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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s799721302
|
p02288
|
u320446607
| 2,000
| 131,072
|
Wrong Answer
| 20
| 7,768
| 447
|
A binary heap which satisfies max-heap property is called max-heap. In a max- heap, for every node $i$ other than the root, $A Write a program which reads an array and constructs a max-heap from the array based on the following pseudo code. $maxHeapify(A, i)$ move the value of $A[i]$ down to leaves to make a sub-tree of node $i$ a max-heap. Here, $H$ is the size of the heap. 1 maxHeapify(A, i) 2 l = left(i) 3 r = right(i) 4 // select the node which has the maximum value 5 if l ≤ H and A[l] > A[i] 6 largest = l 7 else 8 largest = i 9 if r ≤ H and A[r] > A[largest] 10 largest = r 11 12 if largest ≠ i // value of children is larger than that of i 13 swap A[i] and A[largest] 14 maxHeapify(A, largest) // call recursively The following procedure buildMaxHeap(A) makes $A$ a max-heap by performing maxHeapify in a bottom-up manner. 1 buildMaxHeap(A) 2 for i = H/2 downto 1 3 maxHeapify(A, i)
|
H = int(input())
A = [None] + list(map(int, input().split()))
def max_heapify(A, i):
l = i * 2
r = i * 2 + 1
largest = i
if l < H:
if A[l] > A[largest]:
largest = l
if r < H:
if A[r] > A[largest]:
largest = r
if largest != i:
A[i], A[largest] = A[largest], A[i]
max_heapify(A, largest)
def build_max_heap(A):
for i in reversed(range(1, int(H / 2))):
max_heapify(A, i)
build_max_heap(A)
print(" ", end = "")
print(*A[1:])
|
s320131823
|
Accepted
| 870
| 65,520
| 443
|
H = int(input())
A = [None] + list(map(int, input().split()))
def max_heapify(A, i):
l = i * 2
r = i * 2 + 1
largest = i
if l <= H:
if A[l] > A[largest]:
largest = l
if r <= H:
if A[r] > A[largest]:
largest = r
if largest != i:
A[i], A[largest] = A[largest], A[i]
max_heapify(A, largest)
def build_max_heap(A):
for i in range(int(H / 2), 0, -1):
max_heapify(A, i)
build_max_heap(A)
print(" ", end = "")
print(*A[1:])
|
s665684049
|
p02388
|
u217701374
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,456
| 36
|
Write a program which calculates the cube of a given integer x.
|
X = input()
XX = int(X)
print(XX**2)
|
s055602763
|
Accepted
| 30
| 7,436
| 28
|
X = int(input())
print(X**3)
|
s703581102
|
p03672
|
u823885866
| 2,000
| 262,144
|
Wrong Answer
| 123
| 27,232
| 571
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
import sys
import math
import itertools
import collections
import heapq
import re
import numpy as np
from functools import reduce
rr = lambda: sys.stdin.readline().rstrip()
rs = lambda: sys.stdin.readline().split()
ri = lambda: int(sys.stdin.readline())
rm = lambda: map(int, sys.stdin.readline().split())
rl = lambda: list(map(int, sys.stdin.readline().split()))
inf = float('inf')
mod = 10**9 + 7
s = rr()
lens = len(s)
for i in range(2, lens, 2):
if s[:(lens - i) // 2] == s[(lens - i) // 2:lens - i]:
print(s[:(lens - i) // 2])
exit()
|
s349019523
|
Accepted
| 117
| 27,024
| 560
|
import sys
import math
import itertools
import collections
import heapq
import re
import numpy as np
from functools import reduce
rr = lambda: sys.stdin.readline().rstrip()
rs = lambda: sys.stdin.readline().split()
ri = lambda: int(sys.stdin.readline())
rm = lambda: map(int, sys.stdin.readline().split())
rl = lambda: list(map(int, sys.stdin.readline().split()))
inf = float('inf')
mod = 10**9 + 7
s = rr()
lens = len(s)
for i in range(2, lens, 2):
if s[:(lens - i) // 2] == s[(lens - i) // 2:lens - i]:
print(lens - i)
exit()
|
s821724035
|
p02613
|
u121698457
| 2,000
| 1,048,576
|
Wrong Answer
| 159
| 9,220
| 329
|
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())
lis = [0,0,0,0]
for i in range(N):
s = input()
if s == 'AC':
lis[0] += 1
elif s == 'WA':
lis[1] += 1
elif s == 'TLE':
lis[2] += 1
else:
lis[3] += 1
print('AC × '+str(lis[0]))
print('WA × '+str(lis[1]))
print('TLE × '+str(lis[2]))
print('RE × '+str(lis[3]))
|
s817687512
|
Accepted
| 150
| 9,216
| 325
|
N = int(input())
lis = [0,0,0,0]
for i in range(N):
s = input()
if s == 'AC':
lis[0] += 1
elif s == 'WA':
lis[1] += 1
elif s == 'TLE':
lis[2] += 1
else:
lis[3] += 1
print('AC x '+str(lis[0]))
print('WA x '+str(lis[1]))
print('TLE x '+str(lis[2]))
print('RE x '+str(lis[3]))
|
s086701566
|
p03567
|
u289162337
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 110
|
Snuke built an online judge to hold a programming contest. When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.) Determine whether the judge can return the string `AC` as the verdict to a program.
|
l = input()
for i in range(len(l)-1):
if l[i:i+1] == "AC":
print("Yes")
break
else:
print("No")
|
s093320305
|
Accepted
| 17
| 2,940
| 110
|
l = input()
for i in range(len(l)-1):
if l[i:i+2] == "AC":
print("Yes")
break
else:
print("No")
|
s747206541
|
p03827
|
u428855582
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 165
|
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
N = int(input())
tex=list(input())
print(tex)
max=0
cur=0
for i in range(N):
if tex[i]=="I":
cur+=1
if cur>max:
max=cur
else:
cur-=1
print(max)
|
s196415922
|
Accepted
| 17
| 2,940
| 154
|
N = int(input())
tex=list(input())
max=0
cur=0
for i in range(N):
if tex[i]=="I":
cur+=1
if cur>max:
max=cur
else:
cur-=1
print(max)
|
s669016410
|
p03478
|
u086051538
| 2,000
| 262,144
|
Wrong Answer
| 37
| 3,060
| 153
|
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())
counter=0
for m in range(1,n):
s=str(m)
array=list(map(int,s))
if a<=sum(array)<=b:
counter+=m
print(counter)
|
s349476052
|
Accepted
| 39
| 3,060
| 152
|
n,a,b=map(int,input().split())
counter=0
for m in range(1,n+1):
s=str(m)
array=list(map(int,s))
if a<=sum(array)<=b:
counter+=m
print(counter)
|
s182840676
|
p03861
|
u355137116
| 2,000
| 262,144
|
Wrong Answer
| 2,108
| 2,940
| 114
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a, b, x = map(int, input().split())
count = 0
for i in range(a, a+b):
if i % x == 0:
count += 1
print(count)
|
s795591470
|
Accepted
| 17
| 2,940
| 52
|
a,b,x=map(int,input().split())
print(b//x-(a-1)//x)
|
s873237740
|
p03470
|
u128859393
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 56
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
input();print(len(set(list(map(int, input().split())))))
|
s210294486
|
Accepted
| 18
| 2,940
| 78
|
N = int(input());print(len(set(list(map(int, [input() for _ in range(N)])))))
|
s306991627
|
p03673
|
u073775598
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 26,020
| 219
|
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
#ABC066C
n=int(input())
a=list(map(int, input().split()))
front=[]
back=[]
for i in range(n//2):
front.insert(0, a[2*i-(n%2)+1])
back.append(a[2*i+(n%2)])
if n%2==1:
front.insert(0, a[n-1])
print(front+back)
|
s407702260
|
Accepted
| 166
| 31,296
| 264
|
#ABC066C
n=int(input())
a=list(map(int, input().split()))
flg=n%2
half=n//2
front=[0]*(half+flg)
back=[0]*(half)
for i in range(half):
front[half+flg-1-i]=a[2*i-flg+1]
back[i]=a[2*i+flg]
if flg==1:
front[0]=a[n-1]
print(' '.join(map(str, front+back)))
|
s148626339
|
p02259
|
u867503285
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,704
| 377
|
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode.
|
def bubble_sort(target):
flag = True
while flag:
flag = False
for i in range(len(target)-1, 0, -1):
if target[i] < target[i-1]:
target[i-1], target[i] = target[i], target[i-1]
flag = True
return target
N = int(input())
A = [int(s) for s in input().split()]
print(' '.join(str(s) for s in bubble_sort(A)))
|
s234341462
|
Accepted
| 30
| 7,712
| 436
|
def bubble_sort(target):
global cnt
flag = True
while flag:
flag = False
for i in range(len(target)-1, 0, -1):
if target[i] < target[i-1]:
target[i-1], target[i] = target[i], target[i-1]
flag = True
cnt += 1
return target
N = int(input())
A = [int(s) for s in input().split()]
cnt = 0
print(' '.join(str(s) for s in bubble_sort(A)))
print(cnt)
|
s417770317
|
p03997
|
u583507988
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 83
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
ans = (a + b)*h / 2
print(ans)
|
s348464550
|
Accepted
| 27
| 8,876
| 73
|
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
|
s053347144
|
p03436
|
u390694622
| 2,000
| 262,144
|
Wrong Answer
| 31
| 4,972
| 2,033
|
We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j). Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares. Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game. When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares. The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is `.`; if square (i, j) is initially painted by black, s_{i, j} is `#`.
|
# coding: UTF-8
#@document_it
from collections import deque
def document_it(func):
def new_function(*args,**kwargs):
print('Running function:', func.__name__)
print('Positional arguments:', args)
print('Kewword arguments:', kwargs)
result = func(*args,**kwargs)
print('Result:', result)
return result
return new_function
###########################
#@document_it
def nb(i,j,H,W):
ans = []
a = i+1
b = i-1
c = j+1
d = j-1
if a >= 0 and a <= H-1:
ans.append((a,j))
if b >= 0 and b <= H-1:
ans.append((b,j))
if c >= 0 and c <= W-1:
ans.append((i,c))
if d >= 0 and d <= W-1:
ans.append((i,d))
return ans
def e():
dist={}
white = {}
wtotal = 0
H,W = map(int,input().split())
for i in range(H):
row = input()
for j in range(W):
dist[(i,j)] = -1
if row[j] == '.':
white[(i,j)] = True
wtotal += 1
else:
white[(i,j)] = False
#tuple -> [tuple]
link = {}
for i in range(H):
for j in range(W):
link[(i,j)] = []
if not white[(i,j)]:
continue
else:
nlist = nb(i,j,H,W)
for p in nlist:
if white[p]:
link[(i,j)].append(p)
q = deque()
dist[(0,0)] = 1
q.append((0,0))
while q:
v = q.popleft()
for w in link[v]:
if dist[w] == -1:
dist[w] = dist[v] + 1
q.append(w)
#print(w,dist[w])
if dist[(H-1,W-1)] != -1:
ans = wtotal - dist[(H-1,W-1)]
else:
ans = -1
print(wtotal)
print(dist[(H-1,W-1)])
print(ans)
###########################
if __name__ == '__main__':
e()
|
s493427038
|
Accepted
| 32
| 4,972
| 2,035
|
# coding: UTF-8
#@document_it
from collections import deque
def document_it(func):
def new_function(*args,**kwargs):
print('Running function:', func.__name__)
print('Positional arguments:', args)
print('Kewword arguments:', kwargs)
result = func(*args,**kwargs)
print('Result:', result)
return result
return new_function
###########################
#@document_it
def nb(i,j,H,W):
ans = []
a = i+1
b = i-1
c = j+1
d = j-1
if a >= 0 and a <= H-1:
ans.append((a,j))
if b >= 0 and b <= H-1:
ans.append((b,j))
if c >= 0 and c <= W-1:
ans.append((i,c))
if d >= 0 and d <= W-1:
ans.append((i,d))
return ans
def e():
dist={}
white = {}
wtotal = 0
H,W = map(int,input().split())
for i in range(H):
row = input()
for j in range(W):
dist[(i,j)] = -1
if row[j] == '.':
white[(i,j)] = True
wtotal += 1
else:
white[(i,j)] = False
#tuple -> [tuple]
link = {}
for i in range(H):
for j in range(W):
link[(i,j)] = []
if not white[(i,j)]:
continue
else:
nlist = nb(i,j,H,W)
for p in nlist:
if white[p]:
link[(i,j)].append(p)
q = deque()
dist[(0,0)] = 1
q.append((0,0))
while q:
v = q.popleft()
for w in link[v]:
if dist[w] == -1:
dist[w] = dist[v] + 1
q.append(w)
#print(w,dist[w])
if dist[(H-1,W-1)] != -1:
ans = wtotal - dist[(H-1,W-1)]
else:
ans = -1
#print(wtotal)
#print(dist[(H-1,W-1)])
print(ans)
###########################
if __name__ == '__main__':
e()
|
s283701514
|
p03129
|
u529722835
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 214
|
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
N , K = (int(i) for i in input().split())
if N % 2 == 0:
if N/2 >= K:
print('Yes')
else:
print('No')
elif N % 2 == 1:
if N/2 +1 >=K:
print('Yes')
else:
print('No')
|
s644212455
|
Accepted
| 17
| 2,940
| 206
|
N , K = map(int, input().split())
if N % 2 == 0:
if N//2 >= K:
print('YES')
else:
print('NO')
elif N % 2 == 1:
if N//2 +1 >= K:
print('YES')
else:
print('NO')
|
s167669248
|
p02382
|
u175224634
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,640
| 392
|
Your task is to calculate the distance between two $n$ dimensional vectors $x = \\{x_1, x_2, ..., x_n\\}$ and $y = \\{y_1, y_2, ..., y_n\\}$. The Minkowski's distance defined below is a metric which is a generalization of both the Manhattan distance and the Euclidean distance. \\[ D_{xy} = (\sum_{i=1}^n |x_i - y_i|^p)^{\frac{1}{p}} \\] It can be the Manhattan distance \\[ D_{xy} = |x_1 - y_1| + |x_2 - y_2| + ... + |x_n - y_n| \\] where $p = 1 $. It can be the Euclidean distance \\[ D_{xy} = \sqrt{(|x_1 - y_1|)^{2} + (|x_2 - y_2|)^{2} + ... + (|x_n - y_n|)^{2}} \\] where $p = 2 $. Also, it can be the Chebyshev distance \\[ D_{xy} = max_{i=1}^n (|x_i - y_i|) \\] where $p = \infty$ Write a program which reads two $n$ dimensional vectors $x$ and $y$, and calculates Minkowski's distance where $p = 1, 2, 3, \infty$ respectively.
|
num = int(input())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
first = second = third = inf = 0.00000
for i in range(num):
dif = x[i] - y[i]
first = first + abs(dif)
second = second + dif ** 2
third = third + abs(dif) ** 3
inf = max(inf, abs(dif))
print(float(first))
print(float(second ** 0.5))
print(float(third ** 0.3333))
print(float(inf))
|
s511735875
|
Accepted
| 20
| 5,644
| 446
|
num = int(input())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
first = second = third = inf = 0.00000
for i in range(num):
dif = x[i] - y[i]
first = first + abs(dif)
second = second + dif ** 2
third = third + abs(dif) ** 3
inf = max(inf, abs(dif))
print('{:.6f}'.format(first))
print('{:.6f}'.format(second ** 0.5))
print('{:.6f}'.format(third ** 0.333333333333333333))
print('{:.6f}'.format(inf))
|
s360370923
|
p03485
|
u058592821
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 61
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a, b = (int(i) for i in input().split())
print((a+b)//2 + 1)
|
s248305135
|
Accepted
| 17
| 2,940
| 104
|
a, b = (int(i) for i in input().split())
if (a+b) % 2 == 0:
print((a+b)//2)
else:
print((a+b)//2+1)
|
s640976574
|
p03682
|
u010110540
| 2,000
| 262,144
|
Wrong Answer
| 1,775
| 71,624
| 916
|
There are N towns on a plane. The i-th town is located at the coordinates (x_i,y_i). There may be more than one town at the same coordinates. You can build a road between two towns at coordinates (a,b) and (c,d) for a cost of min(|a-c|,|b-d|) yen (the currency of Japan). It is not possible to build other types of roads. Your objective is to build roads so that it will be possible to travel between every pair of towns by traversing roads. At least how much money is necessary to achieve this?
|
from operator import itemgetter
from collections import deque
N = int(input())
xs, ys = [], []
for _ in range(N):
x, y = map(int, input().split())
xs.append(x)
ys.append(y)
xis = sorted(list(enumerate(xs)), key = itemgetter(1))
yis = sorted(list(enumerate(ys)), key = itemgetter(1))
xds, yds = [], []
for i in range(N-1):
xds.append((xis[i+1][1] - xis[i][1], (xis[i][0], xis[i+1][0])))
yds.append((yis[i+1][1] - yis[i][1], (yis[i][0], yis[i+1][0])))
xdq, ydq = deque(sorted(xds)), deque(sorted(yds))
par = list(range(N))
def find(x):
if x == par[x]:
return x
else:
par[x] = find(par[x])
return par[x]
def union(x,y):
rx, ry = find(x), find(y)
par[ry] = rx
ans = 0
while xdq or ydq:
if xdq and (not ydq or xdq[0] < ydq[0]):
d, (i,j) = xdq.popleft()
else:
d, (i,j) = ydq.popleft()
ans += d
union(i,j)
print(ans)
|
s700004986
|
Accepted
| 1,462
| 71,680
| 963
|
from operator import itemgetter
from collections import deque
N = int(input())
xs, ys = [], []
for _ in range(N):
x, y = map(int, input().split())
xs.append(x)
ys.append(y)
xis = sorted(list(enumerate(xs)), key = itemgetter(1))
yis = sorted(list(enumerate(ys)), key = itemgetter(1))
xds, yds = [], []
for i in range(N-1):
xds.append((xis[i+1][1] - xis[i][1], (xis[i][0], xis[i+1][0])))
yds.append((yis[i+1][1] - yis[i][1], (yis[i][0], yis[i+1][0])))
xdq, ydq = deque(sorted(xds)), deque(sorted(yds))
par = list(range(N))
def find(x):
if x == par[x]:
return x
else:
par[x] = find(par[x])
return par[x]
def union(x,y):
rx, ry = find(x), find(y)
par[ry] = rx
ans = 0
while xdq or ydq:
if xdq and (not ydq or xdq[0] < ydq[0]):
d, (i,j) = xdq.popleft()
else:
d, (i,j) = ydq.popleft()
if find(i) != find(j):
ans += d
union(i,j)
print(ans)
|
s453147054
|
p02390
|
u499005012
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,724
| 160
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
sec_input = int(input())
hour = sec_input // (60 * 60)
remain = sec_input % (60 * 60)
min = remain // 60
sec = remain % 60
print('%d %d %d' % (hour, min, sec))
|
s271598879
|
Accepted
| 30
| 6,724
| 160
|
sec_input = int(input())
hour = sec_input // (60 * 60)
remain = sec_input % (60 * 60)
min = remain // 60
sec = remain % 60
print('%d:%d:%d' % (hour, min, sec))
|
s062931461
|
p03369
|
u268977772
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 25
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
print(input().count("o"))
|
s102310484
|
Accepted
| 17
| 2,940
| 33
|
print(700+100*input().count("o"))
|
s039997399
|
p03607
|
u957084285
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 6,244
| 518
|
You are playing the following game with Joisino. * Initially, you have a blank sheet of paper. * Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times. * Then, you are asked a question: How many numbers are written on the sheet now? The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?
|
N = int(input())
l = []
for _ in range(N):
a = int(input())
if len(l) == 0:
l.append(a)
continue
lo = 0
hi = len(l)
mid = (lo+hi) // 2
removed = False
p = (1, 0)
while not p == (lo, hi):
p = (lo, hi)
mid = (lo+hi) // 2
if l[mid] == a:
l.pop(mid)
removed = True
break
if l[mid] < a:
lo = mid
else:
hi = mid
if not removed:
l.insert(mid+1, a)
print(len(l))
|
s502706855
|
Accepted
| 252
| 15,072
| 198
|
N = int(input())
dict = {}
for _ in range(N):
x = int(input())
if x in dict.keys():
dict[x] += 1
else:
dict[x] = 1
print(sum(list(map(lambda x : x%2, dict.values()))))
|
s076845672
|
p03494
|
u877428733
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 135
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
A = list(map(int,input().split()))
for i in A:
p = 0
while i % 2 == 0:
i //= 2
p += 1
print(p)
|
s612139143
|
Accepted
| 18
| 3,060
| 181
|
ans = 10**9
N = int(input())
A = list(map(int,input().split()))
for i in A:
p = 0
while i % 2 == 0:
i //= 2
p += 1
if p < ans:
ans = p
print(ans)
|
s820901937
|
p02406
|
u744506422
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,600
| 301
|
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
import sys
n=int(input())
a=[]
for i in range(1,n+1):
if (i%3==0):
a.append(i)
else:
j=str(i)
for k in range(len(j)-1):
if j[k]=="3":
a.append(i)
break
for i in a:
sys.stdout.write(" {0}".format(i))
sys.stdout.write("\n")
|
s563559353
|
Accepted
| 30
| 5,952
| 299
|
import sys
n=int(input())
a=[]
for i in range(1,n+1):
if (i%3==0):
a.append(i)
else:
j=str(i)
for k in range(len(j)):
if j[k]=="3":
a.append(i)
break
for i in a:
sys.stdout.write(" {0}".format(i))
sys.stdout.write("\n")
|
s313708272
|
p02612
|
u583276018
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,140
| 93
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
a = 0
while(a < n):
a += 1000
print(a)
|
s106221689
|
Accepted
| 26
| 9,148
| 94
|
n = int(input())
a = 0
while(a < n):
a += 1000
print(a-n)
|
s387310568
|
p02536
|
u168416324
| 2,000
| 1,048,576
|
Wrong Answer
| 438
| 26,544
| 486
|
There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M. Road i connects City A_i and City B_i. Snuke can perform the following operation zero or more times: * Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities. After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times). What is the minimum number of roads he must build to achieve the goal?
|
n,m=map(int,input().split())
global fri
fri=[]
global tab
tab=[0]*n
def run(nn):
#print(tab)
global cnt
if tab[nn]:
return
tab[nn]=1
cnt+=1
for i in fri[nn]:
que.append(i)
for i in range(n):
fri.append([])
for i in range(m):
a,b=map(int,input().split())
fri[a-1].append(b-1)
fri[b-1].append(a-1)
i=0
mx=0
cc=0
while i<n:
if tab[i]:
i+=1
continue
que=[i]
cnt=0
while len(que)!=0:
run(que.pop())
mx=max(cnt,mx)
i+=1
cc+=1
print(cc)
|
s806304431
|
Accepted
| 413
| 26,548
| 488
|
n,m=map(int,input().split())
global fri
fri=[]
global tab
tab=[0]*n
def run(nn):
#print(tab)
global cnt
if tab[nn]:
return
tab[nn]=1
cnt+=1
for i in fri[nn]:
que.append(i)
for i in range(n):
fri.append([])
for i in range(m):
a,b=map(int,input().split())
fri[a-1].append(b-1)
fri[b-1].append(a-1)
i=0
mx=0
cc=0
while i<n:
if tab[i]:
i+=1
continue
que=[i]
cnt=0
while len(que)!=0:
run(que.pop())
mx=max(cnt,mx)
i+=1
cc+=1
print(cc-1)
|
s728214782
|
p02612
|
u321492259
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,048
| 40
|
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.
|
i = int(input())
a = i-(i%1000)
print(a)
|
s816376708
|
Accepted
| 26
| 9,000
| 38
|
print((1000-(int(input())%1000))%1000)
|
s052000604
|
p02273
|
u022407960
| 2,000
| 131,072
|
Wrong Answer
| 20
| 7,984
| 1,216
|
Write a program which reads an integer _n_ and draws a Koch curve based on recursive calles of depth _n_. The Koch curve is well known as a kind of You should start (0, 0), (100, 0) as the first segment.
|
# encoding: utf-8
import sys
import math
_input = sys.stdin.readlines()
depth = int(_input[0])
sin_60, cos_60 = math.sin(math.pi / 3), math.cos(math.pi / 3)
class Point(object):
def __init__(self, x=0.0, y=0.0):
"""
initialize the point
"""
self.x = float(x)
self.y = float(y)
def koch_curve(d, p1, p2):
if not d:
return None
s, t, u = Point(), Point, Point()
s.x = (2 * p1.x + p2.x) / 3
s.y = (2 * p1.y + p2.y) / 3
t.x = (p1.x + 2 * p2.x) / 3
t.y = (p1.y + 2 * p2.y) / 3
u.x = (t.x - s.x) * cos_60 - (t.y - s.y) * sin_60 + s.x
u.y = (t.x - s.x) * sin_60 + (t.y - s.y) * cos_60 + s.y
koch_curve(d - 1, p1, s)
print(format(s.x, '.8f'), '', format(s.y, '.8f'))
koch_curve(d - 1, s, u)
print(format(u.x, '.8f'), '', format(u.y, '.8f'))
koch_curve(d - 1, u, t)
print(format(t.x, '.8f'), '', format(t.y, '.8f'))
koch_curve(d - 1, t, p2)
if __name__ == '__main__':
p1_start, p2_end = Point(x=0, y=0), Point(x=100, y=0)
print(format(p1_start.x, '.8f'), '', format(p1_start.y, '.8f'))
koch_curve(depth, p1_start, p2_end)
print(format(p2_end.x, '.8f'), '', format(p2_end.y, '.8f'))
|
s545016471
|
Accepted
| 30
| 8,044
| 1,677
|
#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""
input:
2
output:
0.00000000 0.00000000
11.11111111 0.00000000
16.66666667 9.62250449
22.22222222 0.00000000
33.33333333 0.00000000
38.88888889 9.62250449
33.33333333 19.24500897
44.44444444 19.24500897
50.00000000 28.86751346
55.55555556 19.24500897
66.66666667 19.24500897
61.11111111 9.62250449
66.66666667 0.00000000
77.77777778 0.00000000
83.33333333 9.62250449
88.88888889 0.00000000
100.00000000 0.00000000
"""
import sys
import math
class Point(object):
__slots__ = ('x', 'y')
def __init__(self, x=-1, y=-1):
"""
Initialize the point
"""
self.x = float(x)
self.y = float(y)
def koch_curve(d, p1, p2):
if not d:
return None
s, t, u = (Point() for _ in range(3))
s.x = (2 * p1.x + p2.x) / 3
s.y = (2 * p1.y + p2.y) / 3
t.x = (p1.x + 2 * p2.x) / 3
t.y = (p1.y + 2 * p2.y) / 3
u.x = (t.x - s.x) * COS_60 - (t.y - s.y) * SIN_60 + s.x
u.y = (t.x - s.x) * SIN_60 + (t.y - s.y) * COS_60 + s.y
koch_curve(d - 1, p1, s)
print('{0:.8f} {1:.8f}'.format(s.x, s.y))
koch_curve(d - 1, s, u)
print('{0:.8f} {1:.8f}'.format(u.x, u.y))
koch_curve(d - 1, u, t)
print('{0:.8f} {1:.8f}'.format(t.x, t.y))
koch_curve(d - 1, t, p2)
if __name__ == '__main__':
_input = sys.stdin.readlines()
depth = int(_input[0])
SIN_60, COS_60 = math.sin(math.pi / 3), math.cos(math.pi / 3)
p1_start, p2_end = Point(x=0, y=0), Point(x=100, y=0)
print(format(p1_start.x, '.8f'), format(p1_start.y, '.8f'))
koch_curve(depth, p1_start, p2_end)
print(format(p2_end.x, '.8f'), format(p2_end.y, '.8f'))
|
s677521189
|
p02843
|
u767664985
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 85
|
AtCoder Mart sells 1000000 of each of the six items below: * Riceballs, priced at 100 yen (the currency of Japan) each * Sandwiches, priced at 101 yen each * Cookies, priced at 102 yen each * Cakes, priced at 103 yen each * Candies, priced at 104 yen each * Computers, priced at 105 yen each Takahashi wants to buy some of them that cost exactly X yen in total. Determine whether this is possible. (Ignore consumption tax.)
|
X = int(input())
X %= 105
X %= 104
X %= 103
X %= 102
X %= 101
X %= 100
print(X == 0)
|
s538338553
|
Accepted
| 17
| 2,940
| 209
|
X = int(input())
q, r = X//100, X%100
if r <= 5 * q:
print(1)
else:
print(0)
|
s665456548
|
p02841
|
u912134329
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 148
|
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.
|
xy = [map(int,input().split()) for i in range(2)]
x,y = [list(j) for j in zip(*xy)]
if (x[1] + 1) == y[1] :
ans = 0
else:
ans = 1
print(ans)
|
s324996926
|
Accepted
| 17
| 2,940
| 148
|
xy = [map(int,input().split()) for i in range(2)]
x,y = [list(j) for j in zip(*xy)]
if (y[0] + 1) == y[1] :
ans = 0
else:
ans = 1
print(ans)
|
s381772622
|
p04045
|
u648452607
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 183
|
Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money. Find the amount of money that she will hand to the cashier.
|
[n,k]=[int(_) for _ in input().split()]
d=[int(_) for _ in input().split()]
while True:
t=str(n).split()
for _t in t:
if _t in d:
n+=1
continue
break
print(n)
|
s655351272
|
Accepted
| 69
| 3,188
| 246
|
[n,k]=[int(_) for _ in input().split()]
d=[int(_) for _ in input().split()]
dt=list({_ for _ in range(10)}-set(d))
def c(n,l):
for _c in str(n):
if int(_c) not in l: return False
return True
while True:
if c(n,dt): break
n+=1
print(n)
|
s882911837
|
p02865
|
u061539997
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 24
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
print(int(input())-1//2)
|
s190104738
|
Accepted
| 17
| 2,940
| 26
|
print((int(input())-1)//2)
|
s470153980
|
p03485
|
u202877219
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 47
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b = map(int, input().split())
print (a+b/(2))
|
s677467082
|
Accepted
| 20
| 3,060
| 70
|
import math
a,b = map(int, input().split())
print (math.ceil((a+b)/2))
|
s026888211
|
p02612
|
u024609780
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,140
| 29
|
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.
|
A =int(input())
print(1000-A)
|
s134204846
|
Accepted
| 34
| 9,152
| 73
|
a=int(input())
if(a%1000!=0):
print(1000-(a%1000))
else:
print(0)
|
s651316596
|
p02975
|
u648212584
| 2,000
| 1,048,576
|
Wrong Answer
| 149
| 18,584
| 605
|
Snuke has N hats. The i-th hat has an integer a_i written on it. There are N camels standing in a circle. Snuke will put one of his hats on each of these camels. If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No`. * The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself. What is XOR? The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non- negative integers x_1, x_2, \ldots, x_n is defined as follows: - When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even. For example, 3 \oplus 5 = 6.
|
import collections
N = int(input())
a = list(map(int,input().split()))
a = collections.Counter(a)
a_list = list(a.items())
a_list = sorted(a_list,key = lambda a_list:a_list[0])
print(a_list)
if len(a_list) == 1:
if a_list[0][0] == 0:
print("Yes")
else:
print("No")
elif len(a_list) == 2:
if ((a_list[0][0] == 0) and (a_list[0][1] == int(N/3)) and \
(a_list[1][1] == int(2*N/3))):
print("Yes")
else:
print("No")
elif len(a_list) == 3:
if (a_list[0][1] == int(N/3)) and \
(a_list[1][1] == int(N/3) and \
a_list[2][1] == int(N/3)):
print("Yes")
else:
print("No")
else:
print("No")
|
s630406752
|
Accepted
| 106
| 16,164
| 648
|
import collections
N = int(input())
a = list(map(int,input().split()))
a = collections.Counter(a)
a_list = list(a.items())
a_list = sorted(a_list,key = lambda a_list:a_list[0])
if len(a_list) == 1:
if a_list[0][0] == 0:
print("Yes")
else:
print("No")
elif len(a_list) == 2:
if ((a_list[0][0] == 0) and (a_list[0][1] == int(N/3)) and \
(a_list[1][1] == int(2*N/3))):
print("Yes")
else:
print("No")
elif len(a_list) == 3:
if ((a_list[0][1] == int(N/3)) and \
(a_list[1][1] == int(N/3)) and \
(a_list[2][1] == int(N/3)) and \
(a_list[0][0]^a_list[1][0]^a_list[2][0] == 0)):
print("Yes")
else:
print("No")
else:
print("No")
|
s880346686
|
p03379
|
u951480280
| 2,000
| 262,144
|
Wrong Answer
| 290
| 25,220
| 132
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
n=int(input())
m=n//2
x=sorted(map(int,input().split()))
for i in range(m):
print(x[m-1])
for i in range(m,n):
print(x[m])
|
s133309184
|
Accepted
| 343
| 25,472
| 260
|
n=int(input())
x=list(map(int,input().split()))
x1=sorted(x)
mid=n//2
if n>2:
for i in range(n):
if x[i] < x1[mid]:print(x1[mid])
else:print(x1[mid-1])
else:
for i in range(n):
x2=x.copy()
x2.pop(i)
print(x2[0])
|
s442997465
|
p03719
|
u599547273
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 87
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a, b, c = input().split(" ")
print("Yes" if a[-1] == b[0] and b[-1] == c[0] else "No")
|
s377329033
|
Accepted
| 17
| 2,940
| 77
|
a, b, c = map(int, input().split(" "))
print("Yes" if a <= c <= b else "No")
|
s015158651
|
p02261
|
u922112509
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,608
| 1,256
|
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
# Stable Sort
length = int(input())
cards = input().rstrip().split()
cards1 = [card for card in cards]
cards2 = [card for card in cards]
def value(n):
number = n[1:]
return int(number)
def char(cards):
charSets = [[] for i in range(10)]
for i in range(length):
char = cards[i][:1]
val = value(cards[i])
charSets[val].append(char)
return charSets
def bubble(cards):
for i in range(length):
j = length - 1
while j > i:
if value(cards[j]) < value(cards[j - 1]):
cards[j], cards[j - 1] = cards[j - 1], cards[j]
j -= 1
return cards
def selection(cards):
# count = 0
for i in range(length):
minj = i
for j in range(i, length):
if value(cards[j]) < value(cards[minj]):
minj = j
# count += 1
cards[i], cards[minj] = cards[minj], cards[i]
return cards
def stable(cards1, cards2):
if char(cards1) == char(cards2):
return 'Stable'
else:
return 'Not stable'
c = cards
b = bubble(cards1)
s = selection(cards2)
print(b)
print(stable(c, b))
print(s)
print(stable(c, s))
|
s230943537
|
Accepted
| 20
| 5,612
| 1,276
|
# Stable Sort
length = int(input())
cards = input().rstrip().split()
cards1 = [card for card in cards]
cards2 = [card for card in cards]
def value(n):
number = n[1:]
return int(number)
def char(cards):
charSets = [[] for i in range(10)]
for i in range(length):
char = cards[i][:1]
val = value(cards[i])
charSets[val].append(char)
return charSets
def bubble(cards):
for i in range(length):
j = length - 1
while j > i:
if value(cards[j]) < value(cards[j - 1]):
cards[j], cards[j - 1] = cards[j - 1], cards[j]
j -= 1
return cards
def selection(cards):
# count = 0
for i in range(length):
minj = i
for j in range(i, length):
if value(cards[j]) < value(cards[minj]):
minj = j
# count += 1
cards[i], cards[minj] = cards[minj], cards[i]
return cards
def stable(cards1, cards2):
if char(cards1) == char(cards2):
return 'Stable'
else:
return 'Not stable'
c = cards
b = bubble(cards1)
s = selection(cards2)
print(' '.join(b))
print(stable(c, b))
print(' '.join(s))
print(stable(c, s))
|
s749413787
|
p03719
|
u280512618
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 70
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a,b,c=map(int,input().split())
print('Yes' if a<=b and b<=c else 'No')
|
s496988763
|
Accepted
| 17
| 2,940
| 70
|
a,b,c=map(int,input().split())
print('Yes' if a<=c and c<=b else 'No')
|
s907738523
|
p03698
|
u578501242
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 141
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
a=input()
b=[str(c) for c in a]
b.sort()
c=len(b)
d=0
for i in range(c-1):
if b[i]==b[i+1]:
d=d+1
if d>0:
print('No')
else:
print('Yes')
|
s566512555
|
Accepted
| 17
| 3,060
| 141
|
a=input()
b=[str(c) for c in a]
b.sort()
c=len(b)
d=0
for i in range(c-1):
if b[i]==b[i+1]:
d=d+1
if d>0:
print('no')
else:
print('yes')
|
s483321246
|
p02613
|
u699458609
| 2,000
| 1,048,576
|
Wrong Answer
| 146
| 9,220
| 317
|
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.
|
ans = {"AC": 0, "WA": 0, "TLE": 0, "RE":0}
N = int(input())
for i in range(N):
inpt = input()
if(inpt == "AC"):
ans["AC"] += 1
elif(inpt == "WA"):
ans["WA"] += 1
elif(inpt == "TLE"):
ans["TLE"] += 1
else:
ans["RE"] += 1
for key, value in ans.items():
print("{0} × {1}".format(key, value))
|
s504609997
|
Accepted
| 147
| 9,196
| 317
|
ans = {"AC": 0, "WA": 0, "TLE": 0, "RE":0}
N = int(input())
for i in range(N):
inpt = input()
if(inpt == "AC"):
ans["AC"] += 1
elif(inpt == "WA"):
ans["WA"] += 1
elif(inpt == "TLE"):
ans["TLE"] += 1
else:
ans["RE"] += 1
for key, value in ans.items():
print("{0} x {1}".format(key, value))
|
s189088229
|
p04029
|
u522266410
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 41
|
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())
x = (N*(N+1)/2)
print(x)
|
s247537123
|
Accepted
| 17
| 2,940
| 42
|
N = int(input())
print(round(N*(N+1)/2))
|
s802654231
|
p03352
|
u639592190
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 2,940
| 120
|
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
X=int(input())
ans=0
for b in range(1,int(X**(1/2))+1):
for p in range(2,int(X/b)+1):
ans=max(ans,b**p)
print(ans)
|
s056851537
|
Accepted
| 19
| 2,940
| 138
|
X=int(input())
ans=1
for b in range(1,int(X**(1/2))+1):
for p in range(2,int(X/b)+1):
if b**p<=X:
ans=max(ans,b**p)
print(ans)
|
s353504961
|
p03351
|
u245641078
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 105
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d=map(int, input().split())
if (abs(a-b))>=d and (abs(b-c))>=d:
print("Yes")
else:
print("No")
|
s057263221
|
Accepted
| 17
| 2,940
| 107
|
a,b,c,d = map(int,input().split())
print("Yes" if abs(a-c)<=d or (abs(a-b)<=d and abs(b-c)<=d) else 'No')
|
s457192567
|
p03379
|
u198274496
| 2,000
| 262,144
|
Wrong Answer
| 219
| 25,620
| 135
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
N = int(input())
X = list(map(int, input().split()))
d = N // 2
a = X[d]
b = X[d - 1]
for i in range(N):
print(a if i < d else b)
|
s939585290
|
Accepted
| 303
| 25,556
| 146
|
N = int(input())
X = list(map(int, input().split()))
sX = sorted(X)
d = N // 2
a = sX[d]
b = sX[d - 1]
for c in X:
print(a if c < a else b)
|
s581085867
|
p03605
|
u941884460
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 73
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
N = input()
if N[0] != 9 and N[1] !=9:
print('No')
else:
print('Yes')
|
s406750189
|
Accepted
| 17
| 2,940
| 83
|
N = input()
if (N[0] != '9') and (N[1] != '9'):
print('No')
else:
print('Yes')
|
s207367334
|
p03943
|
u310394471
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 111
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
A,B,C = map(int, input().split())
if A+B == C or A+C ==B or B+C == A:
print('Yec')
else:
print('No')
|
s788674686
|
Accepted
| 17
| 2,940
| 111
|
A,B,C = map(int, input().split())
if A+B == C or A+C ==B or B+C == A:
print('Yes')
else:
print('No')
|
s283013996
|
p03555
|
u344959959
| 2,000
| 262,144
|
Wrong Answer
| 29
| 8,960
| 109
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
a=input()
b=input()
if a[0] == b[2] and a[1] == b[1] and a[2] == b[0]:
print("Yes")
else:
print("No")
|
s009893011
|
Accepted
| 26
| 8,912
| 109
|
a=input()
b=input()
if a[0] == b[2] and a[1] == b[1] and a[2] == b[0]:
print("YES")
else:
print("NO")
|
s054543085
|
p03638
|
u648315264
| 2,000
| 262,144
|
Wrong Answer
| 38
| 9,508
| 977
|
We have a grid with H rows and W columns of squares. Snuke is painting these squares in colors 1, 2, ..., N. Here, the following conditions should be satisfied: * For each i (1 ≤ i ≤ N), there are exactly a_i squares painted in Color i. Here, a_1 + a_2 + ... + a_N = H W. * For each i (1 ≤ i ≤ N), the squares painted in Color i are _4-connected_. That is, every square painted in Color i can be reached from every square painted in Color i by repeatedly traveling to a horizontally or vertically adjacent square painted in Color i. Find a way to paint the squares so that the conditions are satisfied. It can be shown that a solution always exists.
|
import math
from math import gcd,pi,sqrt
INF = float("inf")
import sys
sys.setrecursionlimit(10**6)
import itertools
from collections import Counter,deque
def i_input(): return int(input())
def i_map(): return map(int, input().split())
def i_list(): return list(i_map())
def i_row(N): return [i_input() for _ in range(N)]
def i_row_list(N): return [i_list() for _ in range(N)]
def s_input(): return input()
def s_map(): return input().split()
def s_list(): return list(s_map())
def s_row(N): return [s_input for _ in range(N)]
def s_row_str(N): return [s_list() for _ in range(N)]
def s_row_list(N): return [list(s_input()) for _ in range(N)]
def main():
h,w = i_map()
n = i_input()
a = i_list()
ans = ""
for m,i in enumerate(a):
ans = ans + (str(m+1))*i
for i in range(1,h+1):
if i%2 == 0:
print(ans[w*(i-1):w*(i-1)+w])
else:
print(ans[w*(i-1):w*(i-1)+w][::-1])
if __name__=="__main__":
main()
|
s125573206
|
Accepted
| 37
| 10,100
| 1,016
|
import math
from math import gcd,pi,sqrt
INF = float("inf")
import sys
sys.setrecursionlimit(10**6)
import itertools
from collections import Counter,deque
def i_input(): return int(input())
def i_map(): return map(int, input().split())
def i_list(): return list(i_map())
def i_row(N): return [i_input() for _ in range(N)]
def i_row_list(N): return [i_list() for _ in range(N)]
def s_input(): return input()
def s_map(): return input().split()
def s_list(): return list(s_map())
def s_row(N): return [s_input for _ in range(N)]
def s_row_str(N): return [s_list() for _ in range(N)]
def s_row_list(N): return [list(s_input()) for _ in range(N)]
def main():
h,w = i_map()
n = i_input()
a = i_list()
ans = []
for m,i in enumerate(a):
ans.extend([str(m+1)]*i)
ans = list(ans)
for i in range(1,h+1):
if i%2 == 0:
print(" ".join(ans[w*(i-1):w*(i-1)+w]))
else:
print(" ".join(ans[w*(i-1):w*(i-1)+w][::-1]))
if __name__=="__main__":
main()
|
s542722020
|
p03545
|
u532031128
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 381
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
s = input()
A, B, C, D = int(s[0]), int(s[1]), int(s[2]), int(s[3])
print(A,B,C,D)
N = 3
for i in range(2**N):
sum = int(s[0])
f = s[0]
for j in range(N):
if i>>j & 1:
sum += int(s[j+1])
f += "+" + s[j+1]
else:
sum -= int(s[j+1])
f += "-" + s[j+1]
if sum == 7:
print(f + "=7")
break
|
s600470236
|
Accepted
| 18
| 3,064
| 366
|
s = input()
A, B, C, D = int(s[0]), int(s[1]), int(s[2]), int(s[3])
N = 3
for i in range(2**N):
sum = int(s[0])
f = s[0]
for j in range(N):
if i>>j & 1:
sum += int(s[j+1])
f += "+" + s[j+1]
else:
sum -= int(s[j+1])
f += "-" + s[j+1]
if sum == 7:
print(f + "=7")
break
|
s406518721
|
p03110
|
u547608423
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 180
|
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
N=int(input())
answer=0
for i in range(N):
x,u=input().split()
print(x,u)
if u=="JPY":
answer+=int(x)
else:
answer+=float(x)*380000
print(answer)
|
s347225772
|
Accepted
| 17
| 2,940
| 181
|
N=int(input())
answer=0
for i in range(N):
x,u=input().split()
#print(x,u)
if u=="JPY":
answer+=int(x)
else:
answer+=float(x)*380000
print(answer)
|
s267494997
|
p03162
|
u816631826
| 2,000
| 1,048,576
|
Wrong Answer
| 370
| 3,060
| 128
|
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
a=int(input())
hahaha=0
for i in range(a):
r=str(input()).split()
m=[int(u) for u in r]
hahaha+=max(m)
print(hahaha)
|
s943774572
|
Accepted
| 570
| 24,308
| 372
|
if __name__ == '__main__':
n = int(input())
res = [[0 for i in range(0, 3)] for i in range(0, n+1)]
for i in range(1,n+1):
x, y, z = map(int,input().split())
res[i][0] = max(res[i-1][1], res[i-1][2])
res[i][1] = max(res[i-1][0], res[i-1][2])
res[i][2] = max(res[i-1][0], res[i-1][1])
res[i][0]+=x
res[i][1]+=y
res[i][2]+=z
r = max(res[-1])
print(r)
|
s921831180
|
p03478
|
u865741247
| 2,000
| 262,144
|
Wrong Answer
| 35
| 3,188
| 196
|
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).
|
a,b,c=input().split(" ")
nums=[]
ans=0
b=int(b)
c=int(c)
a=int(a)
for i in range(a):
nums.append(sum(list(map(int,str(i)))))
for j,n in enumerate(nums):
if b<=n and n<=c:
ans+=j
print(ans)
|
s354171412
|
Accepted
| 35
| 3,188
| 198
|
a,b,c=input().split(" ")
nums=[]
ans=0
b=int(b)
c=int(c)
a=int(a)
for i in range(a+1):
nums.append(sum(list(map(int,str(i)))))
for j,n in enumerate(nums):
if b<=n and n<=c:
ans+=j
print(ans)
|
s746065987
|
p02409
|
u179070318
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,604
| 418
|
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
n = int(input())
info = []
for _ in range(n):
temp = [int(x) for x in input().split( )]
info.append(temp)
state = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]
for a in info:
state[a[0]-1][a[1]-1][a[2]-1] += a[3]
for b in range(4):
for f in range(3):
string = ''
for k in range(10):
string += ' '+str(state[b][f][k])
print(string)
print('#'*20)
|
s285361301
|
Accepted
| 20
| 5,608
| 439
|
n = int(input())
info = []
for _ in range(n):
temp = [int(x) for x in input().split( )]
info.append(temp)
state = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]
for a in info:
state[a[0]-1][a[1]-1][a[2]-1] += a[3]
for b in range(4):
for f in range(3):
string = ''
for k in range(10):
string += ' ' + str(state[b][f][k])
print(string)
if b < 3:
print('#'*20)
|
s404858149
|
p04025
|
u733337827
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 321
|
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
sn = input()
xyx = xx = -1
for i in range(1, len(sn)):
if (xx == -1) and (sn[i-1] == sn[i]):
xx = i-1 if xyx == -1 else i
if (xyx == -1) and (i > 2) and (sn[i-2] == sn[i]):
xyx = i-2 if xx == -1 else i
if (xx == -1) or (xyx == -1):
print(-1, -1)
else:
print(min(xx, xyx)+1, max(xx, xyx)+1)
|
s307444356
|
Accepted
| 18
| 2,940
| 172
|
N = int(input())
an = list(map(int, input().split()))
ave = sum(an)/N
rep = int(ave + 0.5) if ave > 0 else int(ave - 0.5)
ans = sum([(a - rep) ** 2 for a in an])
print(ans)
|
s181801980
|
p03943
|
u366644013
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 127
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
na = lambda: list(map(int, input().split()))
a, b, c = na()
if a + b == c or b + c == a:
print("Yes")
else:
print("No")
|
s181489652
|
Accepted
| 17
| 2,940
| 141
|
na = lambda: list(map(int, input().split()))
a, b, c = na()
if a + b == c or b + c == a or c + a == b:
print("Yes")
else:
print("No")
|
s474250156
|
p03854
|
u131464432
| 2,000
| 262,144
|
Wrong Answer
| 40
| 9,584
| 316
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S = input()
A = "".join(list(reversed(S)))
print(A)
i = 0
while i != len(S):
if A[i:i+5] == "maerd":
i += 5
elif A[i:i+5] == "esare":
i += 5
elif A[i:i+7] == "remaerd":
i += 7
elif A[i:i+6] == "resare":
i += 6
else:
print("NO")
exit()
print("YES")
|
s127081809
|
Accepted
| 41
| 9,740
| 307
|
S = input()
A = "".join(list(reversed(S)))
i = 0
while i != len(S):
if A[i:i+5] == "maerd":
i += 5
elif A[i:i+5] == "esare":
i += 5
elif A[i:i+7] == "remaerd":
i += 7
elif A[i:i+6] == "resare":
i += 6
else:
print("NO")
exit()
print("YES")
|
s038042169
|
p03502
|
u655975843
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 129
|
An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10. Given an integer N, determine whether it is a Harshad number.
|
n = input()
N = list(n)
n = int(n)
ans = 0
for i in N:
ans += int(i)
if ans % n == 0:
print('Yes')
else:
print('No')
|
s172231747
|
Accepted
| 18
| 2,940
| 129
|
n = input()
N = list(n)
n = int(n)
ans = 0
for i in N:
ans += int(i)
if n % ans == 0:
print('Yes')
else:
print('No')
|
s447094102
|
p03477
|
u475675023
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 93
|
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
a,b,c,d=map(int,input().split())
print(["Left" if a+b>c+d else "Right","Balanced"][a+b!=c+d])
|
s736467779
|
Accepted
| 17
| 2,940
| 94
|
a,b,c,d=map(int,input().split())
print(["Left" if a+b>c+d else "Right","Balanced"][a+b==c+d])
|
s714748220
|
p03525
|
u780962115
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,064
| 1,436
|
In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi. Takahashi checked and found that the _time gap_ (defined below) between the local times in his city and the i-th person's city was D_i hours. The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours. Here, we are using 24-hour notation. That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example. Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours. Find the maximum possible value of s.
|
#time gap
n=int(input())
lists=list(map(int,input().split()))
if n>24:
print(0)
else:
timelist=[0 for i in range(13)]
for i in lists:
timelist[i]+=1
flag=True
for j in range(13):
if j==0 or j==12:
if timelist[j]>1:
flag=False
else:
if timelist[j]>2:
flag=False
if flag:
uselist1=[]
uselist2_=[]
for i in range(13):
if timelist[i]==1:
uselist1.append((i))
if timelist[i]==2:
uselist2_.append((i))
else:
continue
uselist2=[]
for j in uselist2_:
uselist2.append(j)
uselist2.append(24-j)
lens=len(uselist1)
mini=100
for i in range(2**lens):
sublist=[]
sublist=sublist+uselist2
j=format(i,"0%ib"%lens)
for k in range(lens):
if j[k]=="0":
sublist.append(uselist1[k])
else:
sublist.append(24-uselist1[k])
sublist=sorted(sublist)
lenss=len(sublist)
for k in range(1,lenss):
mini=min(mini,sublist[k]-sublist[k-1])
if n==1:
print(lists[0])
else:
print(mini)
if not flag:
print(0)
|
s256482777
|
Accepted
| 23
| 3,188
| 1,546
|
#time gap
n=int(input())
lists=list(map(int,input().split()))
if n>24:
print(0)
else:
timelist=[0 for i in range(13)]
for i in lists:
timelist[i]+=1
flag=True
for j in range(13):
if j==0:
if timelist[j]>0:
flag=False
else:
if timelist[j]>2:
flag=False
if flag:
uselist1=[]
uselist2_=[]
for i in range(13):
if timelist[i]==1:
uselist1.append((i))
if timelist[i]==2:
uselist2_.append((i))
else:
continue
uselist2=[]
for j in uselist2_:
uselist2.append(j)
uselist2.append(24-j)
uselist2=list(set(uselist2))
lens=len(uselist1)
minilist=[]
for i in range(2**lens):
sublist=[0]
mini=100
sublist=sublist+uselist2
j=format(i,"0%ib"%lens)
for k in range(lens):
if j[k]=="0":
sublist.append(uselist1[k])
else:
sublist.append(24-uselist1[k])
sublists=sorted(sublist)
lenss=len(sublists)
for k in range(1,lenss):
mini=min(mini,sublists[k]-sublists[k-1])
mini=min(mini,24-sublists[-1],sublists[-1])
minilist.append(mini)
else:
print(max(minilist))
if not flag:
print(0)
|
s217271451
|
p03474
|
u366959492
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 168
|
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
a,b=map(int,input().split())
s=input()
f=0
for i in s:
if i=="-":
f+=1
if len(s)==(a+b+1) and s[a+1]=="-" and f==1:
print("Yes")
else:
print("No")
|
s271919856
|
Accepted
| 18
| 2,940
| 165
|
a,b=map(int,input().split())
s=input()
f=0
for i in s:
if i=="-":
f+=1
if len(s)==(a+b+1) and s[a]=="-" and f==1:
print("Yes")
else:
print("No")
|
s053741713
|
p03486
|
u665038048
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 151
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s = list(input().split())
t = list(input().split())
s.sort()
t.reverse()
s = ''.join(s)
t = ''.join(t)
if s < t:
print('Yes')
else:
print('No')
|
s316019867
|
Accepted
| 17
| 2,940
| 114
|
s = list(input())
t = list(input())
s.sort()
t.sort(reverse=True)
if s < t:
print('Yes')
else:
print('No')
|
s825939856
|
p02842
|
u999503965
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,112
| 77
|
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())
num=n//1.08
if num*1.08==n:
print(num)
else:
print(":(")
|
s289644465
|
Accepted
| 31
| 9,032
| 113
|
import math
n=int(input())
num=math.ceil(n/1.08)
if math.floor(num*1.08)==n:
print(num)
else:
print(":(")
|
s385204468
|
p02972
|
u281610856
| 2,000
| 1,048,576
|
Wrong Answer
| 370
| 23,560
| 1,005
|
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.
|
from heapq import heappush, heappop, heapify
from itertools import permutations, accumulate, combinations
from math import pi, ceil, floor
import numpy as np
from collections import defaultdict, deque
from operator import itemgetter
from bisect import bisect_left, bisect_right, insort_left, insort_right
import sys
input = sys.stdin.readline
# MOD = 10 ** 9 + 7
# INF = float("inf")
def main():
n = int(input())
A = list(map(int, input().split()))
cnt = [0] * n
B = deque([])
for i in range(n-1, -1, -1):
if sum(cnt[i::i+1]) % 2 == A[i]:
continue
else:
cnt[i] += 1
print(cnt)
ans = []
for i, v in enumerate(cnt):
if v > 1:
print(-1)
exit()
elif v == 1:
ans.append(i+1)
else:
continue
if len(ans) == 0:
print(len(ans))
else:
print(len(ans))
print(*ans)
if __name__ == '__main__':
main()
|
s584141168
|
Accepted
| 358
| 23,136
| 926
|
from heapq import heappush, heappop, heapify
from itertools import permutations, accumulate, combinations
from math import pi, ceil, floor
import numpy as np
from collections import defaultdict, deque
from operator import itemgetter
from bisect import bisect_left, bisect_right, insort_left, insort_right
import sys
input = sys.stdin.readline
# MOD = 10 ** 9 + 7
# INF = float("inf")
def main():
n = int(input())
A = list(map(int, input().split()))
cnt = [0] * n
B = deque([])
for i in range(n-1, -1, -1):
if sum(cnt[i::i+1]) % 2 == A[i]:
continue
else:
cnt[i] += 1
ans = []
for i, v in enumerate(cnt):
if v > 1:
print(-1)
exit()
elif v == 1:
ans.append(i+1)
else:
continue
print(len(ans))
print(*ans)
if __name__ == '__main__':
main()
|
s607750759
|
p03024
|
u481605952
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 83
|
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s = input()
if s.count('x') >= 8:
print('No')
else:
print('Yes')
|
s441963827
|
Accepted
| 18
| 2,940
| 74
|
s = input()
if s.count('x') >= 8:
print('NO')
else:
print('YES')
|
s311883184
|
p03719
|
u045953894
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 87
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a,b,c=map(int,input().split())
if c >= a and c <= b:
print('YES')
else:
print('NO')
|
s641444839
|
Accepted
| 17
| 2,940
| 77
|
a,b,c=map(int,input().split())
if a<=c<=b:
print('Yes')
else:
print('No')
|
s502053571
|
p02865
|
u546686251
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 2,940
| 136
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
N = int(input())
count = 0
for i in range(N):
for j in range(N):
if i + j == N:
count = count + 1
print(count/2)
|
s506778327
|
Accepted
| 355
| 2,940
| 118
|
N = int(input())
count = 0
for i in range(1, N):
if i != N - i and i < N/2:
count = count + 1
print(count)
|
s659009643
|
p04029
|
u389910364
| 2,000
| 262,144
|
Wrong Answer
| 157
| 13,556
| 302
|
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?
|
from collections import Counter, deque
from fractions import gcd
from functools import lru_cache
from functools import reduce
import functools
import heapq
import itertools
import math
import numpy as np
import sys
sys.setrecursionlimit(10000)
INF = float('inf')
print(math.factorial(int(input())))
|
s454514060
|
Accepted
| 19
| 3,060
| 34
|
print(sum(range(int(input())+1)))
|
s096893028
|
p02393
|
u587193722
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,564
| 54
|
Write a program which reads three integers, and prints them in ascending order.
|
x = sorted([int(i) for i in input().split()])
print(x)
|
s264409777
|
Accepted
| 30
| 7,696
| 89
|
x = sorted([int(i) for i in input().split()])
print("{0} {1} {2}".format(x[0],x[1],x[2]))
|
s129427906
|
p03351
|
u056599756
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 100
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d=map(int, input().split())
print("Yes" if abs(a-c)<d or abs(a-b)<d and abs(b-c)<d else "No")
|
s168471257
|
Accepted
| 17
| 2,940
| 103
|
a,b,c,d=map(int, input().split())
print("Yes" if abs(a-c)<=d or abs(a-b)<=d and abs(b-c)<=d else "No")
|
s103740238
|
p03377
|
u304561065
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 126
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A,B,X=map(int,input().split())
if A<=X:
if A+B<X:
print('No')
else:
print('Yes')
else:
print('No')
|
s962528506
|
Accepted
| 17
| 2,940
| 127
|
A,B,X=map(int,input().split())
if A+B>=X:
if A<=X:
print('YES')
else:
print('NO')
else:
print('NO')
|
s809604206
|
p02393
|
u120602885
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,588
| 127
|
Write a program which reads three integers, and prints them in ascending order.
|
tmp = input()
tmp = tmp.split(" ")
tmp = list(map(int,tmp))
tmp.sort()
print(tmp)
tmp = list(map(str,tmp))
print(" ".join(tmp))
|
s855765528
|
Accepted
| 30
| 7,644
| 116
|
tmp = input()
tmp = tmp.split(" ")
tmp = list(map(int,tmp))
tmp.sort()
tmp = list(map(str,tmp))
print(" ".join(tmp))
|
s052982277
|
p03555
|
u923270446
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 93
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
c1 = list(input())
c2 = list(input())
if c1.reverse == c2:
print("YES")
else:
print("NO")
|
s716355564
|
Accepted
| 17
| 2,940
| 45
|
print("YES"if input()==input()[::-1]else"NO")
|
s152736568
|
p03494
|
u099026234
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,028
| 277
|
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.
|
a = int(input())
b = [int(i) for i in input().split(" ")]
count = 0
iti = 1
while iti == 1:
for i in range(a):
if b[i]%2 == 0:
b[i]=b[i]//2
iti = 1
else :
print(count)
iti = 0
break
count = count+1
|
s646371860
|
Accepted
| 30
| 9,128
| 281
|
a = int(input())
b = [int(i) for i in input().split(" ")]
count = 0
iti = 1
while iti == 1:
for i in range(a):
if b[i]%2 == 0:
b[i]=b[i]//2
iti = 1
else :
print(count)
iti = 0
break
count = count+1
|
s248677300
|
p02841
|
u698849142
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 98
|
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())
print(m1 == m2 if 1 else 0)
|
s022564343
|
Accepted
| 17
| 2,940
| 98
|
m1, d1 = map(int, input().split())
m2, d2 = map(int, input().split())
print(0 if m1 == m2 else 1)
|
s401339636
|
p03251
|
u715355213
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 482
|
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(map(int,input().split()))
y = list(map(int,input().split()))
flag = 0
Z_max = min(y)
Z_min = max(x)
print(Z_min,Z_max)
Z = X + 1
while Z < Y:
if Z_min < Z and Z <= Z_max:
break
Z += 1
print(Z)
for i in range(N):
if x[i] >= Z:
flag = 1
break
if flag == 0:
for i in range(M):
if y[i] < Z:
flag = 1
break
if flag == 0:
print('No War')
else:
print('War')
|
s966338925
|
Accepted
| 17
| 3,064
| 454
|
N,M,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
flag = 0
Z_max = min(y)
Z_min = max(x)
Z = X + 1
while Z < Y:
if Z_min < Z and Z <= Z_max:
break
Z += 1
for i in range(N):
if x[i] >= Z:
flag = 1
break
if flag == 0:
for i in range(M):
if y[i] < Z:
flag = 1
break
if flag == 0:
print('No War')
else:
print('War')
|
s028125401
|
p03433
|
u332420380
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 89
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
n = int(input())
a = int(input())
if n % 500 <= a:
print('YES')
else:
print('NO')
|
s459982635
|
Accepted
| 17
| 2,940
| 89
|
n = int(input())
a = int(input())
if n % 500 <= a:
print('Yes')
else:
print('No')
|
s047178909
|
p04044
|
u666964944
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,160
| 80
|
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())
s = [input() for _ in range(n)]
print(sorted(s))
|
s399562488
|
Accepted
| 26
| 9,104
| 89
|
n,l = map(int, input().split())
s = [input() for _ in range(n)]
print(''.join(sorted(s)))
|
s354554761
|
p03351
|
u672220554
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 158
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a,b,c,d=map(int,input().split())
if abs(c-a)<=d:
print("Yes")
else:
if abs(b-a)<=d and abs(c-a)<=d:
print("Yes")
else:
print("No")
|
s978167593
|
Accepted
| 17
| 2,940
| 158
|
a,b,c,d=map(int,input().split())
if abs(c-a)<=d:
print("Yes")
else:
if abs(b-a)<=d and abs(c-b)<=d:
print("Yes")
else:
print("No")
|
s164186792
|
p03485
|
u150641538
| 2,000
| 262,144
|
Wrong Answer
| 20
| 2,940
| 52
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b = map(int,input().split())
print(int((a+b)/2)+1)
|
s486278643
|
Accepted
| 17
| 2,940
| 63
|
a,b = map(int,input().split())
print(int((a+b+((a+b)%2==1))/2))
|
s577521235
|
p03658
|
u257541375
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 146
|
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 = [int(i) for i in input().split()]
array = [int(i) for i in input().split()]
array.sort()
sum = 0
for i in range(k):
sum += array[-(i+1)]
|
s174315140
|
Accepted
| 18
| 2,940
| 158
|
n,k = [int(i) for i in input().split()]
array = [int(i) for i in input().split()]
array.sort()
sum = 0
for i in range(k):
sum += array[-(i+1)]
print(sum)
|
s488375820
|
p03385
|
u273201018
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 67
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
S = input()
if S == 'abc':
print('Yes')
else:
print('No')
|
s601841069
|
Accepted
| 17
| 2,940
| 107
|
S = input()
if 'a' in S and 'b' in S and 'c' in S and len(S) == 3:
print('Yes')
else:
print('No')
|
s327176041
|
p02612
|
u189188797
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 8,976
| 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.
|
n=int(input())
while n>=1000:
n-=1000
print(abs(n))
|
s573429178
|
Accepted
| 27
| 9,084
| 51
|
n=int(input())
while n>0:
n-=1000
print(abs(n))
|
s839306424
|
p03478
|
u770987902
| 2,000
| 262,144
|
Wrong Answer
| 25
| 2,940
| 241
|
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 = [int(x) for x in input().split()]
total = 0
def func(num):
r = 0
while num > 0:
r = num % 10
num = num // 10
return r
for i in range(1, n+1):
num = func(i)
if num >= a and num <= b:
total += num
print(total)
|
s418528776
|
Accepted
| 25
| 2,940
| 237
|
n,a,b = [int(x) for x in input().split()]
def func(num):
r = 0
while num > 0:
r += num % 10
num = num // 10
return r
total = 0
for i in range(1, n+1):
num = func(i)
if num >= a and num <= b:
total += i
print(total)
|
s135483600
|
p03494
|
u428397309
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 209
|
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.
|
# -*- coding: utf-8 -*-
N = int(input())
A = list(map(int, input().split()))
ans = 0
while 1:
new_a = [a // 2 if a % 2 == 0 else -1 for a in A]
if -1 in new_a:
break
ans += 1
print(ans)
|
s109207327
|
Accepted
| 18
| 2,940
| 250
|
# -*- coding: utf-8 -*-
N = int(input())
A = list(map(int, input().split()))
ans = 0
new_a = A.copy()
while 1:
new_a = [a // 2 if a % 2 == 0 else -1 for a in new_a]
if -1 in new_a:
print(ans)
exit()
ans += 1
print(ans)
|
s282461841
|
p04011
|
u586577600
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 101
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
n, k, x, y = [int(input()) for i in range(4)]
if n <= k: print(x*n)
else: print(x*n + max(0, n-k)*y)
|
s566433652
|
Accepted
| 17
| 2,940
| 101
|
n, k, x, y = [int(input()) for i in range(4)]
if n <= k: print(x*n)
else: print(x*k + max(0, n-k)*y)
|
s855946267
|
p03827
|
u290187182
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 197
|
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
if __name__ == '__main__':
a = int(input())
s = input()
count =0
for i in s:
if i == "I":
count +=1
else:
count -=1
print(count)
|
s498896492
|
Accepted
| 17
| 2,940
| 248
|
if __name__ == '__main__':
a = int(input())
s = input()
count =0
max = 0
for i in s:
if i == "I":
count +=1
else:
count -=1
if max < count:
max = count
print(max)
|
s142845827
|
p03434
|
u738622346
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 385
|
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.
|
def draw_max_card(card_list):
max_card = max(card_list)
card_list.remove(max_card)
return max_card
n = int(input())
card_list = list(map(int, input().split(" ")))
points_alice = 0
points_bob = 0
for count in range(1, n + 1):
print(count)
if count % 2 != 0:
points_alice += draw_max_card(card_list)
else:
points_bob += draw_max_card(card_list)
|
s959671425
|
Accepted
| 20
| 3,188
| 402
|
def draw_max_card(card_list):
max_card = max(card_list)
card_list.remove(max_card)
return max_card
n = int(input())
card_list = list(map(int, input().split(" ")))
points_alice = 0
points_bob = 0
for count in range(1, n + 1):
if count % 2 != 0:
points_alice += draw_max_card(card_list)
else:
points_bob += draw_max_card(card_list)
print(points_alice - points_bob)
|
s847329877
|
p03944
|
u394244719
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,132
| 694
|
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.
|
import sys
import math
inint = lambda: int(sys.stdin.readline())
inintm = lambda: map(int, sys.stdin.readline().split())
inintl = lambda: list(inintm())
instrm = lambda: map(str, sys.stdin.readline().split())
instrl = lambda: list(instrm())
w, h, n = inintm()
ll = [0,0]
lh = [0,h]
rl = [w,0]
rh = [w,h]
for i in range(n):
x, y, a = inintm()
if a == 1:
ll[0] = max(ll[0], x)
lh[0] = max(lh[0], x)
elif a == 2:
rl[0] = min(rl[0], x)
rh[0] = min(rh[0], x)
elif a == 3:
rl[1] = max(rl[1], y)
ll[1] = max(ll[1], y)
else:
rh[1] = min(rh[1], y)
lh[1] = min(lh[1], y)
print(min((rl[0]-ll[0])*(lh[1]-ll[1]), 0))
|
s770813613
|
Accepted
| 31
| 9,180
| 606
|
import sys
import math
inint = lambda: int(sys.stdin.readline())
inintm = lambda: map(int, sys.stdin.readline().split())
inintl = lambda: list(inintm())
instrm = lambda: map(str, sys.stdin.readline().split())
instrl = lambda: list(instrm())
w, h, n = inintm()
ll = [0,0]
rh = [w,h]
for i in range(n):
x, y, a = inintm()
if a == 1:
ll[0] = max(ll[0], x)
elif a == 2:
rh[0] = min(rh[0], x)
elif a == 3:
ll[1] = max(ll[1], y)
else:
rh[1] = min(rh[1], y)
if rh[0]-ll[0] < 0 or rh[1]-ll[1] < 0:
print(0)
else:
print((rh[0]-ll[0])*(rh[1]-ll[1]))
|
s757471046
|
p03435
|
u393253137
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 435
|
We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j. Determine if he is correct.
|
c = [list(map(int, input().split())) for _ in range(3)]
if c[0][1] - c[0][0] == c[1][1] - c[1][0] == c[2][1] - c[2][0]:
if c[0][2] - c[0][1] == c[1][2] - c[1][1] == c[2][2] - c[2][1]:
print('Yes')
print('No')
|
s605037695
|
Accepted
| 18
| 2,940
| 202
|
a, b, c = map(int, input().split())
d, e, f = map(int, input().split())
g, h, i = map(int, input().split())
if a - b == d - e == g - h and b - c == e - f == h - i:
print('Yes')
else:
print('No')
|
s954778587
|
p03853
|
u290187182
| 2,000
| 262,144
|
Wrong Answer
| 36
| 5,144
| 389
|
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
import sys
import copy
import math
import bisect
import pprint
import bisect
from functools import reduce
from copy import deepcopy
from collections import deque
from decimal import *
def lcm(x, y):
return (x * y) // math.gcd(x, y)
if __name__ == '__main__':
b, c = map(int, input().split())
for i in range(b):
s = input()
print(s+s)
print(s+s)
|
s874752808
|
Accepted
| 35
| 5,144
| 385
|
import sys
import copy
import math
import bisect
import pprint
import bisect
from functools import reduce
from copy import deepcopy
from collections import deque
from decimal import *
def lcm(x, y):
return (x * y) // math.gcd(x, y)
if __name__ == '__main__':
b, c = map(int, input().split())
for i in range(b):
s = input()
print(s)
print(s)
|
s110317457
|
p03069
|
u286955577
| 2,000
| 1,048,576
|
Wrong Answer
| 41
| 3,884
| 364
|
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
def solve():
N = int(input())
S = input()
foundBlack = False
black, b, white = 0, 0, 0
for stone in S:
if stone == '#':
if not foundBlack: foundBlack = True
b = b + 1
else:
if not foundBlack: continue
white = white + 1
black = black + b
b = 0
print(white, black)
return min(black, white)
print(solve())
|
s802680111
|
Accepted
| 54
| 5,864
| 473
|
def solve():
N = int(input())
S = input()
stones = []
look = S[0]
cnt = 0
if S[0] == '.': stones.append(0)
for s in S:
if look == s: cnt += 1
else:
stones.append(cnt)
cnt = 1
look = s
stones.append(cnt)
if S[N - 1] == '#': stones.append(0)
count = sum(stones[1::2])
ans = count
for i in range(0, len(stones), 2):
count = count + stones[i] - stones[i + 1]
if ans > count: ans = count
return ans
print(solve())
|
s123948812
|
p03139
|
u303633930
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 319
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
import sys
def r(n, a, b):
both_max = min(a, b)
if 0 < a + b - n:
both_min = a + b - n
else:
both_min = 0
print(both_max, both_min)
def main():
input = sys.stdin.readline
n, a, b = list(map(int, input().split()))
print(r(n, a, b))
if __name__ == '__main__':
main()
|
s193743880
|
Accepted
| 17
| 3,060
| 312
|
import sys
def r(n, a, b):
both_max = min(a, b)
if 0 < a + b - n:
both_min = a + b - n
else:
both_min = 0
print(both_max, both_min)
def main():
input = sys.stdin.readline
n, a, b = list(map(int, input().split()))
r(n, a, b)
if __name__ == '__main__':
main()
|
s427312275
|
p03998
|
u898967808
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 255
|
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
dic_s = {x:input() for x in 'abc'}
dic_i = {x:i for x,i in zip('abc',[0,0,0])}
dic_max = {x:len(dic_s[x]) for x in 'abc'}
nt = 'a'
while True:
dic_i[nt] += 1
if dic_max[nt] == dic_i[nt]:
break
else:
nt = dic_s[nt][dic_i[nt]]
print(nt)
|
s820033458
|
Accepted
| 17
| 3,064
| 273
|
dic_s = {x:input() for x in 'abc'}
dic_i = {x:i for x,i in zip('abc',[0,0,0])}
dic_max = {x:len(dic_s[x]) for x in 'abc'}
nt = 'a'
while True:
if dic_max[nt] == dic_i[nt]:
break
else:
dic_i[nt] += 1
nt = dic_s[nt][dic_i[nt]-1]
print(nt.upper())
|
s437857035
|
p02534
|
u922769680
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,144
| 33
|
You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`.
|
n=int(input())
w="ALC"
print(w*n)
|
s305986103
|
Accepted
| 21
| 9,148
| 33
|
n=int(input())
w="ACL"
print(w*n)
|
s491950617
|
p03679
|
u032222383
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 119
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
x, a, b=map(int, input().split())
if a>=b:
print("delicious")
elif a+x<=b:
print("safe")
else:
print("dangerous")
|
s992679404
|
Accepted
| 18
| 2,940
| 120
|
x, a, b=map(int, input().split())
if a>=b:
print("delicious")
elif a+x>=b:
print("safe")
else:
print("dangerous")
|
s758388166
|
p03401
|
u089032001
| 2,000
| 262,144
|
Wrong Answer
| 138
| 14,048
| 322
|
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.
|
N = int(input())
A = list(map(int, input().split()))
min1 = 0
min2 = 0
max1 = 0
max2 = 0
for a in A:
if(a <= min1):
min2 = min1
min1 = a
elif(a >= max1):
max2 = max1
max1 = a
for a in A:
if(a == min1):
print(max1 - min2)
elif(a == max1):
print(max2 - min1)
else:
print(max1 - min1)
|
s746566339
|
Accepted
| 220
| 14,048
| 291
|
N = int(input())
A = list(map(int, input().split()))
A.append(0)
all = 0
last = 0
diff = []
for i in range(N):
all += abs(A[i] - last)
diff.append(abs(A[i+1] - last) - abs(A[i+1] - A[i]) - abs(A[i] - last))
last = A[i]
all += abs(A[N-1])
for i in range(N):
print(all + diff[i])
|
s808637621
|
p03149
|
u731368968
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 116
|
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
n = list(map(int, input().split()))
n.sort()
if n[0] + n[1] + n[2] + n[3] == 1479:
print('YES')
else:print('NO')
|
s584004112
|
Accepted
| 17
| 2,940
| 102
|
n = input().split()
n.sort()
if n[0] + n[1] + n[2] + n[3] == '1479':
print('YES')
else:print('NO')
|
s599837569
|
p03385
|
u663438907
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 224
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
A = list(input())
l = [0, 0, 0]
for i in range(len(A)):
if A[i] == 'A':
l[0] = 1
elif A[i] == 'B':
l[1] = 1
else:
l[2] = 1
if l[0] == 1 and l[1] == 1 and l[2] == 1:
print('Yes')
else:
print('No')
|
s648690714
|
Accepted
| 17
| 3,064
| 226
|
A = list(input())
l = [0, 0, 0]
for i in range(len(A)):
if A[i] == 'a':
l[0] = 1
elif A[i] == 'b':
l[1] = 1
else:
l[2] = 1
if l[0] == 1 and l[1] == 1 and l[2] == 1:
print('Yes')
else:
print('No')
|
s560320620
|
p03599
|
u201234972
| 3,000
| 262,144
|
Wrong Answer
| 75
| 3,064
| 490
|
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
a, b, c, d, e, f = map( int, input().split())
ans = 0
ansL = 0
ansM = 0
for i in range(30//a+1):
for j in range((30-i)//b+1):
L = a*i + b*j
M = 0
for k in range(f):
if 100*L + k*c <= f:
m = min(((e*L-k*c)//d)*d, ((f-100*L-k*c)//d)*d)
M = max(M, m+k*c)
if M+L != 0 and 100*L + M <= f:
if (M*100)/(M+L*100) > ans:
ansL = L*100
ansM = M
print('{} {}'.format(ansL,ansM))
|
s142685174
|
Accepted
| 70
| 3,064
| 622
|
a, b, c, d, e, f = map( int, input().split())
ans = 0
ansL = 0
ansM = 0
for i in range(30//a+1):
for j in range((30-i)//b+1):
L = a*i + b*j
M = 0
for k in range(f):
if 100*L + k*c <= f and k*c <= L*e:
m = min(((e*L-k*c)//d)*d, ((f-100*L-k*c)//d)*d)
M = max(M, m+k*c)
if M+L != 0 and 100*L + M <= f and M <= L*e:
if (M*100)/(M+L*100) > ans:
ans = (M*100)/(M+L*100)
ansL = L*100
ansM = M
if ansL == 0:
print('{} {}'.format(a*100,0))
else:
print('{} {}'.format(ansL+ansM,ansM))
|
s779158159
|
p02266
|
u462831976
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,304
| 679
|
Your task is to simulate a flood damage. For a given cross-section diagram, reports areas of flooded sections. Assume that rain is falling endlessly in the region and the water overflowing from the region is falling in the sea at the both sides. For example, for the above cross-section diagram, the rain will create floods which have areas of 4, 2, 1, 19 and 9 respectively.
|
# -*- coding: utf-8 -*-
import sys
import os
s = input().strip()
S1 = [] # sum of pond
S2 = [] # each pond
area = 0
for i, c in enumerate(s):
if c == '\\':
S1.append(i)
elif c == '_':
pass
elif c == '/' and S1:
j = S1.pop(-1)
area += i - j
# for each pond
pond_sum = i - j
while S2 and S2[-1][0] > j:
pond_sum += S2[-1][1]
S2.pop(-1)
S2.append([j, pond_sum])
pond_areas = []
for data in S2:
pond_areas.append(data[1])
# answer
print(area)
print(*pond_areas)
|
s844779441
|
Accepted
| 30
| 7,988
| 696
|
# -*- coding: utf-8 -*-
import sys
import os
s = input().strip()
S1 = [] # sum of pond
S2 = [] # each pond
area = 0
for i, c in enumerate(s):
if c == '\\':
S1.append(i)
elif c == '_':
pass
elif c == '/' and S1:
j = S1.pop(-1)
area += i - j
# for each pond
pond_sum = i - j
while S2 and S2[-1][0] > j:
pond_sum += S2[-1][1]
S2.pop(-1)
S2.append([j, pond_sum])
pond_areas = []
for data in S2:
pond_areas.append(data[1])
# answer
print(area)
print(len(pond_areas), *pond_areas)
|
s173523953
|
p03469
|
u368796742
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 65
|
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.
|
a,b,c = map(int,input().split("/"))
print("2018/01/{}".format(c))
|
s991825844
|
Accepted
| 17
| 2,940
| 69
|
a,b,c = map(int,input().split("/"))
print("2018/01/{:02}".format(c))
|
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