
n, m, x = input().split()
n = int(n)
m = int(m)
x = int(x)
c = [0 for _ in range(n+1)]
a = input().split()
for i in range(m):
c[int(a[i])]=1
print(min(sum(c[:x]),sum(c[x:])))
|
s697302095
|
p04044
|
u058850663
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 173
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
# -*- coding: utf-8 -*-
n, l = map(int, input().split())
a = []
b = ''
for i in range(n):
a.append(input())
sorted(a)
for i in range(len(a)):
b += a[i]
print(b)
|
s758341386
|
Accepted
| 18
| 3,060
| 168
|
# -*- coding: utf-8 -*-
n, l = map(int, input().split())
a = []
b = ''
for i in range(n):
a.append(input())
a.sort()
for i in range(len(a)):
b += a[i]
print(b)
|
s052039123
|
p02609
|
u724687935
| 2,000
| 1,048,576
|
Wrong Answer
| 567
| 12,752
| 997
|
Let \mathrm{popcount}(n) be the number of `1`s in the binary representation of n. For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0. Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.) For example, when n=7, it becomes 0 after two operations, as follows: * \mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1. * \mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0. You are given an integer X with N digits in binary. For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted. Find f(X_1), f(X_2), \ldots, f(X_N).
|
from collections import Counter
import sys
sys.setrecursionlimit(10 ** 6)
def dfs(n):
if memo[n] >= 0:
rst = memo[n]
else:
tmp = n
one = 0
while tmp > 0:
one += tmp & 1
tmp >>= 1
m = n % one
rst = 1 if m == 0 else 1 + dfs(m)
memo[n] = rst
return rst
N = int(input())
X = list(map(int, list(input())))
one = Counter(X)[1]
if one > 1:
Sm = 0
mm = one - 1
tm = 1
for i in range(N - 1, -1, -1):
tm %= mm
if X[i] == 1:
Sm += tm
Sm %= mm
tm *= 2
Sp = 0
mp = one + 1
tp = 1
for i in range(N - 1, -1, -1):
tp %= mp
if X[i] == 1:
Sp += tp
Sp %= mp
tp *= 2
memo = [-1] * (2 * (10 ** 5) + 10)
memo[0] = 0
for i in range(N):
if X[i] == 0:
m = (Sp + pow(2, N - 1 - i, mp)) % mp
elif one > 1:
m = (Sm - pow(2, N - 1 - i, mm)) % mm
else:
print(0)
continue
print(dfs(m))
|
s069504706
|
Accepted
| 588
| 12,720
| 1,001
|
from collections import Counter
import sys
sys.setrecursionlimit(10 ** 6)
def dfs(n):
if memo[n] >= 0:
rst = memo[n]
else:
tmp = n
one = 0
while tmp > 0:
one += tmp & 1
tmp >>= 1
m = n % one
rst = 1 if m == 0 else 1 + dfs(m)
memo[n] = rst
return rst
N = int(input())
X = list(map(int, list(input())))
one = Counter(X)[1]
if one > 1:
Sm = 0
mm = one - 1
tm = 1
for i in range(N - 1, -1, -1):
tm %= mm
if X[i] == 1:
Sm += tm
Sm %= mm
tm *= 2
Sp = 0
mp = one + 1
tp = 1
for i in range(N - 1, -1, -1):
tp %= mp
if X[i] == 1:
Sp += tp
Sp %= mp
tp *= 2
memo = [-1] * (2 * (10 ** 5) + 10)
memo[0] = 0
for i in range(N):
if X[i] == 0:
m = (Sp + pow(2, N - 1 - i, mp)) % mp
elif one > 1:
m = (Sm - pow(2, N - 1 - i, mm)) % mm
else:
print(0)
continue
print(1 + dfs(m))
|
s558297469
|
p02866
|
u997322530
| 2,000
| 1,048,576
|
Wrong Answer
| 74
| 14,396
| 211
|
Given is an integer sequence D_1,...,D_N of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition: * For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is D_i.
|
n = int(input())
a = [int(s) for s in input().split(' ')]
counts = [0] * n
for i in a:
counts[i] += 1
ans = 1 if a[0] == 0 else 0
for u, v in zip(counts, counts[1:]):
ans = ans * u * v % 998244353
print(ans)
|
s697395531
|
Accepted
| 95
| 14,396
| 249
|
n = int(input())
a = [int(s) for s in input().split(' ')]
counts = [0] * n
for i in a:
counts[i] += 1
ans = 1 if a[0] == 0 and counts[0] == 1 else 0
for u, v in zip(counts, counts[1:]):
for _ in range(v):
ans = ans * u % 998244353
print(ans)
|
s453480579
|
p04043
|
u698197687
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 397
|
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.
|
def haiku(sentence_length_list):
length_five_cnt=0
length_seven_cnt=0
for s in sentence_length_list:
if s==5:
length_five_cnt=length_five_cnt+1
if s==7:
length_seven_cnt=length_seven_cnt+1
if length_five_cnt==2 and length_seven_cnt==1:
return 'YES'
else:
return 'NO'
if __name__=='__main__':
haiku(input().split())
|
s009494866
|
Accepted
| 17
| 2,940
| 438
|
def haiku(sentence_length_list):
length_five_cnt=0
length_seven_cnt=0
for s in sentence_length_list:
if s=='5':
length_five_cnt=length_five_cnt+1
if s=='7':
length_seven_cnt=length_seven_cnt+1
if length_five_cnt==2 and length_seven_cnt==1:
return 'YES'
else:
return 'NO'
if __name__=='__main__':
length_list = input().split()
print(haiku(length_list))
|
s090653576
|
p04045
|
u063073794
| 2,000
| 262,144
|
Wrong Answer
| 103
| 3,060
| 257
|
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 = map(int,input().split())
d = list(map(str,input().split()))
ans=0
for i in range (n,100000):
s=str(i)
pay=True
for j in range(len(s)):
if s[j] in d:
pay = False
break
if pay:
ans = i
break
print(ans)
|
s788240765
|
Accepted
| 83
| 3,060
| 371
|
n, k = map(int,input().split())
d = list(map(str,input().split()))
ans=0
for i in range (n,100000):
s= str(i)
pay=True
for j in range (len(s)):
if s[j] in d:
pay = False
break
if pay:
ans = i
break
print(ans)
|
s064590984
|
p03814
|
u527993431
| 2,000
| 262,144
|
Wrong Answer
| 39
| 3,516
| 179
|
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
S=input()
ans=0
tmp=0
end=0
for i in range(len(S)):
if S[i]=="A":
tmp=i
break
for i in range(1,len(S)):
if S[-i]=="Z":
end=len(S)-i
break
print(tmp,end)
print(end-tmp+1)
|
s583570747
|
Accepted
| 41
| 3,512
| 164
|
S=input()
ans=0
tmp=0
end=0
for i in range(len(S)):
if S[i]=="A":
tmp=i
break
for i in range(1,len(S)):
if S[-i]=="Z":
end=len(S)-i
break
print(end-tmp+1)
|
s120500898
|
p03067
|
u514299323
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 121
|
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<B:
print('YES')
elif B<C<A:
print('YES')
else:
print('No')
|
s201998737
|
Accepted
| 17
| 2,940
| 121
|
A, B, C = map(int,input().split(" "))
if A<C<B:
print('Yes')
elif B<C<A:
print('Yes')
else:
print('No')
|
s681157343
|
p03645
|
u072717685
| 2,000
| 262,144
|
Wrong Answer
| 375
| 15,572
| 395
|
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
def main():
n, m = input().split()
g1 = []
g2 = []
for _ in range(int(m)):
x, y = input().split()
if x == 1:
g1.append(y)
elif y == n:
g2.append(x)
sx = set(g1)
sy = set(g2)
sxy = sx.intersection(sy)
if len(sxy):
print('POSSIBLE')
else:
print('IMPOSSIBLE')
if __name__ == '__main__':
main()
|
s731895365
|
Accepted
| 379
| 26,540
| 397
|
def main():
n, m = input().split()
g1 = []
g2 = []
for _ in range(int(m)):
x, y = input().split()
if x == '1':
g1.append(y)
elif y == n:
g2.append(x)
sx = set(g1)
sy = set(g2)
sxy = sx.intersection(sy)
if len(sxy):
print('POSSIBLE')
else:
print('IMPOSSIBLE')
if __name__ == '__main__':
main()
|
s509299503
|
p03971
|
u761989513
| 2,000
| 262,144
|
Wrong Answer
| 104
| 4,588
| 259
|
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
n, a, b = map(int, input().split())
s = list(input())
p = 0
ab = 0
for i in s:
if i == "c":
print("No")
elif i == "a":
if p < a + b:
p += 1
print("Yes")
else:
if p < a + b and ab < b:
p += 1
ab += 1
print("Yes")
|
s739757509
|
Accepted
| 109
| 4,712
| 316
|
n, a, b = map(int, input().split())
s = list(input())
p = 0
ab = 0
for i in s:
if i == "c":
print("No")
elif i == "a":
if p < a + b:
p += 1
print("Yes")
else:
print("No")
else:
if p < a + b and ab < b:
p += 1
ab += 1
print("Yes")
else:
print("No")
|
s677588279
|
p03860
|
u582243208
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,064
| 38
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
a,b,c=input().split()
print("A"+b+"C")
|
s600208753
|
Accepted
| 23
| 3,064
| 41
|
a,b,c=input().split()
print("A"+b[0]+"C")
|
s403972085
|
p04045
|
u207241407
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 350
|
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 = map(int,input().split())
d = set(map(int,input().split()))
num = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
usable = num - d
ans = []
n_list = [int(x) for x in list(str(n))]
for i in n_list:
if i in usable:
ans.append(i)
else:
for j in usable:
if i < j:
ans.append(j)
for i in ans:
print(i, end='')
|
s952365356
|
Accepted
| 19
| 3,064
| 735
|
import itertools
N, K = map(int, input().split())
D = set(input().split())
n = [x for x in str(N)]
num = {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9"} - D
judge = "no"
ans = N
for i in n:
if i in D:
judge = "yes"
break
if judge == "yes":
if ans < int(sorted(list(num))[-1] * len(n)):
ans_list = sorted(list(itertools.product(num, repeat=len(n))))
for i in ans_list:
if int("".join(i)) > ans:
ans = int("".join(i))
break
else:
ans_list = sorted(list(itertools.product(num, repeat=len(n) + 1)))
for j in ans_list:
if int("".join(j)) > ans:
ans = int("".join(j))
break
print(ans)
|
s288997092
|
p03386
|
u410118019
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 150
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,k = map(int,input().split())
c = [0] * (2*k)
for i in range(k):
c[i] = a + i
for i in range(k):
c[2*k-1-i] = b - i
for i in set(c):
print(i)
|
s338893584
|
Accepted
| 17
| 3,060
| 157
|
a,b,k = map(int,input().split())
s1 = set(range(a,min(a+k,b+1)))
s2 = set(range(b,max(b-k,a-1),-1))
s0 = sorted(s1|s2)
[print(s0[i]) for i in range(len(s0))]
|
s880100041
|
p00007
|
u915343634
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,648
| 109
|
Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week. Write a program which computes the amount of the debt in n weeks.
|
from math import ceil
n = int(input())
x = 100
for i in range(n):
x = ceil(x * 1.05)
print(x * 1000)
|
s877836732
|
Accepted
| 20
| 7,696
| 113
|
from math import ceil
yen = 100000
for i in range(int(input())):
yen = ceil((yen*1.05)/1000)*1000
print(yen)
|
s800430474
|
p03693
|
u834301346
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,020
| 116
|
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 = list(map(int, input().split(' ')))
num = r*100 + g*10 + b
if num//4==0:
print('YES')
else:
print('NO')
|
s368894743
|
Accepted
| 25
| 9,068
| 115
|
r, g, b = list(map(int, input().split(' ')))
num = r*100 + g*10 + b
if num%4==0:
print('YES')
else:
print('NO')
|
s717532820
|
p03471
|
u706159977
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 481
|
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
n,y = map(int,input().split())
if n*1000>y:
print("-1 -1 -1")
else:
x=n*1000
n10 = 0
n5 = 0
while y-x>=9000:
x = x+9000
n10 = n10+1
if n<n10:
print("-1 -1 -1")
break
print(x)
while y-x>=4000:
x = x+4000
n5 = n5+1
if n10+n5>n:
print("-1 -1 -1")
break
print(x)
if y==x:
l=[n10,n5,n-n5-n10]
print(*l)
else:
print("-1 -1 -1")
|
s978085400
|
Accepted
| 553
| 3,188
| 360
|
n,y = map(int,input().split())
s=[-1,-1,-1]
if n*10000<y:
pass
elif n*10000==y:
s=[n,0,0]
else:
m=1
x_0=n*10000
while m<=n:
for i in range(m+1):
x= x_0-9000*(m-i)-5000*i
if x == y:
s=[n-m,i,m-i]
break
if x==y:
break
m=m+1
print(*s)
|
s772341647
|
p02578
|
u205042576
| 2,000
| 1,048,576
|
Wrong Answer
| 333
| 32,052
| 396
|
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
# coding: utf-8
N = input()
N = int(N)
An = list(map(int, input().split()))
cnt = 0
step = []
step_sum = 0
for i in range(0, N):
if i == 0:
step.append(0)
continue
diff = (An[i - 1] + step[i - 1]) - An[i]
print('i ', i, 'diff ', diff)
if diff <= 0:
step.append(0)
else:
step.append(diff)
for num in step:
step_sum += num
print(step_sum)
|
s411273609
|
Accepted
| 164
| 32,216
| 347
|
N = input()
N = int(N)
An = list(map(int, input().split()))
cnt = 0
step = []
step_sum = 0
for i in range(0, N):
if i == 0:
step.append(0)
continue
diff = (An[i - 1] + step[i - 1]) - An[i]
if diff <= 0:
step.append(0)
else:
step.append(diff)
for num in step:
step_sum += num
print(step_sum)
|
s478892837
|
p03544
|
u021916304
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 87
|
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
n = int(input())
l = [2,1]
for i in range(n):
l.append(l[-1]+l[-2])
print(l[n-1])
|
s817287286
|
Accepted
| 17
| 2,940
| 85
|
n = int(input())
l = [2,1]
for i in range(n):
l.append(l[-1]+l[-2])
print(l[n])
|
s806980486
|
p02613
|
u656801456
| 2,000
| 1,048,576
|
Wrong Answer
| 152
| 16,288
| 476
|
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.
|
import math
import sys
def main():
n = int(input())
s = [input() for i in range(n)]
c0,c1,c2,c3 = 0,0,0,0
for i in range(n):
if "AC" in s[i]:
c0 += 1
if s[i] in 'WA':
c1 += 1
if s[i] in 'TLE':
c2 += 1
if s[i] in 'RE':
c3 += 1
print("AC x "+str(c0))
print("TLE x "+str(c1))
print("TLE x "+str(c2))
print("RE x "+str(c3))
if __name__ == "__main__":
main()
|
s694379848
|
Accepted
| 151
| 16,316
| 476
|
import math
import sys
def main():
n = int(input())
s = [input() for i in range(n)]
c0,c1,c2,c3 = 0,0,0,0
for i in range(n):
if "AC" in s[i]:
c0 += 1
if s[i] in 'WA':
c1 += 1
if s[i] in 'TLE':
c2 += 1
if s[i] in 'RE':
c3 += 1
print("AC x "+str(c0))
print("WA x "+str(c1))
print("TLE x "+str(c2))
print("RE x "+str(c3))
if __name__ == "__main__":
main()
|
s707136127
|
p02399
|
u641357568
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,604
| 78
|
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
i,j= map(int, input().split())
print("{0:d} {1:d} {0:.5f}".format(i//j,i%j))
|
s499529355
|
Accepted
| 20
| 5,600
| 82
|
i,j= map(int, input().split())
print("{0:d} {1:d} {2:.5f}".format(i//j,i%j,i/j))
|
s453149915
|
p03711
|
u243572357
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 110
|
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
x, y = map(int, input().split())
print('YES' if x==y or (x in [4, 6, 9, 11] and y in [4, 6, 9, 11]) else 'NO')
|
s605678607
|
Accepted
| 17
| 2,940
| 219
|
a, b = map(int, input().split())
lst = [4, 6, 9, 11]
lst2 = [1, 3, 5, 7, 8, 10, 12]
if a==b==2:
print('Yes')
elif a in lst and b in lst:
print('Yes')
elif a in lst2 and b in lst2:
print('Yes')
else:
print('No')
|
s601675759
|
p02413
|
u130834228
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,600
| 318
|
Your task is to perform a simple table calculation. Write a program which reads the number of rows r, columns c and a table of r × c elements, and prints a new table, which includes the total sum for each row and column.
|
r, c = map(int, input().split())
sheet = [[0 for i in range (r)] for j in range(c)]
for i in range(r):
sheet[i] = input().split()
k = 0
for j in range(c):
k += int(sheet[i][j])
sheet[i].append(k)
for i in range(r):
for j in range(c+1):
if j != c:
print(sheet[i][j], end=' ')
else:
print(sheet[i][j])
|
s974780493
|
Accepted
| 40
| 8,628
| 476
|
r, c = map(int, input().split())
sheet = [[0 for i in range (c)] for j in range(r)]
for i in range(r):
sheet[i] = input().split()
k = 0
for j in range(c):
k += int(sheet[i][j])
sheet[i].append(k)
l=[0 for i in range (c+1)]
for j in range(c+1):
for i in range(r):
l[j] += int(sheet[i][j])
#print(sheet[i][j])
sheet.append(l)
for i in range(r+1):
for j in range(c+1):
if j != c:
print(sheet[i][j], end=' ')
else:
print(sheet[i][j], end='')
print('')
|
s615831866
|
p00020
|
u548155360
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,540
| 72
|
Write a program which replace all the lower-case letters of a given text with the corresponding captital letters.
|
# coding=utf-8
sentence = input()
sentence.capitalize()
print(sentence)
|
s238271441
|
Accepted
| 20
| 5,544
| 78
|
# coding=utf-8
sentence = input()
sentence = sentence.upper()
print(sentence)
|
s986431696
|
p02744
|
u424768586
| 2,000
| 1,048,576
|
Wrong Answer
| 123
| 18,284
| 416
|
In this problem, we only consider strings consisting of lowercase English letters. Strings s and t are said to be **isomorphic** when the following conditions are satisfied: * |s| = |t| holds. * For every pair i, j, one of the following holds: * s_i = s_j and t_i = t_j. * s_i \neq s_j and t_i \neq t_j. For example, `abcac` and `zyxzx` are isomorphic, while `abcac` and `ppppp` are not. A string s is said to be in **normal form** when the following condition is satisfied: * For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison. For example, `abcac` is in normal form, but `zyxzx` is not since it is isomorphic to `abcac`, which is lexicographically smaller than `zyxzx`. You are given an integer N. Print all strings of length N that are in normal form, in lexicographically ascending order.
|
# -*- coding: utf-8 -*-
N=int(input())
l0=['a']
l1=[]
li=[0]
li1=[]
alp=[chr(i) for i in range(97, 97+26)]
for i in range(1,N):
for j, l in enumerate(l0):
ind=0
while ind<=li[j]+1:
l1.append(l+alp[ind])
if ind==li[j]+1: li1.append(li[j]+1)
else: li1.append(li[j])
ind+=1
l0=l1.copy()
l1=[]
li=li1.copy()
li1=[]
print(l0)
|
s479094460
|
Accepted
| 165
| 15,308
| 432
|
# -*- coding: utf-8 -*-
N=int(input())
l0=['a']
l1=[]
li=[0]
li1=[]
alp=[chr(i) for i in range(97, 97+26)]
for i in range(1,N):
for j, l in enumerate(l0):
ind=0
while ind<=li[j]+1:
l1.append(l+alp[ind])
if ind==li[j]+1: li1.append(li[j]+1)
else: li1.append(li[j])
ind+=1
l0=l1.copy()
l1=[]
li=li1.copy()
li1=[]
for l in l0:
print(l)
|
s695132624
|
p02613
|
u102218630
| 2,000
| 1,048,576
|
Wrong Answer
| 144
| 9,200
| 305
|
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())
AC=0
WA=0
TLE=0
RE=0
for i in range(N):
result=input()
if result=="AC":
AC+=1
elif result == "WA":
WA+=1
elif result == "TLE":
TLE += 1
else:
RE+=1
print("AC x"+str(AC))
print("WA x"+str(WA))
print("TLE x"+str(TLE))
print("RE x"+str(RE))
|
s085988560
|
Accepted
| 147
| 9,200
| 310
|
N=int(input())
AC=0
WA=0
TLE=0
RE=0
for i in range(N):
result=input()
if result=="AC":
AC+=1
elif result == "WA":
WA+=1
elif result == "TLE":
TLE += 1
else:
RE+=1
print("AC x "+str(AC))
print("WA x "+str(WA))
print("TLE x "+str(TLE))
print("RE x "+str(RE))
|
s766199279
|
p03992
|
u970308980
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 34
|
This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.
|
s = input()
print(s[:4+1], s[4:])
|
s078369787
|
Accepted
| 17
| 2,940
| 32
|
s = input()
print(s[:4], s[4:])
|
s908960907
|
p03501
|
u737508101
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
N, A, B = map(int, input().split())
if A*N > B:
print(A*N)
if A*N <= B:
print(B)
|
s862590996
|
Accepted
| 17
| 2,940
| 85
|
N, A, B = map(int, input().split())
if A*N > B:
print(B)
if A*N <= B:
print(A*N)
|
s657469842
|
p03485
|
u179750651
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 92
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b=map(int,input().split())
if 0 == ((a+b)%2):
print((a+b)/2)
else:
print(((a+b)//2)+1)
|
s783550140
|
Accepted
| 17
| 2,940
| 44
|
print(-(-sum(map(int,input().split()))//2))
|
s005354678
|
p02927
|
u717626627
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 3,064
| 280
|
Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have?
|
a,b = map(int, input().split())
ans = 0
for i in range(a+1):
for j in range(b+1):
if len(str(j)) == 2 and i >= 2 and int(str(j)[1]) >= 2 and int(str(j)[0]) >= 2:
if i == int(str(j)[0]) * int(str(j)[1]):
print(str(i) +' '+ str(j))
ans += 1
print(ans)
|
s356029671
|
Accepted
| 32
| 3,064
| 246
|
a,b = map(int, input().split())
ans = 0
for i in range(a+1):
for j in range(b+1):
if len(str(j)) == 2 and i >= 2 and int(str(j)[1]) >= 2 and int(str(j)[0]) >= 2:
if i == int(str(j)[0]) * int(str(j)[1]):
ans += 1
print(ans)
|
s118241999
|
p03470
|
u460129720
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,160
| 63
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
N = list(map(int,input().split()))
S = set(N[1:])
print(len(S))
|
s259457436
|
Accepted
| 25
| 8,948
| 81
|
n = int(input())
mochi = set([int(input()) for i in range(n)])
print(len(mochi))
|
s576316802
|
p02747
|
u746849814
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 112
|
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.
|
s = input()
ans = 0
for i in range(1, len(s)):
if s[i-1] == 'h' and s[i] == 'i':
ans += 1
print(ans)
|
s178637565
|
Accepted
| 17
| 3,060
| 150
|
s = list(input())
if s[::2].count('h') == len(s[::2]) and s[1::2].count('i') == len(s[1::2]) and len(s)%2 == 0:
print('Yes')
else:
print('No')
|
s993929465
|
p03730
|
u763249708
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 128
|
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`.
|
a, b, c = map(int, input().split())
for i in range(a, a*b, a):
if i%b == c:
print("Yes")
exit()
print("No")
|
s627724779
|
Accepted
| 18
| 2,940
| 128
|
a, b, c = map(int, input().split())
for i in range(a, a*b, a):
if i%b == c:
print("YES")
exit()
print("NO")
|
s783413082
|
p03973
|
u794173881
| 2,000
| 262,144
|
Wrong Answer
| 243
| 7,848
| 345
|
N people are waiting in a single line in front of the Takahashi Store. The cash on hand of the i-th person from the front of the line is a positive integer A_i. Mr. Takahashi, the shop owner, has decided on the following scheme: He picks a product, sets a positive integer P indicating its price, and shows this product to customers in order, starting from the front of the line. This step is repeated as described below. At each step, when a product is shown to a customer, if price P is equal to or less than the cash held by that customer at the time, the customer buys the product and Mr. Takahashi ends the current step. That is, the cash held by the first customer in line with cash equal to or greater than P decreases by P, and the next step begins. Mr. Takahashi can set the value of positive integer P independently at each step. He would like to sell as many products as possible. However, if a customer were to end up with 0 cash on hand after a purchase, that person would not have the fare to go home. Customers not being able to go home would be a problem for Mr. Takahashi, so he does not want anyone to end up with 0 cash. Help out Mr. Takahashi by writing a program that determines the maximum number of products he can sell, when the initial cash in possession of each customer is given.
|
n = int(input())
a = [int(input()) for i in range(n)]
a = a[::-1]
ans = 0
num = 1
while a:
if a[-1] == num:
num += 1
elif a[-1] > num:
if a[-1] % num == 0:
a[-1] = a[-1] - (num + 1)
ans += 1
ans += a[-1] // num
else:
ans += a[-1] // num
del a[-1]
print(ans)
|
s154262823
|
Accepted
| 251
| 7,848
| 359
|
n = int(input())
a = [int(input()) for i in range(n)]
a = a[::-1]
ans = 0
num = 1
while a:
if a[-1] < num:
del a[-1]
elif a[-1] == num:
num += 1
elif a[-1] > num:
if a[-1] % num == 0:
ans += a[-1] // num - 1
a[-1] = 1
else:
ans += a[-1] // num
a[-1] = 1
print(ans)
|
s589014754
|
p03549
|
u001024152
| 2,000
| 262,144
|
Wrong Answer
| 32
| 3,700
| 330
|
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).
|
from math import factorial, ceil
def combination(n, r):
return factorial(n) // (factorial(n - r) * factorial(r))
n, m = map(int, input().split())
t = 100*(n-m) + m*1900
rs = [0.5**m * ( 1- 0.5**m)**i for i in range(int(1e4))]
ans = 0.0
for i,r in enumerate(rs):
ans += ((i+1)*t)*r
print(ceil(ans))
|
s611704896
|
Accepted
| 17
| 2,940
| 72
|
N,M = map(int, input().split())
p = 1/2
print((1900*M+100*(N-M))*2**M)
|
s584312783
|
p02742
|
u595893956
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 59
|
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:
|
n,m=map(int,input().split())
print(n*(m//2)+(n%2)*(n//2+1))
|
s732566375
|
Accepted
| 18
| 2,940
| 122
|
n,m=map(int,input().split())
if m==1 or n==1:
print(1)
elif m%2+n%2<2:
print(n*m//2)
else:
print(n*(m//2)+(n//2+1))
|
s816517596
|
p02957
|
u769633956
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 3,964
| 250
|
We have two distinct integers A and B. Print the integer K such that |A - K| = |B - K|. If such an integer does not exist, print `IMPOSSIBLE` instead.
|
import string
while True:
try:
a,b=map(int,input().split())
if a!=b:
if str((a+b)/2.0).isdecimal():
print(int((a+b)/2))
else:print("IMPOSSIBLE")
else:print(int(a+b))
except:break
|
s156762094
|
Accepted
| 32
| 3,956
| 238
|
import string
while True:
try:
a,b=map(int,input().split())
if a!=b:
if ((a+b)%2)==0.0:
print(int((a+b)/2))
else:print("IMPOSSIBLE")
else:print(int(a+b))
except:break
|
s479444569
|
p02401
|
u483716678
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,600
| 290
|
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while True:
a,op,b = input().split()
if(op == '+'):
print(int(a)+int(b))
elif(op == '-'):
print(int(a)-int(b))
elif(op == '*'):
print(int(a)*int(b))
elif(op == '/'):
print(int(a)/int(b))
elif(op == '?'):
break;
|
s880517745
|
Accepted
| 20
| 5,596
| 299
|
while True:
a,op,b = input().split()
if(op == '+'):
print(int(a)+int(b))
elif(op == '-'):
print(int(a)-int(b))
elif(op == '*'):
print((int(a)*int(b)))
elif(op == '/'):
print('%d'%(int(a)/int(b)))
elif(op == '?'):
break;
|
s928751744
|
p04012
|
u637175065
| 2,000
| 262,144
|
Wrong Answer
| 52
| 5,400
| 355
|
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def main():
s = input()
d = collections.defaultdict(int)
for c in s:
d[c] += 1
for c in d.values():
if c % 2 == 1:
return 'No'
return 'YES'
print(main())
|
s059216307
|
Accepted
| 49
| 5,404
| 355
|
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**20
mod = 10**9 + 7
def main():
s = input()
d = collections.defaultdict(int)
for c in s:
d[c] += 1
for c in d.values():
if c % 2 == 1:
return 'No'
return 'Yes'
print(main())
|
s416765790
|
p03711
|
u785066634
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 175
|
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
A=[1,3,5,7,8,10,12]
B=[4,6,9,11]
C=[2]
x,y=map(int,input().split())
if x in A and y in A:
print('YES')
elif x in B and y in B:
print('YES')
else:
print('NO')
|
s193960885
|
Accepted
| 17
| 2,940
| 175
|
A=[1,3,5,7,8,10,12]
B=[4,6,9,11]
C=[2]
x,y=map(int,input().split())
if x in A and y in A:
print('Yes')
elif x in B and y in B:
print('Yes')
else:
print('No')
|
s713664978
|
p04043
|
u580316619
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 89
|
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')
|
s531811700
|
Accepted
| 17
| 2,940
| 89
|
a = input('')
if a.count('5') == 2 and a.count('7') ==1:
print('YES')
else:
print('NO')
|
s496507051
|
p02795
|
u329785342
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 282
|
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.
|
input_list = []
for i in range(0, 3):
input_list.append(int(input()))
h = input_list[0]
w = input_list[1]
n = input_list[2]
temp_num = 0
if w <= h:
temp_num = h
else:
temp_num = w
answer = 0
if n % temp_num == 0:
answer = n / temp_num
else:
answer = (n // temp_num) + 1
|
s138381047
|
Accepted
| 17
| 3,064
| 301
|
input_list = []
for i in range(0, 3):
input_list.append(int(input()))
h = input_list[0]
w = input_list[1]
n = input_list[2]
temp_num = 0
if w <= h:
temp_num = h
else:
temp_num = w
answer = 0
if n % temp_num == 0:
answer = int(n / temp_num)
else:
answer = (n // temp_num) + 1
print(answer)
|
s770900523
|
p03693
|
u771167374
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 66
|
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?
|
s = int(''.join(input().split()))
print('Yes' if s%4==0 else 'No')
|
s679924608
|
Accepted
| 17
| 2,940
| 66
|
s = int(''.join(input().split()))
print('YES' if s%4==0 else 'NO')
|
s449335912
|
p04043
|
u652569315
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 105
|
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.
|
l=sorted(list(map(int,input().split())))
if l[0]==7 and l[1]==l[2]==5:
print('YES')
else:
print('NO')
|
s945231568
|
Accepted
| 17
| 2,940
| 105
|
l=sorted(list(map(int,input().split())))
if l[2]==7 and l[1]==l[0]==5:
print('YES')
else:
print('NO')
|
s314680188
|
p03050
|
u564902833
| 2,000
| 1,048,576
|
Wrong Answer
| 122
| 3,064
| 339
|
Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those.
|
N = int(input())
ans = sum(
N // d - 1
for d in range(1, int(N**0.5))
if N % d == 0 and (N // (N // d - 1)) == (N % (N // d - 1))
)
print(ans)
|
s349713149
|
Accepted
| 123
| 3,060
| 358
|
N = int(input())
ans = sum(
N // d - 1
for d in range(1, int(N**0.5) + 1)
if N % d == 0 and N // d > 1 and (N // (N // d - 1)) == (N % (N // d - 1))
)
print(ans)
|
s019288730
|
p02268
|
u852745524
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,676
| 367
|
You are given a sequence of _n_ integers S and a sequence of different _q_ integers T. Write a program which outputs C, the number of integers in T which are also in the set S.
|
n=int(input())
S=list(map(int,input().split()))
q=int(input())
T=list(map(int,input().split()))
def search(s,T):
left = 0 ; right = len(T)-1
while left < right:
mid = (left + right)//2
if T[mid]==s: return True
elif T[mid]>s: right = mid
else: left = mid+1
count = 0
for s in S:
if search(s,T): count+=1
print(count)
|
s412051378
|
Accepted
| 50
| 21,384
| 120
|
n=int(input())
s=map(int,input().split())
q=int(input())
t=map(int,input().split())
ans=len(set(s)&set(t))
print(ans)
|
s364597940
|
p04029
|
u894440853
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 39
|
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 + 1) * N / 2)
|
s089128805
|
Accepted
| 17
| 2,940
| 44
|
N = int(input())
print(int((N + 1) * N / 2))
|
s965549205
|
p02612
|
u376324032
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,144
| 72
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
a = N - 1000
while a >= 1000:
a = a - 1000
print(a)
|
s486770161
|
Accepted
| 27
| 9,160
| 140
|
N = int(input())
a = 0
if N >= 1000:
a = N - 1000
else:
a = N
while a >= 1000:
a = a - 1000
if a != 0:
a = 1000 - a
print(a)
|
s581418356
|
p03150
|
u865413330
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 188
|
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
S = input()
key = "keyence"
ans = "No"
for i in range(len(S)):
for j in range(i,len(S)):
if S[:i] + S[j:] == key:
ans = "Yes"
print(ans)
|
s646414989
|
Accepted
| 18
| 3,060
| 188
|
S = input()
key = "keyence"
ans = "NO"
for i in range(len(S)):
for j in range(i,len(S)):
if S[:i] + S[j:] == key:
ans = "YES"
print(ans)
|
s494595158
|
p03698
|
u344813796
| 2,000
| 262,144
|
Wrong Answer
| 29
| 8,964
| 79
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s=list(str(input()))
len(s)
print('Yes' if len(list(set(s)))==len(s) else 'No')
|
s401895158
|
Accepted
| 28
| 8,848
| 74
|
s=list(str(input()))
print('yes' if len(list(set(s)))==len(s) else 'no')
|
s061531884
|
p02612
|
u833436666
| 2,000
| 1,048,576
|
Wrong Answer
| 119
| 27,140
| 139
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
import sys
import math
import numpy as np
import functools
import operator
import collections
import itertools
N=int(input())
print(N%1000)
|
s297852942
|
Accepted
| 113
| 27,048
| 227
|
import sys
import math
import numpy as np
import functools
import operator
import collections
import itertools
N=int(input())
count=0
for i in range(N):
count+=1000
if count>=N:
print(count-N)
sys.exit()
|
s578639329
|
p03447
|
u278670845
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 70
|
You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping?
|
x = int(input())
a = int(input())
b = int(input())
print(x-a-b*(x//b))
|
s540066089
|
Accepted
| 18
| 2,940
| 74
|
x = int(input())
a = int(input())
b = int(input())
print(x-a-b*((x-a)//b))
|
s134737198
|
p03575
|
u893063840
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,064
| 951
|
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.
|
WHITE = 0
GRAY = 1
BLACK = 2
def dfs(s, n, m):
color = [WHITE] * n
stack = [s]
color[s] = GRAY
while stack:
u = stack[-1]
for v, flg in enumerate(m[u]):
if flg:
if color[v] == WHITE:
color[v] = GRAY
stack.append(v)
break
else:
stack.pop()
color[u] = BLACK
cnt = color.count(BLACK)
return cnt
def main():
n, m = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(m)]
g = [[0] * n for _ in range(n)]
for a, b in ab:
a -= 1
b -= 1
g[a][b] = 1
g[b][a] = 1
ans = 0
for a, b in ab:
a -= 1
b -= 1
cp = g.copy()
cp[a][b] = 0
cp[b][a] = 0
cnt = dfs(0, n, cp)
if cnt != n:
ans += 1
print(ans)
if __name__ == "__main__":
main()
|
s852167067
|
Accepted
| 121
| 3,572
| 981
|
from copy import deepcopy
WHITE = 0
GRAY = 1
BLACK = 2
def dfs(s, n, m):
color = [WHITE] * n
stack = [s]
color[s] = GRAY
while stack:
u = stack[-1]
for v, flg in enumerate(m[u]):
if flg:
if color[v] == WHITE:
color[v] = GRAY
stack.append(v)
break
else:
stack.pop()
color[u] = BLACK
cnt = color.count(BLACK)
return cnt
def main():
n, m = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(m)]
g = [[0] * n for _ in range(n)]
for a, b in ab:
a -= 1
b -= 1
g[a][b] = 1
g[b][a] = 1
ans = 0
for a, b in ab:
a -= 1
b -= 1
cp = deepcopy(g)
cp[a][b] = 0
cp[b][a] = 0
cnt = dfs(0, n, cp)
if cnt != n:
ans += 1
print(ans)
if __name__ == "__main__":
main()
|
s282244532
|
p03693
|
u559367141
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,148
| 119
|
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 = input().split()
is_four_multiple = g+b
if int(is_four_multiple) % 4 == 0:
print("Yes")
else:
print("No")
|
s410726043
|
Accepted
| 26
| 9,100
| 96
|
r,g,b = map(int,input().split())
if((10*g + b) % 4 == 0):
print('YES')
else:
print('NO')
|
s384990948
|
p00015
|
u957840591
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,612
| 302
|
A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow".
|
N=int(input())
A=[]
B=[]
for i in range(N):
a = int(input())
b = int(input())
A.append(a)
B.append(b)
for i in range(N):
if A[i] >= 10 ** 79 or B[i] >= 10 ** 79:
print("overflow")
elif A[i] + B[i]>= 10 ** 79:
print("overflow")
else:
print(A[i]+B[i])
|
s121335692
|
Accepted
| 20
| 7,532
| 316
|
N = int(input())
A = []
B = []
for i in range(N):
a = int(input())
b = int(input())
A.append(a)
B.append(b)
for i in range(N):
if A[i] >= 10 ** 80 or B[i] >= 10 ** 80:
print("overflow")
elif A[i] + B[i] >= 10 ** 80:
print("overflow")
else:
print(str(A[i] + B[i]))
|
s396144331
|
p02258
|
u311299757
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,564
| 293
|
You can obtain profits from foreign exchange margin transactions. For example, if you buy 1000 dollar at a rate of 100 yen per dollar, and sell them at a rate of 108 yen per dollar, you can obtain (108 - 100) × 1000 = 8000 yen. Write a program which reads values of a currency $R_t$ at a certain time $t$ ($t = 0, 1, 2, ... n-1$), and reports the maximum value of $R_j - R_i$ where $j > i$ .
|
v = []
v = str(input()).splitlines()
dif_max = -200000
dif_max = -200000
for i in range(len(v) ):
v[i] = int(v[i])
for i in range(len(v) - 1):
for j in range(i + 1, len(v), 1):
dif = v[j] - v[i]
dif_max = dif if dif_max < dif else dif_max
print("{}".format(dif_max))
|
s290884790
|
Accepted
| 500
| 7,664
| 243
|
cnt = int(input())
dif_max = -1000000000
v_min = 1000000000
for i in range(cnt):
buf = int(input())
if i > 0:
dif_max = dif_max if buf - v_min < dif_max else buf - v_min
v_min = buf if buf < v_min else v_min
print(dif_max)
|
s547213029
|
p03155
|
u102461423
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 91
|
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
|
N = int(input())
H = int(input())
W = int(input())
answer = (N-H+1) * (H-W+1)
print(answer)
|
s641028955
|
Accepted
| 17
| 2,940
| 91
|
N = int(input())
H = int(input())
W = int(input())
answer = (N-H+1) * (N-W+1)
print(answer)
|
s588555507
|
p03501
|
u971811058
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 98
|
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
n, a, b = map(int, input().split())
print(n, a, b)
if n*a < b :
print(n*a)
else :
print(b)
|
s023338858
|
Accepted
| 17
| 2,940
| 83
|
n, a, b = map(int, input().split())
if n*a < b :
print(n*a)
else :
print(b)
|
s177648780
|
p03827
|
u717001163
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 112
|
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
n=int(input())
s=input()
x=0
ans=0
for i in range(n):
x=x+1 if s[i]=='i' else x-1
ans=max(ans,x)
print(ans)
|
s755656993
|
Accepted
| 18
| 2,940
| 112
|
n=int(input())
s=input()
x=0
ans=0
for i in range(n):
x=x+1 if s[i]=='I' else x-1
ans=max(ans,x)
print(ans)
|
s047651993
|
p02854
|
u328364772
| 2,000
| 1,048,576
|
Wrong Answer
| 72
| 26,764
| 201
|
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters. Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut: * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan). * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen. Find the minimum amount of money needed before cutting the bar into two parts with the same length.
|
n = int(input())
A = list(map(int, input().split()))
l, r = sum(A[:n//2]), sum(A[n//2:])
if n % 2 == 0:
print(abs(l-r))
else:
m = A[n//2]
s, b = min(l, r-m), max(l, r-m)
print(b-s-m)
|
s487266563
|
Accepted
| 179
| 26,024
| 187
|
n = int(input())
A = list(map(int, input().split()))
s = sum(A)
cnt = 0
res = sum(A)
for a in A:
cnt += a
res = min(res, abs(s-cnt*2))
if res == 0:
break
print(res)
|
s640008206
|
p02259
|
u626266743
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,576
| 269
|
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode.
|
N = int(input())
A = list(map(int, input().split()))
count = 0
for i in range(N):
for j in range(i, N):
if(A[j] < A[j-1]):
temp = A[j-1]
A[j-1] = A[j]
A[j] = temp
count += 1
print(*A)
print(count)
|
s366365336
|
Accepted
| 20
| 7,716
| 285
|
N = int(input())
A = list(map(int, input().split()))
count = 0
flag = True
while flag:
flag = False
for i in range(N-1, 0, -1):
if(A[i] < A[i-1]):
A[i], A[i-1] = A[i-1], A[i]
flag = True
count += 1
print(*A)
print(count)
|
s148838478
|
p03044
|
u667469290
| 2,000
| 1,048,576
|
Wrong Answer
| 2,109
| 15,940
| 451
|
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.
|
# -*- coding: utf-8 -*-
import numpy as np
def solve():
N = int(input())
L = np.vstack((np.arange(N+1), np.zeros(N+1, dtype=int)))
for _ in range(N-1):
u, v, w = map(int, input().split())
u, v = sorted([u,v])
I = np.where(L == L[0, u])[1]
L[0, I] = L[0, v]
L[1, I] = L[1, I]^L[1, v]^(w%2)
res = '\n'.join(map(str,L[1][1:]))
return str(res)
if __name__ == '__main__':
print(solve())
|
s295747228
|
Accepted
| 596
| 46,124
| 616
|
# -*- coding: utf-8 -*-
from heapq import heappop, heappush
def solve():
N = int(input())
edges = [[] for _ in range(N+1)]
for _ in range(N-1):
u, v, w = map(int, input().split())
edges[u].append((v, w&1))
edges[v].append((u, w&1))
D = [-1]*(N+1)
hq = [(0,1)]
checked = set()
while hq:
c, u = heappop(hq)
checked.add(u)
D[u] = c
for v, w in edges[u]:
if not v in checked:
heappush(hq, (c^w, v))
res = '\n'.join(map(str, D[1:]))
return str(res)
if __name__ == '__main__':
print(solve())
|
s679538832
|
p03644
|
u361826811
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 272
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
import sys
import itertools
# import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
ans=1
while N>ans:
ans*=2
print(ans)
|
s290816652
|
Accepted
| 20
| 3,188
| 272
|
import sys
import itertools
# import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
ans=1
while N>=ans:
ans*=2
print(ans//2)
|
s734852446
|
p03457
|
u474423089
| 2,000
| 262,144
|
Wrong Answer
| 356
| 27,324
| 181
|
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())
TXY=[list(map(int,input().split(' '))) for i in range(N)]
t,x,y=0,0,0
for i,j,k in TXY:
if i-t!=abs(x-j)+abs(y+k):
print('No')
exit()
print('Yes')
|
s654443569
|
Accepted
| 390
| 27,324
| 210
|
N=int(input())
TXY=[list(map(int,input().split(' '))) for i in range(N)]
t,x,y=0,0,0
for i,j,k in TXY:
chk=(i-t-(abs(x-j)+abs(y+k)))
if chk<0 or chk%2!=0:
print('No')
exit()
print('Yes')
|
s567062983
|
p03359
|
u685244071
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 75
|
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a, b = map(int, input().split())
if a>= b:
print(a - 1)
else:
print(a)
|
s166320416
|
Accepted
| 17
| 2,940
| 76
|
a, b = map(int, input().split())
if a > b:
print(a - 1)
else:
print(a)
|
s863546711
|
p03338
|
u703890795
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 163
|
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()
c = 0
m = 0
for i in range(N):
S1 = S[:i]
S2 = S[i:]
c = 0
for s1 in S1:
if s1 in S2:
c += 1
m = max(c,m)
print(m)
|
s804166502
|
Accepted
| 22
| 3,316
| 259
|
N = int(input())
S = input()
import collections
c = 0
m = 0
for i in range(N):
S1 = S[:i]
S2 = S[i:]
C1 = collections.Counter(S1)
C2 = collections.Counter(S2)
c = 0
for c1 in C1.keys():
if c1 in C2.keys():
c += 1
m = max(m,c)
print(m)
|
s818502565
|
p02268
|
u354053070
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,648
| 409
|
You are given a sequence of _n_ integers S and a sequence of different _q_ integers T. Write a program which outputs C, the number of integers in T which are also in the set S.
|
n = int(input())
S = list(map(int, input().split()))
q = int(input())
T = list(map(int, input().split()))
count = 0
m, M = S[0], S[n - 1]
for x in T:
if x < m or M < x:
continue
l, r = 0, n - 1
while l < r:
m = (l + r) // 2
if S[m] == x:
count += 1
break
elif S[m] > x:
r = m - 1
else:
l = m + 1
print(count)
|
s857029679
|
Accepted
| 590
| 18,668
| 414
|
n = int(input())
S = list(map(int, input().split()))
q = int(input())
T = list(map(int, input().split()))
count = 0
Sm, SM = S[0], S[n - 1]
for x in T:
if x < Sm or SM < x:
continue
l, r = 0, n - 1
while l <= r:
m = (l + r) // 2
if S[m] == x:
count += 1
break
elif S[m] > x:
r = m - 1
else:
l = m + 1
print(count)
|
s638785813
|
p03695
|
u422272120
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,240
| 787
|
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
n = int(input())
a = list(map(int,input().split()))
d = {'gry':0,'brw':0,'grn':0,'miz':0,'blu':0,'yel':0,'dai':0,'red':0,'any':0}
for i in a:
if 1 <= i <= 399:
d['gry'] += 1
elif 400 <= i <= 799:
d['brw'] += 1
elif 800 <= i <= 1199:
d['grn'] += 1
elif 1200 <= i <= 1599:
d['miz'] += 1
elif 1600 <= i <= 1999:
d['blu'] += 1
elif 2000 <= i <= 2399:
d['yel'] += 1
elif 2400 <= i <= 2799:
d['dai'] += 1
elif 2800 <= i <= 3199:
d['red'] += 1
else: #any
d['any'] += 1
ans = 0
zero = 0
for k,v in d.items():
if k == 'any':
pass
else:
if v == 0:
zero += 1
else:
print (k)
ans += 1
print (ans,ans+min(zero,d['any']))
|
s732114775
|
Accepted
| 28
| 9,044
| 726
|
n = int(input())
a = list(map(int,input().split()))
d = {'gry':0,'brw':0,'grn':0,'miz':0,'blu':0,'yel':0,'dai':0,'red':0,'any':0}
for i in a:
if 1 <= i <= 399:
d['gry'] += 1
elif 400 <= i <= 799:
d['brw'] += 1
elif 800 <= i <= 1199:
d['grn'] += 1
elif 1200 <= i <= 1599:
d['miz'] += 1
elif 1600 <= i <= 1999:
d['blu'] += 1
elif 2000 <= i <= 2399:
d['yel'] += 1
elif 2400 <= i <= 2799:
d['dai'] += 1
elif 2800 <= i <= 3199:
d['red'] += 1
else: #any
d['any'] += 1
ans = 0
zero = 0
for k,v in d.items():
if k == 'any':
pass
else:
if v != 0:
ans += 1
print (max(1,ans),ans+d['any'])
|
s877722176
|
p02663
|
u442850625
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 9,304
| 352
|
In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?
|
import datetime
h1, m1, h2, m2, k = map(int, input().split())
one = datetime.timedelta(hours=h1, minutes=m1)
two = datetime.timedelta(hours=h2, minutes=m2)
study = datetime.timedelta(minutes=k)
diff_one_two_seconds = (two.total_seconds() - one.total_seconds()) / 60
study_seconds = study.total_seconds()/60
print(diff_one_two_seconds - study_seconds)
|
s251007652
|
Accepted
| 24
| 9,320
| 357
|
import datetime
h1, m1, h2, m2, k = map(int, input().split())
one = datetime.timedelta(hours=h1, minutes=m1)
two = datetime.timedelta(hours=h2, minutes=m2)
study = datetime.timedelta(minutes=k)
diff_one_two_seconds = (two.total_seconds() - one.total_seconds()) / 60
study_seconds = study.total_seconds()/60
print(int(diff_one_two_seconds - study_seconds))
|
s982927981
|
p03606
|
u366959492
| 2,000
| 262,144
|
Wrong Answer
| 20
| 2,940
| 110
|
Joisino is working as a receptionist at a theater. The theater has 100000 seats, numbered from 1 to 100000. According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive). How many people are sitting at the theater now?
|
N=int(input())
l=0
for i in range(N):
r,l=(int(x) for x in input().split())
l+=(r-l)
print(100000-l)
|
s950567620
|
Accepted
| 20
| 2,940
| 105
|
N=int(input())
m=0
for i in range(N):
l,r=(int(x) for x in input().split())
m+=(r-l+1)
print(m)
|
s729104298
|
p03992
|
u760961723
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 39
|
This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.
|
s = input()
print(s[:4] + " " + s[5:])
|
s331943727
|
Accepted
| 17
| 2,940
| 40
|
s = input()
print(s[:4] + " " + s[4:])
|
s902462338
|
p03047
|
u702582248
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 40
|
Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi. How many ways are there to choose K consecutive integers?
|
n,k=map(int, input().split())
print(n-k)
|
s180936530
|
Accepted
| 17
| 2,940
| 43
|
n,k=map(int, input().split())
print(n-k+1)
|
s776645004
|
p00767
|
u571918510
| 8,000
| 131,072
|
Wrong Answer
| 1,010
| 7,788
| 659
|
Let us consider rectangles whose height, _h_ , and width, _w_ , are both integers. We call such rectangles _integral rectangles_. In this problem, we consider only wide integral rectangles, i.e., those with _w_ > _h_. We define the following ordering of wide integral rectangles. Given two wide integral rectangles, 1. The one shorter in its diagonal line is smaller, and 2. If the two have diagonal lines with the same length, the one shorter in its height is smaller. Given a wide integral rectangle, find the smallest wide integral rectangle bigger than the given one.
|
import sys
import math
def is_square(n):
for i in range(n):
if i*i == n:
return True
return False
def f(h,w):
r = h**2 + w**2
for hh in range(h+1,w):
if is_square(r-hh**2):
return (hh, int(math.sqrt(r-hh**2)))
t1 = h**2+(w+1)**2
t2 = (h+1)**2+w**2
if r+1<t1 and r+1<t2:
return (1,r+1)
elif t1 < t2 or h+1==w:
return (h,w+1)
else:
return (h+1,w)
while True:
line = sys.stdin.readline()
line = line.split(" ")
h = int(line[0])
w = int(line[1])
if h==0 and w==0:
break
ret = f(h,w)
print(str(ret[0])+" "+str(ret[1]))
|
s216242180
|
Accepted
| 1,290
| 11,548
| 911
|
import sys
import math
def f(h,w):
r = h**2 + w**2
ary = []
for hi in range(1,150+1):
for wi in range(hi+1,150+1):
ss = hi**2+wi**2
if ss == r and hi>h:
ary.append({
"ss": ss
,"h": hi
,"w": wi
})
elif ss > r:
ary.append({
"ss": ss
,"h": hi
,"w": wi
})
mino = ary[0]
for i,o in enumerate(ary):
if mino["ss"]>o["ss"]:
mino = o
elif mino["ss"]==o["ss"] and mino["h"]>o["h"]:
mino = o
return (mino["h"],mino["w"])
while True:
line = sys.stdin.readline()
line = line.split(" ")
h = int(line[0])
w = int(line[1])
if h==0 and w==0:
break
ret = f(h,w)
print(str(ret[0])+" "+str(ret[1]))
|
s153634625
|
p03388
|
u547167033
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 245
|
10^{10^{10}} participants, including Takahashi, competed in two programming contests. In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th. The _score_ of a participant is the product of his/her ranks in the two contests. Process the following Q queries: * In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's.
|
q=int(input())
for _ in range(q):
a,b=map(int,input().split())
if a>b:
a,b=b,a
if a==b:
print(2*a-1)
elif a==b+1:
print(2*a-2)
else:
c=int((a*b)**0.5)
if c*(c+1)>=a*b:
print(2*c-2)
else:
print(2*c-1)
|
s185167188
|
Accepted
| 19
| 3,064
| 285
|
q=int(input())
for _ in range(q):
a,b=map(int,input().split())
if a>b:
a,b=b,a
if a==b:
print(2*a-2)
elif b==a+1:
print(2*a-2)
else:
c=int((a*b)**0.5)-1
while (c+1)*(c+1)<a*b:
c+=1
if c*(c+1)>=a*b:
print(2*c-2)
else:
print(2*c-1)
|
s007293514
|
p03067
|
u661576386
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 159
|
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 < B:
if A < C < B:
print("OK")
else:
print("NO")
else:
if B < C < A:
print("OK")
else:
print("NO")
|
s450513424
|
Accepted
| 18
| 2,940
| 162
|
A,B,C =map(int,input().split())
if A < B:
if A < C < B:
print("Yes")
else:
print("No")
else:
if B < C < A:
print("Yes")
else:
print("No")
|
s598616587
|
p03359
|
u440904221
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 94
|
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a,b = map(int,input().split())
count = 0
count = a-1
if(a < b):
count=count+1
print(count)
|
s829117170
|
Accepted
| 17
| 2,940
| 95
|
a,b = map(int,input().split())
count = 0
count = a-1
if(a <= b):
count=count+1
print(count)
|
s305768287
|
p03457
|
u558242240
| 2,000
| 262,144
|
Wrong Answer
| 247
| 3,064
| 421
|
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.
|
import sys
sys.setrecursionlimit(10**6)
input = sys.stdin.readline
n = int(input())
#n = 2
prev = [0,0,0]
for i in range(n):
txy = list(map(int, input().split()))
#txy = list(map(int, ["5 1 1", "100 1 1"][i].split()))
t = txy[0] - prev[0]
x = abs(txy[1] - prev[1])
y = abs(txy[2] - prev[2])
d = x + y
if not( (d - t) % 2 == 0):
print('NO')
exit()
prev = txy
print('YES')
|
s172294203
|
Accepted
| 248
| 3,064
| 429
|
import sys
sys.setrecursionlimit(10**6)
input = sys.stdin.readline
n = int(input())
#n = 2
prev = [0,0,0]
for i in range(n):
txy = list(map(int, input().split()))
#txy = list(map(int, ["3 1 2", "6 1 1"][i].split()))
t = txy[0] - prev[0]
x = abs(txy[1] - prev[1])
y = abs(txy[2] - prev[2])
d = x + y
if not(d <= t and (d - t) % 2 == 0):
print('No')
exit()
prev = txy
print('Yes')
|
s713276301
|
p04031
|
u598229387
| 2,000
| 262,144
|
Wrong Answer
| 25
| 3,064
| 218
|
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
n=int(input())
a=[int(i) for i in input().split()]
check=float('inf')
for i in range(-100,101):
check2=0
for j in a:
check2+=(i-j)**2
if check>check2:
check=check2
ans=i
print(ans)
|
s271446619
|
Accepted
| 24
| 3,060
| 194
|
n=int(input())
a=[int(i) for i in input().split()]
check=float('inf')
for i in range(-100,101):
ans=0
for j in a:
ans+=(i-j)**2
if check>ans:
check=ans
print(check)
|
s091920404
|
p03457
|
u962803758
| 2,000
| 262,144
|
Wrong Answer
| 1,308
| 38,584
| 574
|
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.
|
t = []
N = input()
for i in range(int(N)):
t.append(list(map(int, input().split())))
t.sort()
co, co_ = [0, 0, 0], [0, 0, 0]
flag = 1
for i in range(len(t)):
co_[0] = t[i][0] - co[0]
print(co,co_)
co_[1] = t[i][1] - co[1]
print(co,co_)
co_[2] = t[i][2] - co[2]
print(co,co_)
dis = abs(co_[1]) + abs(co_[2])
print(co,co_,dis)
if dis % 2 != (t[i][0]-co[0]) % 2:
flag = 0
break
elif dis > co_[0]:
flag = 0
break
co = t[i]
if flag == 0:
print('No')
else:
print('Yes')
|
s751825713
|
Accepted
| 464
| 27,312
| 502
|
t = []
N = input()
for i in range(int(N)):
t.append(list(map(int, input().split())))
t.sort()
co, co_ = [0, 0, 0], [0, 0, 0]
flag = 1
for i in range(len(t)):
co_[0] = t[i][0] - co[0]
co_[1] = t[i][1] - co[1]
co_[2] = t[i][2] - co[2]
dis = abs(co_[1]) + abs(co_[2])
if dis % 2 != (t[i][0]-co[0]) % 2:
flag = 0
break
elif dis > co_[0]:
flag = 0
break
co = t[i]
if flag == 0:
print('No')
else:
print('Yes')
|
s123555184
|
p03556
|
u633548583
| 2,000
| 262,144
|
Wrong Answer
| 56
| 2,940
| 91
|
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
n=int(input())
m=1
for i in range(100000):
if i**2<=n:
m=max(i,m)
print(m)
|
s804458395
|
Accepted
| 60
| 2,940
| 95
|
n=int(input())
m=1
for i in range(100000):
if i**2<=n:
m=max(i,m)
print(m**2)
|
s361726841
|
p03150
|
u903460784
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 3,064
| 412
|
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
s=input()
ans='keyence'
index=0
isCut=0
isKeyence=0
for si in s:
if isCut==1:
if si!=ans[index]:
isKeyence=0
break
elif isCut==-1:
if si==ans[index]:
isCut=1
else:
if si==ans[index]:
isKeyence=1
else:
isCut=-1
index+=1
if index>6:
break
if isKeyence:
print('Yes')
else:
print('NO')
|
s521754785
|
Accepted
| 17
| 3,064
| 1,202
|
s=input()
ans='keyence'
i=0 # index of s
index=0 # index of ans
isCut=0
canRestart=0
restart=0
lenS=len(s)
while i<lenS:
# print(i,index,end='=>')
if isCut==1:
if index>6: # 7=len(ans)
if canRestart:
canRestart=0
isCut=-1
i=restartI
index=restartIndex
else:
break
else:
if s[i]==ans[index]:
index+=1
else:
if canRestart:
canRestart=0
isCut=-1
i=restartI
index=restartIndex
else:
break
elif isCut==-1:
if s[i]==ans[index]:
restartIndex=index
restartI=i
canRestart=1
isCut=1
index+=1
else:
if s[i]==ans[index]:
index+=1
else:
isCut=-1
i+=1
# print(i,index,isCut)
if (index<=6)|(i<lenS)|(isCut==-1): # 7=len(ans)
print('NO')
else:
print('YES')
|
s658743080
|
p03494
|
u840958781
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 145
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n=int(input())
a=list(map(int,input().split()))
s=[]
cnt=0
for i in range(n):
while a[i]%2==0:
cnt+=1
s.append(cnt)
print(min(s))
|
s186708828
|
Accepted
| 18
| 3,060
| 223
|
n=int(input())
a=list(map(int,input().split()))
cnt=-1
while True:
cnt+=1
for i in range(n):
if a[i]%2==0:
a[i]//=2
else:
break
else:
continue
break
print(cnt)
|
s620591086
|
p03351
|
u374802266
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 106
|
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,x=map(int,input().split())
d=sorted([a,b,c])
if d[2]-d[0]<=x:
print('Yes')
else:
print('No')
|
s875187115
|
Accepted
| 18
| 2,940
| 101
|
a,b,c,x=map(int,input().split())
print('Yes' if abs(c-a)<=x or abs(b-a)<=x and abs(c-b)<=x else 'No')
|
s815781516
|
p03605
|
u992910889
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 78
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
N=list(input())
if N[0]==9 or N[1]==9:
print('Yes')
else:
print('No')
|
s536674424
|
Accepted
| 17
| 3,068
| 82
|
N=list(input())
if N[0]=='9' or N[1]=='9':
print('Yes')
else:
print('No')
|
s747017940
|
p00007
|
u424041287
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,672
| 87
|
Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week. Write a program which computes the amount of the debt in n weeks.
|
import math
a = 100000
for n in range(int(input())):
a = math.ceil(a*105/100)
print(a)
|
s292319759
|
Accepted
| 20
| 7,580
| 100
|
import math
a = 100000
for n in range(int(input())):
a = math.ceil(a*105/100000)*1000
print(int(a))
|
s312735291
|
p03713
|
u106297876
| 2,000
| 262,144
|
Wrong Answer
| 82
| 3,064
| 381
|
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 = 0
if H % 3 == 0 or W % 3 == 0:
ans = 0
if H % 2 == 0 or W % 2 == 0:
ans = H * W
if W % 2 == 0:
H, W = W, H
for i in range(W):
aaa = abs(((H//2) * (i + 1)) - (W - (i + 1)) * H)
ans = min(ans, aaa)
if H % 2 == 1 and W % 2 == 1:
ans = min((H // 2) * W - (H // 2 + 1) * (W // 2), (W // 2) * H - (W // 2 + 1) * (H // 2))
print(ans)
|
s502082813
|
Accepted
| 495
| 3,188
| 507
|
H, W = map(int,input().split( ))
ans = H * W
for i in range(H):
l_1 = [(i+1) * W, (H - i - 1) * (W//2), (H - i - 1) * (W-(W//2))]
l_2 = [(i+1) * W, ((H - i - 1)//2) * W, (H-i-1 -((H - i - 1)//2)) * W]
ans = min(ans, max(l_1) - min(l_1), max(l_2) - min(l_2))
H, W = W, H
for i in range(H):
l_1 = [(i+1) * W, (H - i - 1) * (W//2), (H - i - 1) * (W-(W//2))]
l_2 = [(i+1) * W, ((H - i - 1)//2) * W, (H-i-1 -((H - i - 1)//2)) * W]
ans = min(ans, max(l_1) - min(l_1), max(l_2) - min(l_2))
print(ans)
|
s514135406
|
p00028
|
u184989919
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,632
| 278
|
Your task is to write a program which reads a sequence of integers and prints mode values of the sequence. The mode value is the element which occurs most frequently.
|
import sys
def ModeValue():
mode=[0 for i in range(0,101)]
try:
for n in sys.stdin:
mode[int(n)]+=1
except EOFError:
a=1
maxN=max(mode)
for i in mode:
if mode[i]>maxN:
print(i)
ModeValue()
|
s716987253
|
Accepted
| 30
| 7,632
| 237
|
import sys
def ModeValue():
mode=[0 for i in range(0,101)]
for n in sys.stdin:
mode[int(n)]+=1
maxN=max(mode)
for i in range(0,101):
if mode[i]==maxN:
print(i)
ModeValue()
|
s085479317
|
p03719
|
u467307100
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 79
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a, b, c = map(int,input().split())
print("YES" if a <= c and c <= b else "NO" )
|
s966507338
|
Accepted
| 17
| 2,940
| 69
|
a, b, c = map(int,input().split())
print("Yes" if a<=c<=b else "No" )
|
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