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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s431452434
|
p03163
|
u269235541
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 108,220
| 274
|
There are N items, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i. Taro has decided to choose some of the N items and carry them home in a knapsack. The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W. Find the maximum possible sum of the values of items that Taro takes home.
|
N,W = map(int,input().split())
V = sorted([[int(x) for x in input().split()] for _ in range(N)])
dp = [0]*(W+1)
s = 0
for w,v in V:
s += w
for j in range(w-1,min(W,s)):
if dp[j-w] + v > dp[j]:
dp[j] = dp[j-w] + v
print(dp)
print(max(dp))
|
s198586929
|
Accepted
| 212
| 15,456
| 224
|
N, W = map(int, input().split())
import numpy as np
dp = np.zeros(W + 1, dtype='int64')
for i in range(1, N + 1):
w, v = map(int, input().split())
dp[w:W+1] = np.maximum(dp[w:W+1], dp[:W-w+1] + v)
print(dp[-1])
|
s565425487
|
p03501
|
u353855427
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 106
|
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.
|
NAB = list(map(int,input().split()))
if(NAB[0]*NAB[1]>=NAB[2]):
print(NAB[0]*NAB[1])
else:
print(NAB[2])
|
s161307769
|
Accepted
| 17
| 2,940
| 106
|
NAB = list(map(int,input().split()))
if(NAB[0]*NAB[1]<=NAB[2]):
print(NAB[0]*NAB[1])
else:
print(NAB[2])
|
s185447855
|
p03471
|
u912650255
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 269
|
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 = 1000
Y = 1234000
##N,Y = map(int,input().split())
ans = -1,-1,-1
for x in range(N+1):
for y in range(N+1-x):
z = N - x - y
if 10000*x + 5000*y + 1000*z == Y:
ans = x,y,z
break
else:
break
print(*ans)
|
s241789796
|
Accepted
| 774
| 3,060
| 241
|
N,Y = map(int,input().split())
ans = -1,-1,-1
for x in range(N+1):
for y in range(N+1-x):
z = N - x - y
if 10000*x + 5000*y + 1000*z == Y:
ans = x,y,z
break
else:
continue
print(*ans)
|
s606793469
|
p03455
|
u819135704
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 100
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
c = a * b
if c % 2 == 0:
print('even')
else:
print('odd')
|
s986327033
|
Accepted
| 17
| 2,940
| 100
|
a, b = map(int, input().split())
c = a * b
if c % 2 == 0:
print('Even')
else:
print('Odd')
|
s998032863
|
p03079
|
u894258749
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 114
|
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
inpl = lambda: list(map(int,input().split()))
A, B, C = inpl()
if A==B==C:
print('YES')
else:
print('NO')
|
s710331356
|
Accepted
| 17
| 2,940
| 114
|
inpl = lambda: list(map(int,input().split()))
A, B, C = inpl()
if A==B==C:
print('Yes')
else:
print('No')
|
s821145541
|
p02742
|
u313317027
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 102
|
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:
|
W, R = map(int,input().split())
seki = W*R
if seki%2 == 0:
print(seki/2)
else:
print((seki+1)/2)
|
s056943888
|
Accepted
| 17
| 3,060
| 168
|
import sys
W, R = map(int,input().split())
seki = W*R
if(W==1 or R==1):
print(1)
sys.exit()
if seki%2 == 0:
print(int(seki/2))
else:
print(int((seki+1)/2))
|
s284013504
|
p03730
|
u276115223
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 195
|
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())
canSelect = 'No'
for i in range(1, b + 1):
if (i * a) % b == c:
canSelect = 'Yes'
break
print(canSelect)
|
s760479521
|
Accepted
| 19
| 2,940
| 195
|
a, b, c = map(int, input().split())
canSelect = 'NO'
for i in range(1, b + 1):
if (i * a) % b == c:
canSelect = 'YES'
break
print(canSelect)
|
s967147431
|
p03597
|
u618373524
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 3
|
We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black?
|
3
4
|
s043838998
|
Accepted
| 17
| 2,940
| 47
|
N = int(input())
A = int(input())
print(N**2-A)
|
s911078563
|
p03129
|
u581040514
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 157
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
N, K = map(int, input().split())
if N == 1:
print('No')
elif N == 2:
print('No')
elif (int(N**2+N)/2)+1 <= K:
print('Yes')
else:
print('No')
|
s692904299
|
Accepted
| 17
| 2,940
| 91
|
N, K = map(int, input().split())
if (K-1)*2+1 <= N:
print('YES')
else:
print('NO')
|
s165490597
|
p02388
|
u656153606
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,512
| 75
|
Write a program which calculates the cube of a given integer x.
|
InputNunber = input('Input: ')
x = int(InputNunber) ** 3
print('Output=',x)
|
s596363016
|
Accepted
| 20
| 7,632
| 58
|
InputNunber = input('')
x = int(InputNunber) ** 3
print(x)
|
s566494853
|
p03738
|
u820351940
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 77
|
You are given two positive integers A and B. Compare the magnitudes of these numbers.
|
a = input()
b = input()
a > b and "GREATER" or (a == b and "EQUAL" or "LESS")
|
s093661703
|
Accepted
| 17
| 2,940
| 95
|
a = int(input())
b = int(input())
print(a > b and "GREATER" or (a == b and "EQUAL" or "LESS"))
|
s815263843
|
p03739
|
u882620594
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 112,768
| 301
|
You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through i-th term is not zero. * For every i (1≤i≤n-1), the sign of the sum of the terms from the 1-st through i-th term, is different from the sign of the sum of the terms from the 1-st through (i+1)-th term.
|
n=int(input())
a=list(map(int,input().split()))
sum=[0,]*n
sum[0]=a[0]
zoubun=0
cnt=0
for i in range(1,n):
sum[i]=sum[i-1]+a[i]
if(sum[i]*sum[i-1]>=0):
cnt+=abs(sum[i])+1
if sum[i]>=0:
sum[i]=-1
else:
sum[i]=1
print(sum)
print(str(cnt))
|
s525562311
|
Accepted
| 240
| 14,888
| 1,691
|
import copy
import sys
write = sys.stdout.write
n = int(input())
A = list(map(int,input().split())) # +, -, +, ...
B = copy.deepcopy(A) #-, +, -, ...
sumA = []
sumB = []
cntA = 0
cntB = 0
if A[0] == 0:
A[0] += 1
cntA += 1
B[0] -= 1
cntB += 1
elif A[0] > 0:
cntB += (B[0]+1)
B[0] = -1
else:
cntA += abs(A[0])+1
A[0] = 1
sumA.append(A[0])
sumB.append(B[0])
for i in range(1, n):
tempA = sumA[i-1] + A[i]
tempB = sumB[i-1] + B[i]
if i%2 == 1:
if tempA == 0:
#A[i] -= 1
cntA += 1
sumA.append(-1)
elif tempA > 0:
#A[i] -= abs(tempA) + 1
cntA += (abs(tempA) + 1)
sumA.append(-1)
else:
sumA.append(tempA)
if tempB == 0:
#B[i] += 1
cntB += 1
sumB.append(1)
elif tempB < 0:
cntB += (abs(tempB) + 1)
sumB.append(1)
else:
sumB.append(tempB)
else:
if tempA == 0:
cntA += 1
sumA.append(1)
elif tempA < 0:
cntA += (abs(tempA) + 1)
sumA.append(1)
else:
sumA.append(tempA)
if tempB == 0:
#B[i] -= 1
cntB += 1
sumB.append(-1)
elif tempB > 0:
cntB += (abs(tempB) + 1)
sumB.append(-1)
else:
sumB.append(tempB)
print(str(min(cntA, cntB)))
|
s390399732
|
p03359
|
u785205215
| 2,000
| 262,144
|
Wrong Answer
| 108
| 3,572
| 874
|
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
import math
import itertools
from heapq import heapify, heappop, heappush
from sys import stdin, stdout, setrecursionlimit
from bisect import bisect, bisect_left, bisect_right
from collections import defaultdict, deque
# inf = float("inf")
def LM(t, r): return list(map(t, r))
def R(): return stdin.readline()
def RS(): return R().split()
def I(): return int(R())
def F(): return float(R())
def LI(): return LM(int,RS())
def LF(): return LM(float,RS())
def ONE_SL(): return list(input())
def ONE_IL(): return LM(int, ONE_SL())
def ALL_IL(): return LM(int,stdin)
##### tools #####
def ap(f): return f.append
def pll(li): print('\n'.join(LM(str,li)))
def pljoin(li, s): print(s.join(li))
##### main #####
def main():
m,d = LI()
print(m-1 if m >= d else m)
if __name__ == '__main__':
main()
|
s539841071
|
Accepted
| 55
| 3,444
| 874
|
import math
import itertools
from heapq import heapify, heappop, heappush
from sys import stdin, stdout, setrecursionlimit
from bisect import bisect, bisect_left, bisect_right
from collections import defaultdict, deque
# inf = float("inf")
def LM(t, r): return list(map(t, r))
def R(): return stdin.readline()
def RS(): return R().split()
def I(): return int(R())
def F(): return float(R())
def LI(): return LM(int,RS())
def LF(): return LM(float,RS())
def ONE_SL(): return list(input())
def ONE_IL(): return LM(int, ONE_SL())
def ALL_IL(): return LM(int,stdin)
##### tools #####
def ap(f): return f.append
def pll(li): print('\n'.join(LM(str,li)))
def pljoin(li, s): print(s.join(li))
##### main #####
def main():
m,d = LI()
print(m if m <= d else m-1)
if __name__ == '__main__':
main()
|
s579538878
|
p03565
|
u521020719
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 367
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
sd = input()
t = input()
n = len(sd)
m = len(t)
s = []
for i in range(n - m, -1, -1):
t_kamo = sd[i:i + m]
for j in range(m + 1):
if j == m:
print((sd[:i] + t + sd[i + len(t):]).replace("?", "a"))
exit()
if t_kamo[j] == "?":
continue
elif t_kamo != t[j]:
break
print("UNRESTORABLE")
|
s627845164
|
Accepted
| 17
| 3,064
| 370
|
sd = input()
t = input()
n = len(sd)
m = len(t)
s = []
for i in range(n - m, -1, -1):
t_kamo = sd[i:i + m]
for j in range(m + 1):
if j == m:
print((sd[:i] + t + sd[i + len(t):]).replace("?", "a"))
exit()
if t_kamo[j] == "?":
continue
elif t_kamo[j] != t[j]:
break
print("UNRESTORABLE")
|
s002118204
|
p03543
|
u463775490
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 97
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
n=str(input())
if n[0]==n[1]==n[2] or n[1]==n[2]==n[-1:]:
print("YES")
else:
print("NO")
|
s319365192
|
Accepted
| 17
| 2,940
| 96
|
n=str(input())
if n[0]==n[1]==n[2] or n[1]==n[2]==n[-1:]:
print("Yes")
else:
print("No")
|
s941473970
|
p03455
|
u472899240
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 100
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
c = (a + b) % 2
if c == 1:
print("Odd")
else:
print("Even")
|
s365191973
|
Accepted
| 17
| 2,940
| 100
|
a, b = map(int, input().split())
c = (a * b) % 2
if c == 1:
print("Odd")
else:
print("Even")
|
s755566090
|
p02646
|
u933929042
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 9,112
| 162
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
if v-w != 0 and abs(a-b)/abs(v-w) > t:
print("Yes")
else:
print("No")
|
s428178142
|
Accepted
| 22
| 9,172
| 159
|
a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
if v-w > 0 and abs(a-b) <= t*(v-w):
print("YES")
else:
print("NO")
|
s365840646
|
p03360
|
u019584841
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 151
|
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?
|
a,b,c = map(int,input().split())
k = int(input())
if max(a,b,c) == a:
print(a**k+b+c)
elif max(a,b,c) == b:
print(b**k+a+c)
else:
print(a+b+c**k)
|
s447212037
|
Accepted
| 19
| 3,316
| 165
|
a,b,c = map(int,input().split())
k = int(input())
if max(a,b,c) == a:
print(a*(2**k)+b+c)
elif max(a,b,c) == b:
print(b*(2**k)+a+c)
else:
print(a+b+c*(2**k))
|
s970502816
|
p03044
|
u016128476
| 2,000
| 1,048,576
|
Wrong Answer
| 599
| 25,764
| 739
|
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
N = int(input())
U, V, W = [], [], []
T = [[] for _ in range(N)]
S = set()
for _ in range(N-1):
u, v, w = map(int, input().split())
if w % 2 == 1:
T[u-1].append(v)
T[v-1].append(u)
S.add(u)
S.add(v)
# U.append(u)
# V.append(v)
# W.append(w)
ans = [None for _ in range(N)]
while len(S) > 0:
s = S.pop()
search = set()
search.add(s)
ans[s-1] = '0'
while len(search) > 0:
p = search.pop()
if p in S:
S.remove(p)
for t in T[p-1]:
if t in S:
search.add(t)
ans[t-1] = '0' if ans[p-1] == '1' else '1'
for i in range(N):
if ans[i] is None:
ans[i] = '0'
print(' '.join(ans))
|
s048975915
|
Accepted
| 807
| 66,736
| 1,255
|
N = int(input())
U, V, W = [], [], []
T = [[] for _ in range(N)]
# tree
class Tree(object):
def __init__(self, us, vs, ws):
self.con_table = [[] for _ in range(len(us) + 1)]
self.weights = dict()
for (u, v, w) in zip(us, vs, ws):
self.con_table[u-1].append(v-1)
self.con_table[v-1].append(u-1)
self.weights[(u-1, v-1)] = w
self.init_search()
def init_search(self):
self.visited = [False for _ in range(len(self.con_table))]
dists = [None for _ in range(N)]
from collections import deque
def dfs(tree, s):
tree.init_search()
dists[s] = 0
while not all(tree.visited):
search = deque()
tree.visited[s] = True
search.append(s)
while len(search) > 0:
v = search.pop()
for u in tree.con_table[v]:
if not tree.visited[u]:
dists[u] = dists[v] + tree.weights[(min(u, v), max(u, v))]
tree.visited[u] = True
search.append(u)
for _ in range(N-1):
u, v, w = map(int, input().split())
U.append(u)
V.append(v)
W.append(w)
tree = Tree(U, V, W)
dfs(tree, 0)
print('\n'.join(('0' if d % 2 == 0 else '1') for d in dists))
|
s163869109
|
p03720
|
u437215432
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 194
|
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
n, m = map(int, input().split())
ab = [0] * m
for i in range(m):
ab[i] = list(map(int, input().split()))
roads = [0] * n
for a, b in ab:
roads[a-1] += 1
roads[b-1] += 1
print(roads)
|
s958103615
|
Accepted
| 18
| 2,940
| 210
|
n, m = map(int, input().split())
ab = [0] * m
for i in range(m):
ab[i] = list(map(int, input().split()))
roads = [0] * n
for a, b in ab:
roads[a-1] += 1
roads[b-1] += 1
for i in roads:
print(i)
|
s360527463
|
p03860
|
u221888299
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 31
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
s = input()
print('A'+s[0]+'C')
|
s377956351
|
Accepted
| 17
| 2,940
| 31
|
s = input()
print('A'+s[8]+'C')
|
s656053524
|
p03970
|
u333945892
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 110
|
CODE FESTIVAL 2016 is going to be held. For the occasion, Mr. Takahashi decided to make a signboard. He intended to write `CODEFESTIVAL2016` on it, but he mistakenly wrote a different string S. Fortunately, the string he wrote was the correct length. So Mr. Takahashi decided to perform an operation that replaces a certain character with another in the minimum number of iterations, changing the string to `CODEFESTIVAL2016`. Find the minimum number of iterations for the rewrite operation.
|
S1 = input()
S2 = "CODEFESTIVAL2016"
ans = 0
for i in range(16):
if S1[i] == S2[i]:
ans+=1
print(ans)
|
s697407903
|
Accepted
| 17
| 2,940
| 111
|
S1 = input()
S2 = "CODEFESTIVAL2016"
ans = 0
for i in range(16):
if S1[i] != S2[i]:
ans+=1
print(ans)
|
s764831024
|
p01225
|
u316584871
| 5,000
| 131,072
|
Wrong Answer
| 30
| 5,604
| 1,123
|
あなたの友達は最近 UT-Rummy というカードゲームを思いついた． このゲームで使うカードには赤・緑・青のいずれかの色と1から9までのいずれかの番号が つけられている． このゲームのプレイヤーはそれぞれ9枚の手札を持ち， 自分のターンに手札から1枚選んで捨てて， 代わりに山札から1枚引いてくるということを繰り返す． このように順番にターンを進めていき， 最初に手持ちのカードに3枚ずつ3つの「セット」を作ったプレイヤーが勝ちとなる． セットとは，同じ色の3枚のカードからなる組で，すべて同じ数を持っているか 連番をなしているもののことを言う． 連番に関しては，番号の巡回は認められない． 例えば，7, 8, 9は連番であるが 9, 1, 2は連番ではない． あなたの友達はこのゲームをコンピュータゲームとして売り出すという計画を立てて， その一環としてあなたに勝利条件の判定部分を作成して欲しいと頼んできた． あなたの仕事は，手札が勝利条件を満たしているかどうかを判定する プログラムを書くことである．
|
n = int(input())
def check(nl):
for i in nl:
if ((i+1 in nl) and (i+2 in nl)):
for k in range(i,i+3):
nl.remove(k)
elif(nl.count(i)%3 == 0):
for k in range(nl.count(i)):
nl.remove(i)
for i in range(n):
num_list = list(map(int, input().split()))
c_list = list(map(str, input().split()))
r = c_list.count('R')
g = c_list.count('G')
b = c_list.count('B')
if (r%3 == 0 and g % 3 == 0 and b % 3 == 0):
rlist = []
glist = []
blist = []
for k in range(9):
if(c_list[k] == 'R'):
rlist.append(num_list[k])
elif(c_list[k] == 'G'):
glist.append(num_list[k])
elif(c_list[k] == 'B'):
blist.append(num_list[k])
rlist.sort()
blist.sort()
glist.sort()
check(rlist)
check(glist)
check(blist)
if(rlist == [] and glist == [] and blist == []):
print(1)
else:
print(0)
else:
print(0)
|
s332321375
|
Accepted
| 20
| 5,604
| 1,246
|
n = int(input())
def check(nl):
fl = list(nl)
for w in range(len(nl)):
if (fl[w] in nl):
i = fl[w]
if ((i+1 in nl) and (i+2 in nl)):
for k in range(i,i+3):
nl.remove(k)
elif(nl.count(i)%3 == 0):
for k in range(nl.count(i)):
nl.remove(i)
for i in range(n):
num_list = list(map(int, input().split()))
c_list = list(map(str, input().split()))
r = c_list.count('R')
g = c_list.count('G')
b = c_list.count('B')
if (r%3 == 0 and g % 3 == 0 and b % 3 == 0):
rlist = []
glist = []
blist = []
for k in range(9):
if(c_list[k] == 'R'):
rlist.append(num_list[k])
elif(c_list[k] == 'G'):
glist.append(num_list[k])
elif(c_list[k] == 'B'):
blist.append(num_list[k])
rlist.sort()
glist.sort()
blist.sort()
check(rlist)
check(glist)
check(blist)
if(len(rlist) == 0 and len(glist) == 0 and len(blist) == 0):
print(1)
else:
print(0)
else:
print(0)
|
s904902987
|
p02841
|
u094948011
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 117
|
In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month.
|
M1,D1 = map(int, input().split())
M2,D2 = map(int, input().split())
if M1 == M2:
print('1')
else:
print('0')
|
s454187623
|
Accepted
| 17
| 2,940
| 117
|
M1,D1 = map(int, input().split())
M2,D2 = map(int, input().split())
if M1 == M2:
print('0')
else:
print('1')
|
s976755369
|
p03434
|
u247680229
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 270
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
N=int(input())
a=list(map(int, input().split()))
a.sort(reverse=True)
A=[]
B=[]
for i,k in enumerate(a):
if i%2 == 0:
A.append(a[i])
elif i%2 == 1:
B.append(a[i])
print(i)
print(sum(A)-sum(B))
|
s926681021
|
Accepted
| 17
| 3,060
| 260
|
N=int(input())
a=list(map(int, input().split()))
a.sort(reverse=True)
A=[]
B=[]
for i,k in enumerate(a):
if i%2 == 0:
A.append(a[i])
elif i%2 == 1:
B.append(a[i])
print(sum(A)-sum(B))
|
s103551212
|
p02663
|
u282908818
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,072
| 289
|
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?
|
T = list(str(input()))
for i in range(len(T)):
if T[i] == '?':
if T[i - 1] == 'P':
T[i] = 'D'
elif T[i + 1] == '?':
T[i] = 'P'
T[i + 1] = 'D'
i= i+1
else:
T[i] = 'D'
TL=''.join(T)
print(TL)
|
s169430962
|
Accepted
| 21
| 9,164
| 112
|
H,M,HH,MM,K=map(int,input().split())
hour = HH - H -1
minute=60+MM-M
time = hour * 60 + minute - K
print(time)
|
s451202415
|
p02600
|
u468972478
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,076
| 37
|
M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have?
|
n = int(input())
print(10 - (n // 2))
|
s563500780
|
Accepted
| 27
| 9,156
| 39
|
n = int(input())
print(10 - (n // 200))
|
s242892629
|
p03597
|
u669057361
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 47
|
We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black?
|
N = int(input())
A = int(input())
print(N-A**2)
|
s442253301
|
Accepted
| 17
| 2,940
| 47
|
N = int(input())
A = int(input())
print(N**2-A)
|
s332176709
|
p02608
|
u733866054
| 2,000
| 1,048,576
|
Wrong Answer
| 459
| 9,464
| 236
|
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
N=int(input())
ans=[0]*(N+5)
for i in range(100) :
for j in range(100) :
for k in range(100) :
n=i*i+j*j+k*k+i*j+j*k+k*i
if N>=n :
ans[n]+=1
for i in range(1,N+1) :
print(ans[i])
|
s319442260
|
Accepted
| 444
| 9,296
| 241
|
N=int(input())
ans=[0]*(N+5)
for i in range(1,100) :
for j in range(1,100) :
for k in range(1,100) :
n=i*i+j*j+k*k+i*j+j*k+k*i
if N>=n :
ans[n]+=1
for i in range(1,N+1) :
print(ans[i])
|
s710140125
|
p03574
|
u734749411
| 2,000
| 262,144
|
Wrong Answer
| 53
| 3,924
| 592
|
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
H, W = map(int, input().split())
table = []
table.append(list("_" * (W + 2)))
for _ in range(H):
table.append(list("_" + input() + "_"))
table.append(list("_" * (W + 2)))
for h in range(1, H + 1):
for w in range(1, W + 1):
if table[h][w] == "#":
continue
count = 0
for _h in [-1, 0, +1]:
for _w in [-1, 0, +1]:
print(h + _h, w + _w)
if table[h + _h][w + _w] == "#":
count += 1
table[h][w] = str(count)
for h in range(1, H + 1):
print("".join(table[h]).replace("_", ""))
|
s133166994
|
Accepted
| 24
| 3,188
| 554
|
H, W = map(int, input().split())
table = []
table.append(list("_" * (W + 2)))
for _ in range(H):
table.append(list("_" + input() + "_"))
table.append(list("_" * (W + 2)))
for h in range(1, H + 1):
for w in range(1, W + 1):
if table[h][w] == "#":
continue
count = 0
for _h in [-1, 0, +1]:
for _w in [-1, 0, +1]:
if table[h + _h][w + _w] == "#":
count += 1
table[h][w] = str(count)
for h in range(1, H + 1):
print("".join(table[h]).replace("_", ""))
|
s773660422
|
p02612
|
u268402865
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,140
| 71
|
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 math
a = int(input())
a_ceil = math.ceil(a/1000)*1000
a_ceil - a
|
s453383024
|
Accepted
| 26
| 9,108
| 83
|
N = int(input())
tmp = N%1000
if tmp == 0:
print(0)
else:
print(1000 - tmp)
|
s307426514
|
p03636
|
u102242691
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 81
|
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
s = input()
number = len(s) - 2
print(s[0])
print(s[0] + str(number) + s[-1])
|
s085919126
|
Accepted
| 17
| 2,940
| 112
|
s = list(input())
ans = []
ans.append(s[0])
ans.append(str((len(s)-2)))
ans.append(s[-1])
print("".join(ans))
|
s034403581
|
p03478
|
u814986259
| 2,000
| 262,144
|
Wrong Answer
| 34
| 3,060
| 141
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
N,A,B=map(int,input().split())
ans=0
for i in range(1,N+1):
s=list(map(int,str(i)))
s=sum(s)
if A<=s and s>=B:
ans+=1
print(ans)
|
s190159758
|
Accepted
| 29
| 2,940
| 207
|
N, A, B = map(int, input().split())
ans = 0
for i in range(1, N+1):
k = i
tmp = 0
while k > 0:
tmp += k % 10
k = k // 10
if tmp >= A and tmp <= B:
ans += i
print(ans)
|
s554939749
|
p03657
|
u496821919
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 244
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
# -*- coding: utf-8 -*-
"""
Created on Thu May 14 12:29:30 2020
@author: shinba
"""
a,b = map(int,input().split())
if a % 3 == 0:
print("Yes")
elif b % 3 == 0:
print("Yes")
elif (a+b) % 3 == 0:
print("Yes")
else:
print("No")
|
s758239516
|
Accepted
| 18
| 3,064
| 267
|
# -*- coding: utf-8 -*-
"""
Created on Thu May 14 12:29:30 2020
@author: shinba
"""
a,b = map(int,input().split())
if a % 3 == 0:
print("Possible")
elif b % 3 == 0:
print("Possible")
elif (a+b) % 3 == 0:
print("Possible")
else:
print("Impossible")
|
s232771116
|
p03433
|
u631429391
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 90
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N=int(input())
A=int(input())
if N%500 <= A:
print("YES")
elif N%500 > A:
print("NO")
|
s470375197
|
Accepted
| 17
| 2,940
| 90
|
N=int(input())
A=int(input())
if N%500 <= A:
print("Yes")
elif N%500 > A:
print("No")
|
s164420471
|
p02742
|
u933717615
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 100
|
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
H,W=map(int, input().split())
if W%2==0:
ans = int(H*W/2)
else:
ans = int(H*W/2)+1
print(ans)
|
s840889734
|
Accepted
| 18
| 3,060
| 138
|
H,W=map(int, input().split())
if H==1 or W==1:
ans=1
elif W%2!=0 and H%2!=0:
ans = int(H*W/2)+1
else:
ans = int(H*W/2)
print(ans)
|
s113349700
|
p04044
|
u999331208
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 176
|
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.
|
n,l=map(int,input().split())
print(n)
print(l)
input_str = []
for i in range(n):
input_str.append(input())
input_str.sort()
mojiretu = ''.join(input_str)
print(mojiretu)
|
s863109977
|
Accepted
| 17
| 3,060
| 158
|
n,l=map(int,input().split())
input_str = []
for i in range(n):
input_str.append(input())
input_str.sort()
mojiretu = ''.join(input_str)
print(mojiretu)
|
s578996227
|
p02972
|
u609061751
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 109,884
| 333
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
N=int(input())
a=list(map(int,input().split()))
M=0
b=[]
M=0
for i in range(N):
b.append(0)
for j in range(N,0,-1):
cnt=0
for k in range(j,N,j):
cnt+=b[k-1]
if cnt%2!=a[j-1]:
b[j-1]=1
print(b)
M+=1
c=[l+1 for l, x in enumerate(b) if x==1]
c=list(map(str,c))
print(M)
print(' '.join(c))
|
s272302167
|
Accepted
| 648
| 19,004
| 318
|
N=int(input())
a=list(map(int,input().split()))
M=0
b=[]
M=0
for i in range(N):
b.append(0)
for j in range(N,0,-1):
cnt=0
for k in range(j,N+1,j):
cnt+=b[k-1]
if cnt%2!=a[j-1]:
b[j-1]=1
M+=1
c=[l+1 for l, x in enumerate(b) if x==1]
c=list(map(str,c))
print(M)
print(' '.join(c))
|
s139727944
|
p03370
|
u806403461
| 2,000
| 262,144
|
Wrong Answer
| 325
| 4,536
| 219
|
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
n, x = map(int, input().split())
m = list()
for a in range(0, n):
m.append(int(input()))
sm = sum(m)
ans = n
while sm <= x:
sm += min(m)
ans += 1
print(sm, ans)
if sm < x:
ans -= 1
print(ans)
|
s670347878
|
Accepted
| 17
| 2,940
| 137
|
n, x = map(int, input().split())
m = list()
for a in range(0, n):
m.append(int(input()))
s = sum(m)
print(n + (x - s) // min(m))
|
s123806570
|
p03658
|
u396391104
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 137
|
Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.
|
n,k = map(int,input().split())
l = list(map(int,input().split()))
l.sort(reverse=True)
ans = 0
for i in range(k+1):
ans += i
print(ans)
|
s033371683
|
Accepted
| 17
| 2,940
| 139
|
n,k = map(int,input().split())
l = list(map(int,input().split()))
l.sort(reverse=True)
ans = 0
for i in range(k):
ans += l[i]
print(ans)
|
s239181736
|
p03457
|
u231189826
| 2,000
| 262,144
|
Wrong Answer
| 469
| 30,668
| 440
|
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())
T = []
Z = []
time = []
T.append([0,0,0])
for i in range(N):
t = list(map(int, input().split()))
T.append(t)
goal = True
for l in range(len(T)-1):
Z.append(abs(T[l+1][1]-T[l][1]) + abs(T[l+1][2]-T[l][2]))
time.append(T[l+1][0] - T[l][0])
for j in range(len(T)-1):
if Z[j] > time[j] or Z[j]%2 != time[j]%2:
goal = False
break
if goal == False:
print('NO')
else:
print('YES')
|
s742840342
|
Accepted
| 100
| 26,632
| 109
|
_,*l=map(int,open(0).read().split());print("YNeos"[any((t+x+y)%2+(t<x+y)for t,x,y in zip(*[iter(l)]*3))::2])
|
s849469462
|
p02821
|
u967136506
| 2,000
| 1,048,576
|
Wrong Answer
| 989
| 14,340
| 712
|
Takahashi has come to a party as a special guest. There are N ordinary guests at the party. The i-th ordinary guest has a _power_ of A_i. Takahashi has decided to perform M _handshakes_ to increase the _happiness_ of the party (let the current happiness be 0). A handshake will be performed as follows: * Takahashi chooses one (ordinary) guest x for his left hand and another guest y for his right hand (x and y can be the same). * Then, he shakes the left hand of Guest x and the right hand of Guest y simultaneously to increase the happiness by A_x+A_y. However, Takahashi should not perform the same handshake more than once. Formally, the following condition must hold: * Assume that, in the k-th handshake, Takahashi shakes the left hand of Guest x_k and the right hand of Guest y_k. Then, there is no pair p, q (1 \leq p < q \leq M) such that (x_p,y_p)=(x_q,y_q). What is the maximum possible happiness after M handshakes?
|
def ints():
return [int(x) for x in input().split()]
def ii():
return int(input())
N, M = ints()
A = ints()
A.sort()
def combinations(x):
s = 0
i = 0
j = N-1
while j>=0:
while i<N and A[i]+A[j]<x:
i += 1
s += N-i
j -= 1
return s
def koufukudo(x):
s = 0
si = 0
j = 0
i = N-1
while j<N:
while i>=0 and A[i]+A[j]>=x:
si += A[i]
i -= 1
s += si + A[j]*(N-1-i)
j += 1
return s
def bsearch(lower, upper):
l = lower
u = upper
m = (l+u)//2
c = combinations(m)
if c<M:
return bsearch(l, m)
else:
if l==m:
return (l, c-M)
return bsearch(m, u)
x, dm = bsearch(0, A[-1]*2+1)
print(x)
print(koufukudo(x)-dm*x)
|
s526223414
|
Accepted
| 1,263
| 13,972
| 560
|
def ints():
return [int(x) for x in input().split()]
def ii():
return int(input())
N, M = ints()
A = ints()
A.sort()
A.reverse()
def combinations_and_kofukudo(x):
c = 0
k = 0
si = 0
i = 0
for j in reversed(range(N)):
while i<N and A[i]+A[j]>=x:
si += A[i]
i += 1
c += i
k += si + A[j]*i
return (c, k)
def bsearch(l, u):
m = (l+u)//2
c, k = combinations_and_kofukudo(m)
if c<M:
return bsearch(l, m)
else:
if l==m:
print(k-(c-M)*l)
exit()
return bsearch(m, u)
bsearch(0, A[0]*2+1)
|
s226022796
|
p04029
|
u667694979
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 50
|
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())
sum=0
for i in range(N):
sum+=i+1
|
s358709753
|
Accepted
| 17
| 2,940
| 61
|
N=int(input())
sum=0
for i in range(N):
sum+=i+1
print(sum)
|
s913709009
|
p03730
|
u663767599
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 314
|
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`.
|
from itertools import count
A, B, C = map(int, input().split())
reminders = []
for n in count(start=A // C):
r = (n * A - C) % B
if r == 0:
print("Yes")
break
else:
if r not in reminders:
reminders.append(r)
else:
print("No")
break
|
s966819323
|
Accepted
| 18
| 3,060
| 314
|
from itertools import count
A, B, C = map(int, input().split())
reminders = []
for n in count(start=A // C):
r = (n * A - C) % B
if r == 0:
print("YES")
break
else:
if r not in reminders:
reminders.append(r)
else:
print("NO")
break
|
s570128704
|
p03067
|
u498202416
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 81
|
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 < C:
print("YES")
else:
print("No")
|
s595588327
|
Accepted
| 17
| 2,940
| 94
|
A,B,C = map(int,input().split())
if A < C < B or B < C < A:
print("Yes")
else:
print("No")
|
s188226376
|
p03597
|
u339922532
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black?
|
n = int(input())
a = int(input())
print(n * 2 - a)
|
s214282167
|
Accepted
| 17
| 2,940
| 52
|
n = int(input())
a = int(input())
print(n ** 2 - a)
|
s829742262
|
p04029
|
u364027015
| 2,000
| 262,144
|
Wrong Answer
| 28
| 8,960
| 31
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N=int(input())
print(N*(N+1)/2)
|
s877423570
|
Accepted
| 25
| 9,144
| 33
|
N=int(input())
print(N*(N+1)//2)
|
s220960085
|
p03657
|
u136647933
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 129
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a,b= map(int,input().split())
if a%3==0 or b%3==0 or (a*b)%3==0:
result='Possible'
else:
result='Impossible'
print(result)
|
s748422394
|
Accepted
| 17
| 2,940
| 129
|
a,b= map(int,input().split())
if a%3==0 or b%3==0 or (a+b)%3==0:
result='Possible'
else:
result='Impossible'
print(result)
|
s156780901
|
p03417
|
u936985471
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 57
|
There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region. The front and back sides of these cards can be distinguished, and initially every card faces up. We will perform the following operation once for each square contains a card: * For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square. It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed. Find the number of cards that face down after all the operations.
|
N,M=map(int,input().split())
print(max(M-2,1)*max(N-2,1))
|
s733313944
|
Accepted
| 17
| 2,940
| 53
|
N,M=map(int,input().split())
print(abs((M-2)*(N-2)))
|
s160354784
|
p03501
|
u357751375
| 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 n * a >= b:
print(n * a)
else:
print(b)
|
s934844818
|
Accepted
| 17
| 2,940
| 85
|
n,a,b = map(int,input().split())
if n * a <= b:
print(n * a)
else:
print(b)
|
s247644431
|
p03693
|
u568711768
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 97
|
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?
|
A,B,C = list(map(int,input().split()))
if 10*B + C % 4 == 0:
print("YES")
else:
print("NO")
|
s928822304
|
Accepted
| 17
| 2,940
| 99
|
A,B,C = list(map(int,input().split()))
if (10*B + C) % 4 == 0:
print("YES")
else:
print("NO")
|
s949163246
|
p03457
|
u743383679
| 2,000
| 262,144
|
Wrong Answer
| 245
| 9,200
| 294
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
# coding: utf-8
import math
n = int(input())
t = 0
x = 0
y = 0
for i in range(n):
ti, xi, yi = list(map(int, input().split()))
if xi - x + yi - y > ti - t:
print("NO")
exit()
elif (xi - x + yi - y - (ti - t)) % 2 != 0:
print("NO")
exit()
print("YES")
|
s916899402
|
Accepted
| 252
| 9,128
| 486
|
# coding: utf-8
import math
n = int(input())
t = 0
x = 0
y = 0
# txy = []
# ti, xi, yi = list(map(int, input().split()))
# txy.append((ti, xi, yi))
for i in range(n):
ti, xi, yi = map(int, input().split())
if abs(xi - x) + abs(yi - y) > abs(ti - t):
print("No")
exit()
elif (abs(xi - x) + abs(yi - y) - abs(ti - t)) % 2 != 0:
print("No")
exit()
else:
t = ti
x = xi
y = yi
print("Yes")
|
s530611218
|
p03469
|
u806855121
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 34
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
S = input()
print('2017' + S[4:])
|
s736314620
|
Accepted
| 17
| 2,940
| 34
|
S = input()
print('2018' + S[4:])
|
s850711508
|
p03644
|
u417835834
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 205
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n = int(input())
listn = list(range(1,n+1))
def n2(n):
counter = 0
while n % 2 == 0:
n = n / 2
counter += 1
return counter
lists_n2 = list(map(n2, listn))
print(max(lists_n2))
|
s398385439
|
Accepted
| 17
| 3,060
| 288
|
n = int(input())
def n2(n):
counter = 0
while n % 2 == 0:
n = n / 2
counter += 1
return counter
max_resistor = 0
max_n2 = 0
for i in range(1,n+1):
i_n2 = n2(i)
if max_n2 <= i_n2:
max_n2 = i_n2
max_resistor = i
print(max_resistor)
|
s928615015
|
p03577
|
u970308980
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 26
|
Rng is going to a festival. The name of the festival is given to you as a string S, which ends with `FESTIVAL`, from input. Answer the question: "Rng is going to a festival of what?" Output the answer. Here, assume that the name of "a festival of s" is a string obtained by appending `FESTIVAL` to the end of s. For example, `CODEFESTIVAL` is a festival of `CODE`.
|
s = input()
print(s[:-9])
|
s455411381
|
Accepted
| 17
| 2,940
| 25
|
s = input()
print(s[:-8])
|
s655166374
|
p02613
|
u652034825
| 2,000
| 1,048,576
|
Wrong Answer
| 176
| 16,232
| 226
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
s = [input() for i in range(N)]
res = {
'AC' : 0,
'TLE' : 0,
'WA' : 0,
'RE' : 0
}
for k in s:
print(res[k])
res[k] = res[k] + 1
for k, v in res.items():
print(k + ' x ' + str(v))
|
s924091247
|
Accepted
| 146
| 16,276
| 208
|
N = int(input())
s = [input() for i in range(N)]
res = {
'AC' : 0,
'WA' : 0,
'TLE' : 0,
'RE' : 0
}
for k in s:
res[k] = res[k] + 1
for k, v in res.items():
print(k + ' x ' + str(v))
|
s228846121
|
p02265
|
u387437217
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,732
| 523
|
Your task is to implement a double linked list. Write a program which performs the following operations: * insert x: insert an element with key x into the front of the list. * delete x: delete the first element which has the key of x from the list. If there is not such element, you need not do anything. * deleteFirst: delete the first element from the list. * deleteLast: delete the last element from the list.
|
n=int(input())
commands=[input().split() for i in range(n)]
print(commands)
List=[]
def command(list_command):
if 'insert' in list_command:
List.insert(0,list_command[1])
elif 'delete' in list_command:
try:
t=List.index(list_command[1])
del List[t]
except:
pass
elif 'deleteFirst' in list_command:
del List[0]
elif 'deleteLast' in list_command:
del List[-1]
return True
for i in commands:
command(i)
print(*List)
|
s176083786
|
Accepted
| 4,650
| 557,404
| 470
|
from collections import deque
n=int(input())
commands=[input().split() for i in range(n)]
List=deque()
for list_command in commands:
if 'insert' in list_command:
List.appendleft(list_command[1])
elif 'delete' in list_command:
try:
List.remove(list_command[1])
except:
pass
elif 'deleteFirst' in list_command:
List.popleft()
elif 'deleteLast' in list_command:
List.pop()
print(*List)
|
s004740842
|
p03944
|
u870518235
| 2,000
| 262,144
|
Wrong Answer
| 74
| 3,440
| 731
|
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
#Beginner 047 B
W, H, N = map(int, input().split())
s = [input() for i in range(N)]
S = []
for i in range(N):
S.append(list(map(int, s[i].split())))
lst = [[1 for i in range(W)] for j in range(H)]
print(lst)
for i in range(N):
if S[i][2] == 1:
for j in range(0,S[i][0]):
for k in range(H):
lst[k][j] = 0
if S[i][2] == 2:
for j in range(S[i][0],W):
for k in range(H):
lst[k][j] = 0
if S[i][2] == 3:
for j in range(0,S[i][1]):
for k in range(W):
lst[j][k] = 0
if S[i][2] == 4:
for j in range(S[i][1],H):
for k in range(W):
lst[j][k] = 0
print(sum(lst,[]).count(1))
|
s068881798
|
Accepted
| 71
| 3,520
| 720
|
#Beginner 047 B
W, H, N = map(int, input().split())
s = [input() for i in range(N)]
S = []
for i in range(N):
S.append(list(map(int, s[i].split())))
lst = [[1 for i in range(W)] for j in range(H)]
for i in range(N):
if S[i][2] == 1:
for j in range(0,S[i][0]):
for k in range(H):
lst[k][j] = 0
if S[i][2] == 2:
for j in range(S[i][0],W):
for k in range(H):
lst[k][j] = 0
if S[i][2] == 3:
for j in range(0,S[i][1]):
for k in range(W):
lst[j][k] = 0
if S[i][2] == 4:
for j in range(S[i][1],H):
for k in range(W):
lst[j][k] = 0
print(sum(lst,[]).count(1))
|
s893897059
|
p02833
|
u697690147
| 2,000
| 1,048,576
|
Wrong Answer
| 2,205
| 9,192
| 324
|
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
n = int(input())
if n % 2 == 0:
n5 = 1
y = 10
while n % y == 0:
n5 += 1
y = 2*(5**n5)
n5 -= 1
c = [n//(2*(5**i)) for i in range(1, n5+1)]
cnt = 0
for i in range(n5):
if i != n5-1:
c[i] -= c[i+1]
cnt += c[i]*(i+1)
print(cnt)
else:
print(0)
|
s410735882
|
Accepted
| 28
| 9,064
| 274
|
n = int(input())
if n % 2 == 0:
n5 = 1
y = 10
c = []
while y <= n:
c.append(n//y)
if len(c) > 1:
c[-2] -= c[-1]
y *= 5
cnt = 0
for i in range(len(c)):
cnt += c[i]*(i+1)
print(cnt)
else:
print(0)
|
s221824058
|
p03610
|
u905615376
| 2,000
| 262,144
|
Wrong Answer
| 29
| 3,188
| 143
|
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
def getStrangeString(s):
res = ""
for i in range(len(s)):
if i%2 != 0:
res += s[i]
print(getStrangeString(input()))
|
s288224374
|
Accepted
| 30
| 3,188
| 159
|
def getStrangeString(s):
res = ""
for i in range(len(s)):
if i%2 == 0:
res += s[i]
return res
print(getStrangeString(input()))
|
s045868642
|
p02388
|
u070117804
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,644
| 55
|
Write a program which calculates the cube of a given integer x.
|
x = input("input x: ")
x = int(x)
x = pow(x,3)
print(x)
|
s822787285
|
Accepted
| 30
| 7,584
| 26
|
x=int(input())
print(x**3)
|
s973066340
|
p02409
|
u236295012
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,616
| 322
|
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.
|
house = [[[0 for i in range(10)] for i in range(3)] for i in range(4)]
n = int(input())
input_4 = list(map(int,input().split()))
for i in range(n):
house[input_4[0]-1][input_4[1]-1][input_4[2]-1] = input_4[3]
for i in range(4):
for j in range(3):
print('',''.join(map(str,house[i][j])))
print('#'*20)
|
s909117450
|
Accepted
| 50
| 7,764
| 345
|
house = [[[0 for i in range(10)] for i in range(3)] for i in range(4)]
n = int(input())
for i in range(n):
input_4 = list(map(int,input().split()))
house[input_4[0]-1][input_4[1]-1][input_4[2]-1] += input_4[3]
for i in range(4):
for j in range(3):
print('',' '.join(map(str,house[i][j])))
if i != 3:
print('#'*20)
|
s680702533
|
p04029
|
u102930666
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 67
|
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())
i = 0
ans=0
for i in range(N+1):
ans += i
ans
|
s447970578
|
Accepted
| 18
| 2,940
| 74
|
N = int(input())
i = 0
ans=0
for i in range(N+1):
ans += i
print(ans)
|
s150794606
|
p03251
|
u224983328
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 3,064
| 470
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
N,M,X,Y = map(int, input().split(" "))
xList = input().split(" ")
yList = input().split(" ")
for i in range(len(xList)):
xList[i] = int(xList[i])
for j in range(len(yList)):
yList[j] = int(yList[j])
xList.sort()
yList.sort()
if yList[0] - 1 >= xList[len(xList) - 1]:
index = yList[0] - 1
while index > xList[len(xList) - 1]:
if X < index and index <= Y:
print("No War")
break
else:
index -= 1
print("War")
else:
print("War")
|
s978843471
|
Accepted
| 17
| 3,064
| 354
|
N,M,X,Y = map(int, input().split(" "))
xList = input().split(" ")
yList = input().split(" ")
for i in range(len(xList)):
xList[i] = int(xList[i])
for j in range(len(yList)):
yList[j] = int(yList[j])
xList.sort()
yList.sort()
if yList[0] > xList[len(xList) - 1] and yList[0] > X and Y > xList[len(xList) - 1]:
print("No War")
else:
print("War")
|
s360531200
|
p02972
|
u669173971
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 7,148
| 262
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
n=int(input())
a=list(map(int, input().split()))
b=[]
for i in reversed(range(n)):
count=0
for j in range(1,n+1):
if j%(i+1)==0:
count+=a[j-1]
if count%2==1:
b.append(1)
for _ in range(len(b)):
print("1")
|
s907236254
|
Accepted
| 626
| 14,132
| 296
|
n=int(input())
a=list(map(int, input().split()))
b=[0]*n
for i in range(n,0,-1):
count=0
for j in range(i,n+1,i):
count+=b[j-1]
if (count%2)!=a[i-1]:
b[i-1]=1
c=[i+1 for i in range(n) if b[i]==1]
if len(c)==0:
print("0")
else:
print(len(c))
print(*c)
|
s197669025
|
p03599
|
u198336369
| 3,000
| 262,144
|
Time Limit Exceeded
| 3,317
| 298,080
| 699
|
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
a,b,c,d,e,f = map(int, input().split())
con = list()
s = list()
for x in range(31):
for y in range(31):
for v in range(1501):
if x == y == 0:
continue
else:
w1 = (f-(100*a*x + 100*b*y + c*v))//d
w2 = (100*a*x + 100*b*y -c*v)//d
w3 = ((100*a*x + 100*b*y)*e -c*v*100)//(100*d)
w = min(w1,w2,w3)
con1 = 100*(c*v + d*w)/(100*a*x + 100*b*y +c*v +d*w)
con.append(con1)
s.append([x,y,v,w])
conm = max(con)
conin =con.index(conm)
t = s[conin]
# print(t)
ans1 = 100*a*t[0] + 100*b*t[1] + c*t[2] + d*t[3]
ans2 = c*t[2] + d*t[3]
print(ans1,ans2)
|
s197555597
|
Accepted
| 1,957
| 9,192
| 726
|
a,b,c,d,e,f = map(int, input().split())
s = [0,0,0,0]
con1 = 0
for x in range(31):
for y in range(31):
for v in range(1501):
if x == y == 0:
continue
else:
w1 = (f-(100*a*x + 100*b*y + c*v))//d
# w2 = (100*a*x + 100*b*y -c*v)//d
w3 = ((100*a*x + 100*b*y)*e -c*v*100)//(100*d)
w = min(w1,w3)
con2 = 100*(c*v + d*w)/(100*a*x + 100*b*y +c*v +d*w)
if con1 <= con2 and w >= 0:
con1 = con2
s = [x,y,v,w]
else:
continue
ans1 = 100*a*s[0] + 100*b*s[1] + c*s[2] + d*s[3]
ans2 = c*s[2] + d*s[3]
print(ans1,ans2)
|
s880624722
|
p03044
|
u187516587
| 2,000
| 1,048,576
|
Wrong Answer
| 485
| 64,644
| 479
|
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.
|
import sys
def input():
return sys.stdin.readline()[:-1]
sys.setrecursionlimit(200000)
N=int(input())
s=[[]for i in range(N)]
c=[None]*N
for i in range(N-1):
u,v,w=map(int,input().split())
s[v-1].append(u)
s[u-1].append(v)
def mu(n,l,co):
for i in l:
if c[i-1]==None:
if co==1:
c[i-1]=0
mu(i,s[i-1],0)
else:
c[i-1]=1
mu(i,s[i-1],1)
mu(1,s[0],0)
print(*c,sep="\n")
|
s288101065
|
Accepted
| 552
| 76,464
| 511
|
import sys
def input():
return sys.stdin.readline()[:-1]
sys.setrecursionlimit(200000)
N=int(input())
s=[[]for i in range(N)]
c=[None]*N
for i in range(N-1):
u,v,w=map(int,input().split())
s[v-1].append((u,w%2))
s[u-1].append((v,w%2))
def mu(n,l,co):
for i in l:
if c[i[0]-1]==None:
if co==i[1]%2:
c[i[0]-1]=0
mu(i,s[i[0]-1],0)
else:
c[i[0]-1]=1
mu(i,s[i[0]-1],1)
mu(1,s[0],0)
print(*c,sep="\n")
|
s893448170
|
p03494
|
u298297089
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 3,060
| 271
|
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())
x = list([int(i) for i in input().split()])
cnt = 0
while True:
flag = False
for i in x:
if i % 2:
flag = True
break
else:
i /= 2
else:
cnt += 1
if flag:
break
print(cnt)
|
s474595588
|
Accepted
| 18
| 3,060
| 190
|
N = int(input())
x = list(map(int, input().split()))
mn_lst = []
for i in x:
cnt = 0
while i % 2 == 0:
cnt += 1
i >>= 1
mn_lst.append(cnt)
print(min(mn_lst))
|
s277483481
|
p03387
|
u089032001
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 276
|
You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
A = list(map(int, input().split()))
ans = 0
even = 0
odd = 0
for a in A:
if(a%2==0):
even += 1
for i in range(len(A)):
if(A[i]%2==0 and even == 2):
A[i] += 1
elif(A[i]%2==1 and even == 1):
A[i] += 1
for a in A:
ans += (max(A) - a) // 2
print(a)
|
s658969775
|
Accepted
| 18
| 3,064
| 322
|
A = list(map(int, input().split()))
even = 0
odd = 0
for a in A:
if(a%2==0):
even += 1
for i in range(len(A)):
if(A[i]%2==0 and even == 2):
A[i] += 1
elif(A[i]%2==1 and even == 1):
A[i] += 1
if(even == 2 or even == 1):
ans = 1
else:
ans = 0
for a in A:
ans += (max(A) - a) // 2
print(ans)
|
s080134231
|
p02257
|
u909075105
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 254
|
A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.
|
n=int(input())
A=[0]*n
for m in range(n):
A[m]=int(input())
k=0
for i in range(n):
for j in range(1,A[n-1]):
if j==1:
continue
if j==(A[n-1]-1):
k+=1
if (A[i-1]%j)==0:
break
print(j)
|
s694812435
|
Accepted
| 2,620
| 5,996
| 298
|
import math as mh
a=int(input())
A=[0]*a
x=0
for i in range(a):
A[i]=int(input())
for j in range(1,A[i]):
if A[i]==2:
x+=1
continue
if j==1:
continue
elif (A[i]%j)==0:
break
elif j>=mh.sqrt(A[i]):
x+=1
break
print(x)
|
s495463446
|
p02744
|
u083960235
| 2,000
| 1,048,576
|
Wrong Answer
| 39
| 5,204
| 1,198
|
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.
|
import sys, re, os
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import permutations, combinations, product, accumulate
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from fractions import gcd
from bisect import bisect, bisect_left, bisect_right
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def S_MAP(): return map(str, input().split())
def LIST(): return list(map(int, input().split()))
def S_LIST(): return list(map(str, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
l = 'abcdefghijklmnopqrstuvwxyz'
N = INT()
# ans = (N-1) * "a" +
# p1 = "a" * N
q = ["a"]
# while q:
a = "a"
k = [n for n in l[:N]]
# print(k)
kazu = []
for i in range(1, N):
y = k[:i]
u = k[:i+1]
B = list(product(y, u))
for b in B:
# for c in b:
kazu.append("".join(b))
kazu.sort()
print(*kazu, sep="\n")
|
s265431192
|
Accepted
| 173
| 5,156
| 976
|
import sys, re, os
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import permutations, combinations, product, accumulate
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def S_MAP(): return map(str, input().split())
def LIST(): return list(map(int, input().split()))
def S_LIST(): return list(map(str, input().split()))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
N = INT()
S = "abcdefghijk"
ans = 0
def dfs(a, mx):
# while q:
if len(a) == N:
print(a)
return
for c in range(len(S[:mx+2])):
t = a
t += S[:mx+2][c]
dfs(t, max(c, mx))
dfs("", -1)
|
s655254860
|
p03251
|
u155236040
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 220
|
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
n,m,x,y = map(int,input().split())
x_list = sorted(list(map(int,input().split())))
y_list = sorted(list(map(int,input().split())))
if x < y and x < y_list[0] and x_list[-1] < y:
print('No war')
else:
print('War')
|
s095657223
|
Accepted
| 17
| 2,940
| 222
|
n,m,x,y = map(int,input().split())
x_list = list(map(int,input().split()))
y_list = list(map(int,input().split()))
x_list.append(x)
y_list.append(y)
if max(x_list) < min(y_list):
print('No War')
else:
print('War')
|
s386477503
|
p03962
|
u733774002
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 150
|
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.
|
a, b, c = map(int, input().split())
if a == b and b == c:
print(1)
elif (a != b and b == c) or (a == b or b != c):
print(2)
else:
print(3)
|
s630843722
|
Accepted
| 17
| 3,060
| 174
|
a, b, c = map(int, input().split())
if a == b and b == c:
print(1)
elif (a != b and b == c) or (a == b and b != c) or (a == c and a != b):
print(2)
else:
print(3)
|
s205903715
|
p03712
|
u079022116
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 138
|
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
h,w = map(int,input().split())
a = [input() for i in range(h)]
print(a)
print('#'*(w+2))
for i in a:
print('#'+i+'#')
print('#'*(w+2))
|
s074230968
|
Accepted
| 17
| 3,060
| 129
|
h,w = map(int,input().split())
a = [input() for i in range(h)]
print('#'*(w+2))
for i in a:
print('#'+i+'#')
print('#'*(w+2))
|
s260538281
|
p03713
|
u940102677
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 167
|
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())
s = [h//2+w//3+1, h//1+w//2+1]
if h%3 == 0 and w%3 == 0:
s += [0]
if h%2 == 0:
s += [h//2]
if w%2 == 0:
s += [w//2]
print(min(s))
|
s118707537
|
Accepted
| 17
| 3,064
| 172
|
h,w = map(int, input().split())
s = [h//2+w//3+1, h//3+w//2+1, h, w]
if h%3 == 0 or w%3 == 0:
s += [0]
if h%2 == 0:
s += [h//2]
if w%2 == 0:
s += [w//2]
print(min(s))
|
s278486458
|
p02612
|
u440478998
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,152
| 55
|
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 math
n = int(input())
math.ceil(n/1000)*1000 - n
|
s240191319
|
Accepted
| 31
| 9,148
| 62
|
import math
n = int(input())
print(math.ceil(n/1000)*1000 - n)
|
s830145981
|
p03380
|
u281303342
| 2,000
| 262,144
|
Wrong Answer
| 350
| 14,060
| 214
|
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
N = int(input())
A = list(map(int,input().split()))
maxA = max(A)
half = maxA/2
t = 100000009
for i in range(N):
print("i",i,"t",t,"Ai",A[i],)
if abs(A[i]-half) < abs(t-half):
t = A[i]
print(maxA,t)
|
s471397781
|
Accepted
| 92
| 14,040
| 688
|
# Python3 (3.4.3)
import sys
input = sys.stdin.readline
# -------------------------------------------------------------
# function
# -------------------------------------------------------------
# -------------------------------------------------------------
# main
# -------------------------------------------------------------
N = int(input())
A = list(map(int,input().split()))
maxA = max(A)
half = maxA/2
t = 10**10
for i in range(N):
if A[i] != maxA and abs(A[i]-half) < abs(t-half):
t = A[i]
print(maxA,t)
|
s369222265
|
p04044
|
u679089074
| 2,000
| 262,144
|
Wrong Answer
| 23
| 9,168
| 144
|
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.
|
N,L = map(int,input().split())
wordlist = []
for i in range(N):
x = input()
wordlist.append(x)
print(sorted(wordlist))
|
s523473675
|
Accepted
| 22
| 9,076
| 169
|
N,L = map(int,input().split())
wordlist = []
for i in range(N):
x = input()
wordlist.append(x)
answer = "".join(sorted(wordlist))
print(answer)
|
s374996003
|
p02612
|
u293326264
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,140
| 30
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(n%1000)
|
s962488845
|
Accepted
| 30
| 9,152
| 89
|
n = int(input())
if n % 1000 == 0:
print(n % 1000)
else:
print(1000 - (n % 1000))
|
s973295245
|
p03730
|
u651803486
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 134
|
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+1, a):
if i % b == c:
print('Yes')
exit()
print('No')
|
s935402124
|
Accepted
| 18
| 2,940
| 134
|
a, b, c = map(int, input().split())
for i in range(a, a*b+1, a):
if i % b == c:
print('YES')
exit()
print('NO')
|
s823291725
|
p03854
|
u344959886
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 150
|
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="dreamerer"
s=s.replace('eraser','').replace('erase','').replace('dreamer','').replace('dream','')
if len(s)==0:print("YES")
else:print("NO")
|
s971670437
|
Accepted
| 19
| 3,188
| 168
|
s=input()
s=s.replace('eraser','0').replace('erase','0').replace('dreamer','0').replace('dream','0')
s=s.replace('0','')
if len(s)==0:print("YES")
else:print("NO")
|
s863451872
|
p02613
|
u569776981
| 2,000
| 1,048,576
|
Wrong Answer
| 147
| 9,212
| 307
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N = int(input())
a = 0
w = 0
t = 0
r = 0
for i in range(N):
i = input()
if i == 'AC':
a += 1
elif i == 'WA':
w += 1
elif i == 'TLE':
t += 1
else:
r += 1
print('AC × ' + str(a))
print('WA × ' + str(w))
print('TLE × ' + str(t))
print('RE × ' + str(r))
|
s794050779
|
Accepted
| 142
| 9,168
| 301
|
N = int(input())
a, w, t, r = 0, 0, 0, 0
for _ in range(N):
S = input()
if S == 'AC':
a += 1
elif S == 'WA':
w += 1
elif S == 'TLE':
t += 1
else:
r += 1
print('AC x ' + str(a))
print('WA x ' + str(w))
print('TLE x ' + str(t))
print('RE x ' + str(r))
|
s418244296
|
p03155
|
u106181248
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 85
|
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())
a = int(input())
b = int(input())
ans = (n-a+1)+(n-b+1)
print(ans)
|
s269123860
|
Accepted
| 17
| 2,940
| 85
|
n = int(input())
a = int(input())
b = int(input())
ans = (n-a+1)*(n-b+1)
print(ans)
|
s437806201
|
p02646
|
u537905693
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 9,124
| 352
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
#!/usr/bin/env python
# coding: utf-8
def ri():
return int(input())
def rl():
return list(input().split())
def rli():
return list(map(int, input().split()))
def main():
a, v = rli()
b, w = rli()
t = ri()
if t*(v-w) >= abs(a-b):
print("Yes")
else:
print("No")
if __name__ == '__main__':
main()
|
s091382696
|
Accepted
| 23
| 9,180
| 352
|
#!/usr/bin/env python
# coding: utf-8
def ri():
return int(input())
def rl():
return list(input().split())
def rli():
return list(map(int, input().split()))
def main():
a, v = rli()
b, w = rli()
t = ri()
if t*(v-w) >= abs(a-b):
print("YES")
else:
print("NO")
if __name__ == '__main__':
main()
|
s048045066
|
p04029
|
u141419468
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 70
|
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())
P = 0
for i in range(1,n):
P += i
print(P)
|
s179605605
|
Accepted
| 17
| 2,940
| 72
|
n = int(input())
P = 0
for i in range(1,n+1):
P += i
print(P)
|
s522303271
|
p04043
|
u770035520
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 141
|
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(check,syllables):
check.sort()
syllables.sort()
if check==syllables:
return "YES"
else:
return "NO"
|
s114783954
|
Accepted
| 17
| 2,940
| 152
|
syllables=[5,5,7]
check=list(map(int,input().split()))
check.sort()
syllables.sort()
if check==syllables:
print("YES")
else:
print("NO")
|
s612969625
|
p03408
|
u386089355
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,064
| 382
|
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
n = int(input())
ls_s = [input() for _ in range(n)]
m = int(input())
ls_t = [input() for _ in range(m)]
ls_u =ls_s + ls_t
p = m + n
ans = 0
for i in range(n):
cnt = 0
for j in range(p):
if ls_u[i] == ls_u[j] and j < n - 1:
cnt += 1
elif ls_u[i] == ls_u[j] and n - 1 <= j:
cnt += -1
if cnt >= ans:
ans = cnt
print(ans)
|
s028907843
|
Accepted
| 21
| 3,188
| 383
|
n = int(input())
ls_s = [input() for _ in range(n)]
m = int(input())
ls_t = [input() for _ in range(m)]
ls_u =ls_s + ls_t
p = m + n
ans = 0
for i in range(n):
cnt = 0
for j in range(p):
if ls_u[i] == ls_u[j] and j <= n - 1:
cnt += 1
elif ls_u[i] == ls_u[j] and n - 1 < j:
cnt += -1
if cnt >= ans:
ans = cnt
print(ans)
|
s047309008
|
p02612
|
u225463683
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,140
| 33
|
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.
|
d = int(input())
print(d % 1000)
|
s692611149
|
Accepted
| 25
| 9,112
| 81
|
d = int(input())
if d % 1000 == 0:
print(0)
else:
print(1000 - d % 1000)
|
s223062187
|
p03378
|
u687044304
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 215
|
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.
|
# -*- coding:utf-8 -*-
if __name__ == "__main__":
N, M, X = list(map(int, input().split()))
As = list(map(int, input().split()))
cost = 0
for A in As:
if X > A:
break
cost += 1
print(max(cost, M-cost))
|
s728720585
|
Accepted
| 17
| 2,940
| 215
|
# -*- coding:utf-8 -*-
if __name__ == "__main__":
N, M, X = list(map(int, input().split()))
As = list(map(int, input().split()))
cost = 0
for A in As:
if A > X:
break
cost += 1
print(min(cost, M-cost))
|
s223874815
|
p03455
|
u933622697
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
if a*b % 2:
print('Even')
else:
print('Odd')
|
s618940390
|
Accepted
| 17
| 2,940
| 174
|
a, b = map(int, input().split())
if (a*b) % 2:
print('Odd')
else:
print('Even')
|
s147536993
|
p02694
|
u244737745
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,168
| 131
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
#2020-05-02
import math
a = int(input())
may = 100
ans = 0
while may <= a:
may = math.floor(may * 1.01)
ans += 1
print(ans)
|
s745275534
|
Accepted
| 21
| 9,168
| 118
|
import math
may = 100
a = int(input())
ans = 0
while a > may:
may = math.floor(may * 1.01)
ans += 1
print(ans)
|
s432543346
|
p03081
|
u261646994
| 2,000
| 1,048,576
|
Wrong Answer
| 1,006
| 6,804
| 1,141
|
There are N squares numbered 1 to N from left to right. Each square has a character written on it, and Square i has a letter s_i. Besides, there is initially one golem on each square. Snuke cast Q spells to move the golems. The i-th spell consisted of two characters t_i and d_i, where d_i is `L` or `R`. When Snuke cast this spell, for each square with the character t_i, all golems on that square moved to the square adjacent to the left if d_i is `L`, and moved to the square adjacent to the right if d_i is `R`. However, when a golem tried to move left from Square 1 or move right from Square N, it disappeared. Find the number of golems remaining after Snuke cast the Q spells.
|
import re
N, Q = [int(item) for item in re.split(r"\s", input())]
s = input()
t = []
d = []
for _ in range(Q):
t_in, d_in = [item for item in re.split(r"\s", input())]
t.append(t_in)
d.append(d_in)
l_most = -1
r_most = N
for q in range(Q-1, -1, -1):
# find last s[0], L
if t[q] == s[0] and d[q] == "L":
l_most = 0
for i in range(q-1, -1, -1):
# look_l = l_most-1
look_r = l_most+1
# if look_l >= 0 and s[look_l] == t[i] and d[i] == "R":
# l_most = look_l
if look_r <= N-1 and s[look_r] == t[i] and d[i] == "L":
l_most = look_r
break
for q in range(Q-1, -1, -1):
# find last s[N-1], R
if t[q] == s[N-1] and d[q] == "R":
r_most = N-1
for i in range(q-1, -1, -1):
look_l = r_most-1
# look_r = l_most+1
if look_l >= 0 and s[look_l] == t[i] and d[i] == "R":
r_most = look_l
# elif look_r <= N-1 and s[look_r] == t[i] and d[i] == "L":
# r_most = look_r
break
print(l_most, r_most)
print(r_most-l_most-1)
|
s792349691
|
Accepted
| 1,963
| 41,296
| 997
|
N, Q = map(int, input().split())
s = input()
TD = [input().split() for i in range(Q)]
def check(i):
if i == -1:
return -1
if i == N:
return 1
for q in range(Q):
t, d = TD[q]
if t == s[i]:
if d == "L":
i -= 1
else:
i += 1
if i == -1:
return -1
if i == N:
return 1
return 0
s_index = -1 # include
e_index = N # include
while s_index+1 < e_index:
c_index = (s_index+e_index)//2
if check(c_index) == -1:
s_index = c_index
else:
e_index = c_index
left = s_index
s_index = -1 # include
e_index = N # include
while s_index+1 < e_index:
c_index = (s_index+e_index)//2
if check(c_index) == 1:
e_index = c_index
else:
s_index = c_index
right = e_index
print(right-left-1)
|
s227930009
|
p03371
|
u761989513
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 271
|
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
a, b, c, x, y = map(int, input().split())
ans = 0
if a > 2 * c and b > 2 * c:
ans += max(x, y) * 2 * c
else:
if a + b > 2 * c:
n = min(x, y)
ans += n * 2 * c
x -= n
y -= n
print(ans)
ans += a * x
ans += b * y
print(ans)
|
s230604586
|
Accepted
| 17
| 2,940
| 189
|
a, b, c, x, y = map(int, input().split())
if x > y:
print(min(a * x + b * y, a * (x - y) + 2 * c * y, 2 * c * x))
else:
print(min(a * x + b * y, b * (y - x) + 2 * c * x, 2 * c * y))
|
s726698591
|
p02613
|
u161693347
| 2,000
| 1,048,576
|
Wrong Answer
| 80
| 17,264
| 1,093
|
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 sys, re, os
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import permutations, combinations, product, accumulate
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from functools import reduce
from bisect import bisect_left, insort_left
from heapq import heapify, heappush, heappop
INPUT = lambda: sys.stdin.readline().rstrip()
INT = lambda: int(INPUT())
MAP = lambda: map(int, INPUT().split())
S_MAP = lambda: map(str, INPUT().split())
LIST = lambda: list(map(int, INPUT().split()))
S_LIST = lambda: list(map(str, INPUT().split()))
ZIP = lambda: zip(*(MAP() for _ in range(n)))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
def main():
N = INT()
S = [INPUT() for _ in range(N)]
c = Counter(S)
print(c)
print("AC x " + str(c['AC']))
print("WA x " + str(c['WA']))
print("TLE x " + str(c['TLE']))
print("RE x " + str(c['RE']))
if __name__ == '__main__':
main()
|
s275547664
|
Accepted
| 73
| 17,256
| 1,080
|
import sys, re, os
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import permutations, combinations, product, accumulate
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from functools import reduce
from bisect import bisect_left, insort_left
from heapq import heapify, heappush, heappop
INPUT = lambda: sys.stdin.readline().rstrip()
INT = lambda: int(INPUT())
MAP = lambda: map(int, INPUT().split())
S_MAP = lambda: map(str, INPUT().split())
LIST = lambda: list(map(int, INPUT().split()))
S_LIST = lambda: list(map(str, INPUT().split()))
ZIP = lambda: zip(*(MAP() for _ in range(n)))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
def main():
N = INT()
S = [INPUT() for _ in range(N)]
c = Counter(S)
print("AC x " + str(c['AC']))
print("WA x " + str(c['WA']))
print("TLE x " + str(c['TLE']))
print("RE x " + str(c['RE']))
if __name__ == '__main__':
main()
|
s777848483
|
p02402
|
u539789745
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,588
| 136
|
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
def main():
nums = list(map(int, input().split()))
print(min(nums), max(nums), sum(nums))
if __name__ == "__main__":
main()
|
s159466830
|
Accepted
| 20
| 6,552
| 152
|
def main():
n = input()
nums = list(map(int, input().split()))
print(min(nums), max(nums), sum(nums))
if __name__ == "__main__":
main()
|
s103156964
|
p02744
|
u952467214
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 2,940
| 262
|
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.
|
import sys
sys.setrecursionlimit(10 ** 7)
n = int(input())
def rec(before, depth):
if depth == n:
return before+1
else:
ret = 0
for i in range(before+1):
ret += rec(i+1, depth+1)
return ret
print(rec(0,1))
|
s557129886
|
Accepted
| 241
| 22,904
| 468
|
import sys
sys.setrecursionlimit(10 ** 7)
from functools import lru_cache
n = int(input())
ans = []
if n == 1:
print('a')
exit()
@lru_cache(None)
def rec(before, depth, MAX):
for i in range(MAX-ord('a') +2):
tmp = before + chr(ord('a') + i)
if len(tmp) >= n:
ans.append(tmp)
else:
MAX = max(MAX, ord(tmp[-1]) )
rec(tmp, depth+1, MAX)
rec('a',1, ord('a'))
ans.sort()
print(*ans, sep ='\n')
|
s503000677
|
p03170
|
u703474183
| 2,000
| 1,048,576
|
Wrong Answer
| 1,851
| 9,924
| 243
|
There is a set A = \\{ a_1, a_2, \ldots, a_N \\} consisting of N positive integers. Taro and Jiro will play the following game against each other. Initially, we have a pile consisting of K stones. The two players perform the following operation alternately, starting from Taro: * Choose an element x in A, and remove exactly x stones from the pile. A player loses when he becomes unable to play. Assuming that both players play optimally, determine the winner.
|
N, K = map(int,input().split())
A = list(map(int,input().split()))
dp = [None for _ in range(K+1)]
dp[0] = 0
for k in range(1,K+1):
for a in A:
if k-a>=0:
if dp[k-a]==0:
dp[k]=1
if dp[K]==1:
print('First')
else:
print('Second')
|
s790609151
|
Accepted
| 1,442
| 9,920
| 246
|
N, K = list(map(int,input().split()))
A = list(map(int,input().split()))
dp = [0 for _ in range(K+1)]
dp[0] = 0
for k in range(1,K+1):
for a in A:
if k-a>=0:
if dp[k-a]==0:
dp[k]=1
if dp[K]==1:
print('First')
else:
print('Second')
|
s737554710
|
p03711
|
u676264453
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,192
| 1,661
|
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.
|
import math
import sys
T = list(map(int, input().split()))
S = T[0] * T[1]
if ((T[0] % 3 == 0) or (T[1] % 3 == 0)):
print(0)
sys.exit()
minS = 100000000000
for i in range(1, math.ceil(T[0]/2)):
A = i * T[1]
F = math.floor(T[1] / 2) * (T[0] - i)
C = S - (A + F)
if (F == C):
minS = min(minS, abs(A - F))
continue
else:
if (A > C):
minS = min(minS, abs(A - F))
elif(C >= A and A >= F):
minS = min(minS, abs(C - F))
else:
minS = min(minS, abs(C - A))
A = i * T[0]
F = math.floor(T[0] / 2) * (T[1] - i)
C = S - (A + F)
if (F == C):
minS = min(minS, abs(A - F))
continue
else:
if (A > C):
minS = min(minS, abs(A - F))
elif(C >= A and A >= F):
minS = min(minS, abs(C - F))
else:
minS = min(minS, abs(C - A))
for i in range(1, math.ceil(T[1]/2)):
A = i * T[1]
F = math.floor(T[1] / 2) * (T[0] - i)
C = S - (A + F)
if (F == C):
minS = min(minS, abs(A - F))
continue
else:
if (A > C):
minS = min(minS, abs(A - F))
elif(C >= A and A >= F):
minS = min(minS, abs(C - F))
else:
minS = min(minS, abs(C - A))
A = i * T[0]
F = math.floor(T[0] / 2) * (T[1] - i)
C = S - (A + F)
if (F == C):
minS = min(minS, abs(A - F))
continue
else:
if (A > C):
minS = min(minS, abs(A - F))
elif(C >= A and A >= F):
minS = min(minS, abs(C - F))
else:
minS = min(minS, abs(C - A))
print(minS)
|
s735467510
|
Accepted
| 19
| 3,188
| 211
|
x,y = list(map(int, input().split()))
a = set([2])
b = set([4,6,9,11])
c = set([1,3,5,7,8,10,12])
if (x in a and y in a) or (x in b and y in b) or (x in c and y in c):
print('Yes')
else:
print('No')
|
s967040949
|
p03377
|
u811967730
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 112
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A, B, X = map(int, input().split())
if X >= A and X <= A + B:
ans = "Yes"
else:
ans = "No"
print(ans)
|
s904098849
|
Accepted
| 17
| 2,940
| 112
|
A, B, X = map(int, input().split())
if X >= A and X <= A + B:
ans = "YES"
else:
ans = "NO"
print(ans)
|
s592324941
|
p02842
|
u847165882
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 185
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
N=int(input())
Len=int(N-N*0.1)
Ans=[]
for i in range(Len,N):
j=i*1.08
if int(j)==N:
Ans.append(int(j))
if len(Ans)!=0:
print(Ans[0])
else:
print(":(")
|
s873302917
|
Accepted
| 32
| 2,940
| 161
|
N=int(input())
Ans=[]
for i in range(N+1):
j=i*1.08
if int(j)==N:
Ans.append(i)
if len(Ans)!=0:
print(Ans[0])
else:
print(":(")
|
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