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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s384148001
|
p00005
|
u777299405
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,720
| 200
|
Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b.
|
def gcd(a, b):
while b:
a, b = b, a % b
return a
while True:
try:
a, b = map(int, (input.split()))
except:
break
g = gcd(a, b)
print(g, a * b // g)
|
s592393420
|
Accepted
| 30
| 6,724
| 198
|
def gcd(a, b):
while b:
a, b = b, a % b
return a
while True:
try:
a, b = map(int, input().split())
print(gcd(a, b), a * b // gcd(a, b))
except:
break
|
s305169511
|
p03457
|
u488737015
| 2,000
| 262,144
|
Wrong Answer
| 1,091
| 4,468
| 455
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
n = int(input())
previous_t = 0
for i in range(n):
t, x, y = list(map(int, input().split()))
print(t)
print(x)
print(y)
if x + y > t:
print("No")
break
if (x + y) % 2 != t % 2:
print("No")
break
if previous_t > 0 and (abs(x - previous_x) + abs(y - previous_y) > (t - previous_t)):
print("No")
break
previous_t = t
previous_x = x
previous_y = y
else:
print("Yes")
|
s596635541
|
Accepted
| 438
| 3,060
| 414
|
n = int(input())
previous_t = 0
ans = "Yes"
for i in range(n):
t, x, y = list(map(int, input().split()))
if x + y > t:
ans = "No"
break
if (x + y) % 2 != t % 2:
ans = "No"
break
if previous_t > 0 and (abs(x - previous_x) + abs(y - previous_y) > (t - previous_t)):
ans = "No"
break
previous_t = t
previous_x = x
previous_y = y
print(ans)
|
s241509275
|
p03635
|
u119655368
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?
|
s = input()
print(s[0] + str(len(s[1:-1])) + s[-1])
|
s256531185
|
Accepted
| 17
| 2,940
| 62
|
a,b = map(int,input().split())
print((int(a)-1) * (int(b)-1) )
|
s863553661
|
p03139
|
u214866184
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 83
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
n, a, b = map(int, input().split())
print('{} {}'.format(max(a, b), max(0, n-a-b)))
|
s765125762
|
Accepted
| 18
| 2,940
| 83
|
n, a, b = map(int, input().split())
print('{} {}'.format(min(a, b), max(0, a+b-n)))
|
s222582781
|
p03720
|
u497040007
| 2,000
| 262,144
|
Wrong Answer
| 28
| 3,444
| 288
|
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
from collections import OrderedDict
n, m = map(int, input().split())
info = OrderedDict()
for i in range(0, n + 1):
info[i] = 0
for i in range(0, m):
a, b = map(int, input().split())
info[a] = info[a] + 1
info[b] = info[b] + 1
for value in info.values():
print(value)
|
s149245847
|
Accepted
| 17
| 2,940
| 142
|
n, m = map(int, input().split())
l = list()
[l.extend(input().split()) for i in range(0, m)]
[print(l.count(str(i))) for i in range(1, n + 1)]
|
s644895335
|
p02694
|
u265118937
| 2,000
| 1,048,576
|
Time Limit Exceeded
| 2,205
| 9,076
| 86
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
x = int(input())
t = 0
i = 0
while t < x:
t = int(t*(1.01))
i += 1
print(ans)
|
s918061320
|
Accepted
| 20
| 9,156
| 87
|
x = int(input())
t = 100
i = 0
while t < x:
t = int(t*(1.01))
i += 1
print(i)
|
s974322487
|
p02399
|
u009101629
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,616
| 123
|
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
lis = [int(x) for x in input().split()]
print(str(lis[0]//lis[1]) + " " + str(lis[0] % lis[1]) + " " + str(lis[0]/lis[1]))
|
s429781446
|
Accepted
| 20
| 5,620
| 148
|
lis = [int(x) for x in input().split()]
print(lis[0]//lis[1],end = " ")
print(lis[0]%lis[1],end = " ")
a = lis[0]/lis[1]
print("{0:.5f}".format(a))
|
s886618397
|
p03588
|
u813174766
| 2,000
| 262,144
|
Wrong Answer
| 285
| 3,060
| 117
|
A group of people played a game. All players had distinct scores, which are positive integers. Takahashi knows N facts on the players' scores. The i-th fact is as follows: the A_i-th highest score among the players is B_i. Find the maximum possible number of players in the game.
|
n=int(input())
a=0
ans=0
for i in range(n):
x,y=map(int,input().split())
if a<x:
ans=x+y-1
a=x
print(ans)
|
s956961612
|
Accepted
| 285
| 3,060
| 116
|
n=int(input())
a=0
ans=0
for i in range(n):
x,y=map(int,input().split())
if a<x:
ans=x+y
a=x
print(ans)
|
s642751579
|
p03352
|
u410118019
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 35
|
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())
print(int(x**0.5))
|
s886226242
|
Accepted
| 18
| 2,940
| 138
|
x = int(input())
m = 1
for i in range(2,(x+1)//2):
tmp = 0
k = 0
while tmp <= x:
m = max(m,tmp)
tmp = i**k
k+=1
print(m)
|
s663318655
|
p03433
|
u221345507
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 87
|
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-A)%500==0:
print("Yes")
else:
print ("No")
|
s000402389
|
Accepted
| 79
| 2,940
| 176
|
N=int(input())
A=int(input())
import sys
for i in range (0,1000):
for j in range(A+1):
if N==500*i+j:
print ('Yes')
sys.exit()
print ('No')
|
s280215949
|
p03695
|
u757584836
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 214
|
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
n = int(input())
v = list(map(int, input().split()))
c = [0] * 9
for i in range(0, n):
if v[i] >= 3200:
c[8] += 1
else:
c[v[i]//400] += 1
a = 0
for i in range(0, 8):
a += c[i]
print(a, min(a, a+c[8]))
|
s523526845
|
Accepted
| 17
| 3,064
| 228
|
n = int(input())
v = list(map(int, input().split()))
c = [0] * 9
for i in range(0, n):
if v[i] >= 3200:
c[8] += 1
else:
c[v[i]//400] += 1
a = 0
for i in range(0, 8):
if c[i] > 0:
a += 1
print(max(a, 1), a+c[8])
|
s311966411
|
p03360
|
u163449343
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 201
|
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?
|
x= list(map(int,input().split()))
k = int(input())
max = 0
for f in range(3):
temp = 0
for i in range(k):
temp = temp + x[f] * 2
if temp > max:
max = temp
print(max)
|
s356948065
|
Accepted
| 20
| 2,940
| 83
|
a = list(map(int, input().split()))
print((sum(a)-max(a))+max(a)*(2**int(input())))
|
s213375657
|
p03480
|
u505830998
| 2,000
| 262,144
|
Wrong Answer
| 26
| 4,776
| 674
|
You are given a string S consisting of `0` and `1`. Find the maximum integer K not greater than |S| such that we can turn all the characters of S into `0` by repeating the following operation some number of times. * Choose a contiguous segment [l,r] in S whose length is at least K (that is, r-l+1\geq K must be satisfied). For each integer i such that l\leq i\leq r, do the following: if S_i is `0`, replace it with `1`; if S_i is `1`, replace it with `0`.
|
# -*- coding: utf-8 -*-
import sys
input = sys.stdin.readline
def io_generator():
return input()
#+++++++++++++++++++
def main(io):
s=list(io())
if len(s)% 2==1:
center=len(s)//2
mm=s[center]
ret=center
for ca,cb in zip(s[:center-1:-1],s[center+1:]):
#print(ca,cb)
if ca == mm and cb == mm:
ret += 1
else:
break
else:
center=len(s)//2
ret=center
mm=-1
la=s[center:-1]
lb=s[center:]
for ca, cb in zip(la, lb):
#print(mm,ca,cb, center)
if (mm == ca or mm == -1) and ca == cb:
ret+=1
mm=ca
else:
break
return ret
#++++++++++++++++++++
if __name__ == "__main__":
io= lambda : io_generator()
print (main(io))
|
s786459784
|
Accepted
| 26
| 4,776
| 723
|
# -*- coding: utf-8 -*-
import sys
input = sys.stdin.readline
def io_generator():
return input()
#+++++++++++++++++++
def main(io):
s=list(io())[:-1]
if len(s)% 2==1:
center=len(s)//2
mm=s[center]
ret=center+1
la=s[center-1::-1]
lb=s[center+1:]
for ca,cb in zip(la,lb):
#print(ca,cb,mm)
if ca == mm and cb == mm:
ret += 1
else:
break
else:
center=len(s)//2
ret=center
mm=-1
la=s[center-1::-1]
lb=s[center:]
for ca, cb in zip(la, lb):
if (mm == ca or mm == -1) and ca == cb:
ret+=1
mm=ca
else:
break
if 0:
print(la)
print(lb)
print(mm)
return ret
#++++++++++++++++++++
if __name__ == "__main__":
io= lambda : io_generator()
print (main(io))
|
s134187014
|
p03543
|
u198062737
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 76
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
if len(set(int(i) for i in input())) < 2:
print("Yes")
else:
print("No")
|
s086515570
|
Accepted
| 18
| 3,060
| 117
|
N = input()
if (N[0] == N[1] and N[1] == N[2]) or (N[1] == N[2] and N[2] == N[3]):
print("Yes")
else:
print("No")
|
s612505707
|
p02255
|
u487861672
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,604
| 347
|
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
#! /usr/local/bin/python3
# coding: utf-8
def insertion_sort(a):
for i in range(1, len(a)):
w = a[i]
j = i - 1
while j >= 0 and w < a[j]:
a[j + 1] = a[j]
j -= 1
a[j + 1] = w
print(" ".join(map(str, a)))
n = int(input())
a = [int(i) for i in input().split()]
insertion_sort(a)
|
s088287509
|
Accepted
| 20
| 5,612
| 376
|
#! /usr/local/bin/python3
# coding: utf-8
def insertion_sort(a):
for i in range(1, len(a)):
print(" ".join(map(str, a)))
w = a[i]
j = i - 1
while j >= 0 and w < a[j]:
a[j + 1] = a[j]
j -= 1
a[j + 1] = w
n = int(input())
a = [int(i) for i in input().split()]
insertion_sort(a)
print(" ".join(map(str, a)))
|
s311040169
|
p03469
|
u791527495
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 33
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
s=list(input())
s[3]="8"
print(s)
|
s567651798
|
Accepted
| 17
| 2,940
| 42
|
s=list(input())
s[3]="8"
print(''.join(s))
|
s997073202
|
p03455
|
u600261652
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 73
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
print("Even" if (a+b)%2 == 0 else "Odd")
|
s438477788
|
Accepted
| 17
| 2,940
| 73
|
a, b = map(int, input().split())
print("Even" if (a*b)%2 == 0 else "Odd")
|
s347281721
|
p02402
|
u825994660
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,676
| 62
|
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
a = list(map(int,input().split()))
print(min(a),max(a),sum(a))
|
s171388371
|
Accepted
| 60
| 8,608
| 73
|
input()
a = list(map(int, input().split()))
print(min(a) ,max(a) ,sum(a))
|
s964199971
|
p03494
|
u158865915
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 239
|
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.
|
input()
strs = input().split(' ')
nums = [int(i) for i in strs]
ct = 0
b_flag = False
while(1):
for num in nums:
if num % 2 == 0:
continue
else:
b_flag = True
break
if b_flag:
break
ct += 1
print(ct)
|
s539952071
|
Accepted
| 20
| 3,060
| 277
|
input()
strs = input().split(' ')
nums = [int(i) for i in strs]
ct = 0
b_flag = False
while(1):
for i, num in enumerate(nums):
if num % 2 == 0:
nums[i] = num/2
continue
else:
b_flag = True
break
if b_flag:
break
ct += 1
print(ct)
|
s852955306
|
p03469
|
u672898046
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 32
|
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.
|
i = input()
i[3] == "8"
print(i)
|
s526750352
|
Accepted
| 18
| 2,940
| 34
|
i = input()
print(i[:3]+"8"+i[4:])
|
s690497826
|
p02273
|
u890722286
| 2,000
| 131,072
|
Wrong Answer
| 20
| 7,780
| 834
|
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.
|
import math
n = int(input())
def three(p1x, p1y, p2x, p2y):
dx = (p2x - p1x)
dy = (p2y - p1y)
sx = dx / 3 + p1x
sy = dy / 3 + p1y
tx = p2x - (dx / 3)
ty = p2y - (dy / 3)
mtx = tx - sx
mty = ty - sy
rad = math.radians(60)
mux = mtx * math.cos(rad) - (mty * math.sin(rad))
muy = mtx * math.sin(rad) + (mty * math.cos(rad))
ux = mux + sx
uy = muy + sy
return [sx, sy, ux, uy, tx, ty]
def koch(p1x, p1y, p2x, p2y, n):
if 0 == n:
print(*[p1x, p1y])
if 0 < n:
sx, sy, ux, uy, tx, ty = three(p1x, p1y, p2x, p2y)
koch(p1x, p1y, sx, sy, n - 1)
koch(sx, sy, ux, uy, n - 1)
koch(ux, uy, tx, ty, n - 1)
koch(tx, ty, p2x, p2y, n - 1)
koch(0.00000000, 0.00000000, 100.00000000, 0.00000000, n)
print(*[100.00000000, 100.00000000])
|
s691779051
|
Accepted
| 30
| 8,044
| 874
|
import math
n = int(input())
def three(p1x, p1y, p2x, p2y):
dx = (p2x - p1x)
dy = (p2y - p1y)
sx = dx / 3 + p1x
sy = dy / 3 + p1y
tx = p2x - (dx / 3)
ty = p2y - (dy / 3)
mtx = tx - sx
mty = ty - sy
rad = math.radians(60)
mux = mtx * math.cos(rad) - (mty * math.sin(rad))
muy = mtx * math.sin(rad) + (mty * math.cos(rad))
ux = mux + sx
uy = muy + sy
return [sx, sy, ux, uy, tx, ty]
def koch(p1x, p1y, p2x, p2y, n):
if 0 == n:
print('{:.8f} {:.8f}'.format(p1x, p1y))
if 0 < n:
sx, sy, ux, uy, tx, ty = three(p1x, p1y, p2x, p2y)
koch(p1x, p1y, sx, sy, n - 1)
koch(sx, sy, ux, uy, n - 1)
koch(ux, uy, tx, ty, n - 1)
koch(tx, ty, p2x, p2y, n - 1)
koch(0.00000000, 0.00000000, 100.00000000, 0.00000000, n)
print('{:.8f} {:.8f}'.format(100.00000000, 0.00000000))
|
s488216494
|
p04012
|
u979418645
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 215
|
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
w=input()
w_list=list(w)
w_unique=list(set(w_list))
c=0
for x in w_unique:
if w_list.count(x)%2==0:
c=c
else:
c+=1
if c==0:
print('YES')
else:
print("NO")
|
s954386929
|
Accepted
| 17
| 2,940
| 224
|
w=input()
w_list=list(w)
w_unique=list(set(w_list))
c=0
for x in w_unique:
if w_list.count(x)%2==0:
c=c
else:
c+=1
if c==0:
print('Yes')
else:
print("No")
|
s560289023
|
p03696
|
u626881915
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,228
| 20,464
| 232
|
You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.
|
n = int(input())
s = input()
while True:
c = 0
for i in range(len(s)):
if s[i] == "(":
c += 1
else:
c -= 1
if c < 0:
s = "(" + s
break
if c > 0:
s = s + ")"
if c == 0:
print(s)
|
s699558484
|
Accepted
| 24
| 9,168
| 242
|
n = int(input())
s = input()
while True:
c = 0
for i in range(len(s)):
if s[i] == "(":
c += 1
else:
c -= 1
if c < 0:
s = "(" + s
break
if c > 0:
s = s + ")"
if c == 0:
print(s)
break
|
s163690300
|
p00026
|
u358919705
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,684
| 561
|
As shown in the following figure, there is a paper consisting of a grid structure where each cell is indicated by (x, y) coordinate system. We are going to put drops of ink on the paper. A drop comes in three different sizes: Large, Medium, and Small. From the point of fall, the ink sinks into surrounding cells as shown in the figure depending on its size. In the figure, a star denotes the point of fall and a circle denotes the surrounding cells. Originally, the paper is white that means for each cell the value of density is 0. The value of density is increased by 1 when the ink sinks into the corresponding cells. For example, if we put a drop of Small ink at (1, 2) and a drop of Medium ink at (3, 2), the ink will sink as shown in the following figure (left side): In the figure, density values of empty cells are 0. The ink sinking into out of the paper should be ignored as shown in the figure (top side). We can put several drops of ink at the same point. Your task is to write a program which reads a sequence of points of fall (x, y) with its size (Small = 1, Medium = 2, Large = 3), and prints the number of cells whose density value is 0. The program must also print the maximum value of density. You may assume that the paper always consists of 10 × 10, and 0 ≤ x < 10, 0 ≤ y < 10\.
|
a = [[0] * 14 for _ in range(14)]
while True:
try:
x, y, s = map(int, input().split(','))
except:
break
x += 2
y += 2
for d in [(0, 0), (0, -1), (0, 1), (-1, 0), (1, 0)]:
a[x + d[0]][y + d[1]] += 1
if s >= 2:
for d in [(1, 1), (1, -1), (-1, 1), (-1, -1)]:
a[x + d[0]][y + d[1]] += 1
if s == 3:
for d in [(0, 2), (0, -2), (2, 0), (-2, 0)]:
a[x + d[0]][y + d[1]] += 1
print(sum(a[i][2:12].count(0) for i in range(2, 12)))
print(max(max(a[i]) for i in range(14)))
print(a)
|
s115588064
|
Accepted
| 40
| 7,684
| 561
|
a = [[0] * 14 for _ in range(14)]
while True:
try:
x, y, s = map(int, input().split(','))
except:
break
x += 2
y += 2
for d in [(0, 0), (0, -1), (0, 1), (-1, 0), (1, 0)]:
a[x + d[0]][y + d[1]] += 1
if s >= 2:
for d in [(1, 1), (1, -1), (-1, 1), (-1, -1)]:
a[x + d[0]][y + d[1]] += 1
if s == 3:
for d in [(0, 2), (0, -2), (2, 0), (-2, 0)]:
a[x + d[0]][y + d[1]] += 1
print(sum(a[i][2:12].count(0) for i in range(2, 12)))
print(max(max(a[i][2:12]) for i in range(2, 12)))
|
s268358573
|
p03386
|
u887207211
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 168
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
A, B, K = map(int,input().split())
ans = []
for i in range(A, B+1)[:K]:
ans.append(i)
for i in range(A, B+1)[-K:]:
ans.append(i)
for a in list(set(ans)):
print(a)
|
s787769706
|
Accepted
| 17
| 3,060
| 156
|
A, B, K = map(int,input().split())
x = set()
for i in range(A, B+1)[:K]:
x.add(i)
for i in range(A, B+1)[-K:]:
x.add(i)
for xx in sorted(x):
print(xx)
|
s399931870
|
p03610
|
u210295876
| 2,000
| 262,144
|
Wrong Answer
| 49
| 9,132
| 89
|
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s = input()
t=''
for i in range (len(s)):
if i % 2 == 0:
t = s[i]
print(t)
|
s138033064
|
Accepted
| 44
| 9,188
| 83
|
s = input()
t=''
for i in range (len(s)):
if i % 2 == 0:
t = t+s[i]
print(t)
|
s343962969
|
p03890
|
u128914900
| 2,000
| 262,144
|
Wrong Answer
| 696
| 18,720
| 406
|
_Kode Festival_ is an anual contest where the hardest stone in the world is determined. (Kode is a Japanese word for "hardness".) This year, 2^N stones participated. The hardness of the i-th stone is A_i. In the contest, stones are thrown at each other in a knockout tournament. When two stones with hardness X and Y are thrown at each other, the following will happen: * When X > Y: The stone with hardness Y will be destroyed and eliminated. The hardness of the stone with hardness X will become X-Y. * When X = Y: One of the stones will be destroyed and eliminated. The hardness of the other stone will remain the same. * When X < Y: The stone with hardness X will be destroyed and eliminated. The hardness of the stone with hardness Y will become Y-X. The 2^N stones will fight in a knockout tournament as follows: 1. The following pairs will fight: (the 1-st stone versus the 2-nd stone), (the 3-rd stone versus the 4-th stone), ... 2. The following pairs will fight: (the winner of (1-st versus 2-nd) versus the winner of (3-rd versus 4-th)), (the winner of (5-th versus 6-th) versus the winner of (7-th versus 8-th)), ... 3. And so forth, until there is only one stone remaining. Determine the eventual hardness of the last stone remaining.
|
def setGroup(a):
l = len(a)
i = 0
b = []
while i < l:
if a[i] < a[i+1]:
b.append(a[i + 1] - a[i])
elif a[i] > a[i+1]:
b.append(a[i] - a[i+1])
else:
b.append(a[i])
i += 2
return b
n = int(input())
i = 0
a =[]
while i < 2**n:
a.append(int(input()))
i += 1
while len(a) > 1:
a = setGroup(a)
print(a)
|
s417178243
|
Accepted
| 668
| 18,720
| 409
|
def setGroup(a):
l = len(a)
i = 0
b = []
while i < l:
if a[i] < a[i+1]:
b.append(a[i + 1] - a[i])
elif a[i] > a[i+1]:
b.append(a[i] - a[i+1])
else:
b.append(a[i])
i += 2
return b
n = int(input())
i = 0
a =[]
while i < 2**n:
a.append(int(input()))
i += 1
while len(a) > 1:
a = setGroup(a)
print(a[0])
|
s248350588
|
p03407
|
u440478998
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,016
| 75
|
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
a,b,c = map(int, input().split())
print("Yes") if c >= a+b else print("No")
|
s589055230
|
Accepted
| 27
| 8,988
| 75
|
a,b,c = map(int, input().split())
print("Yes") if c <= a+b else print("No")
|
s455317301
|
p02806
|
u969190727
| 2,525
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 240
|
Niwango created a playlist of N songs. The title and the duration of the i-th song are s_i and t_i seconds, respectively. It is guaranteed that s_1,\ldots,s_N are all distinct. Niwango was doing some work while playing this playlist. (That is, all the songs were played once, in the order they appear in the playlist, without any pause in between.) However, he fell asleep during his work, and he woke up after all the songs were played. According to his record, it turned out that he fell asleep at the very end of the song titled X. Find the duration of time when some song was played while Niwango was asleep.
|
import sys
input=sys.stdin.readline
n=int(input())
ST=[]
for i in range(n):
s,t=map(str,input().split())
ST.append((s,t))
chk=False
x=input()
ans=0
for s,t in ST:
if chk:
ans+=int(t)
else:
if s==x:
chk=True
print(ans)
|
s702430145
|
Accepted
| 17
| 3,060
| 204
|
n=int(input())
ST=[]
for i in range(n):
s,t=map(str,input().split())
ST.append((s,t))
chk=False
x=input()
ans=0
for s,t in ST:
if chk:
ans+=int(t)
else:
if s==x:
chk=True
print(ans)
|
s067770501
|
p02612
|
u486773779
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,148
| 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.
|
n=int(input())
print(n%1000)
|
s169426799
|
Accepted
| 30
| 9,012
| 74
|
n=int(input())
if n%1000==0:
print(0)
else:
print(1000-(n%1000))
|
s929813105
|
p02257
|
u227438830
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,676
| 392
|
A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.
|
import math
n = int(input())
a = [int(i) for i in input().split("\n")]
def isprime(x):
if x == 2:
return True
elif x % 2 == 0:
return False
i = 3
while i < math.sqrt(x):
if x % i == 0:
return False
i += 2
else:
return True
counter = 0
for i in a:
if isprime(i) == True:
counter += 1
print(counter)
|
s180671279
|
Accepted
| 830
| 8,144
| 398
|
import math
n = int(input())
a = []
for i in range(n):
a.append(int(input()))
def isprime(x):
if x == 2:
return True
elif x % 2 == 0:
return False
i = 3
while i <= math.sqrt(x):
if x % i == 0:
return False
i += 2
else:
return True
counter = 0
for i in a:
if isprime(i) == True:
counter += 1
print(counter)
|
s967410845
|
p03827
|
u729133443
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 68
|
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).
|
input();s=input();print(max(i-2*s[:i].count('D')for i in range(99)))
|
s977857192
|
Accepted
| 17
| 2,940
| 61
|
_,t=open(0);a=b=0
for x in t:b+=x>'D'or-1;a=max(a,b)
print(a)
|
s909330692
|
p04043
|
u077816564
| 2,000
| 262,144
|
Wrong Answer
| 21
| 9,044
| 62
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
print(['Yes','No'][sum(list(map(int,input().split()))) != 17])
|
s648305711
|
Accepted
| 26
| 8,968
| 62
|
print(['YES','NO'][sum(list(map(int,input().split()))) != 17])
|
s616376974
|
p03130
|
u932465688
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,064
| 449
|
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
L = list(map(int,input().split()))
M = list(map(int,input().split()))
N = list(map(int,input().split()))
O = [L,M,N]
P = L.count(1)+M.count(1)+N.count(1)
Q = L.count(2)+M.count(2)+N.count(2)
R = L.count(3)+M.count(3)+N.count(3)
S = L.count(4)+M.count(4)+N.count(4)
T = [P,Q,R,S]
T.sort()
if T[0] == T[1] == 1:
if T[2] == T[3] == 2:
if [T[1],T[2]] in O:
print('YES')
else:
print('NO')
else:
print('NO')
else:
print('NO')
|
s445319418
|
Accepted
| 17
| 3,064
| 464
|
L = list(map(int,input().split()))
M = list(map(int,input().split()))
N = list(map(int,input().split()))
O = [L,M,N]
P = L.count(1)+M.count(1)+N.count(1)
Q = L.count(2)+M.count(2)+N.count(2)
R = L.count(3)+M.count(3)+N.count(3)
S = L.count(4)+M.count(4)+N.count(4)
T = [P,Q,R,S]
T.sort()
if T[0] == T[1] == 1:
if T[2] == T[3] == 2:
if [T[1],T[2]] or [T[2],T[1]] in O:
print('YES')
else:
print('NO')
else:
print('NO')
else:
print('NO')
|
s679257323
|
p03711
|
u728566015
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 226
|
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
x, y = map(int, input().split())
group_A = [1, 3, 5, 7, 8, 10, 12]
group_B = [4, 6, 9, 11]
group_C = [2]
for i in [group_A, group_B, group_C]:
if x in i and y in i:
print("YES")
break
else:
print("NO")
|
s618221440
|
Accepted
| 17
| 2,940
| 239
|
x, y = map(int, input().split())
group_A = [1, 3, 5, 7, 8, 10, 12]
group_B = [4, 6, 9, 11]
group_C = [2]
for group in [group_A, group_B, group_C]:
if x in group and y in group:
print("Yes")
break
else:
print("No")
|
s847962199
|
p03779
|
u151785909
| 2,000
| 262,144
|
Wrong Answer
| 2,113
| 129,916
| 308
|
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
import math
x = int(input())
n = math.ceil(math.sqrt(x*2))
s = [list(map(int,bin(x)[2:].zfill(n))) for x in range(2**n)]
t = 0
c = 0
min = x
for i in range(2**n):
for j in range(len(s[i])):
t+=s[i][j]*(j+1)
if s[i][j]==1:
c =j+1
if t==x and c<min:
min =c
print(c)
|
s675526740
|
Accepted
| 28
| 2,940
| 67
|
x = int(input())
s= 0
i=1
while s<x:
s=s+i
i+=1
print(i-1)
|
s822173720
|
p03438
|
u222841610
| 2,000
| 262,144
|
Wrong Answer
| 60
| 4,600
| 355
|
You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal. Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions **simultaneously** : * Add 2 to a_i. * Add 1 to b_j.
|
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
sum = 0
for i in range(n):
if a[i] - b[i] > 0:
sum += a[i]- b[i]
else:
sum += (b[i] - a[i]) / 2
sum_a = 0
sum_b = 0
for i in range(n):
sum_a += a.pop(0)
sum_b += b.pop(0)
if sum <= sum_b - sum_a:
print ('YES')
else:
print ('NO')
|
s388088579
|
Accepted
| 61
| 4,596
| 369
|
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
a = a.copy()
b = b.copy()
sum_c = 0
for i in range(n):
if (a[i] - b[i]) < 0:
sum_c += int(((b[i] - a[i]) / 2 ) + 0.5)
sum_a = 0
sum_b = 0
for i in range(n):
sum_a += a.pop(0)
sum_b += b.pop(0)
if sum_c <= (sum_b - sum_a):
print ('Yes')
else:
print ('No')
|
s533773353
|
p02401
|
u941509088
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,672
| 281
|
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while True:
a,op,b = input().split()
if op == "?":
break
elif op == "+":
print(int(a) + int(b))
elif op == "-":
print(int(a) - int(b))
elif op == "*":
print(int(a) * int(b))
elif op == "/":
print(int(a) / int(b))
|
s498982503
|
Accepted
| 50
| 7,664
| 282
|
while True:
a,op,b = input().split()
if op == "?":
break
elif op == "+":
print(int(a) + int(b))
elif op == "-":
print(int(a) - int(b))
elif op == "*":
print(int(a) * int(b))
elif op == "/":
print(int(a) // int(b))
|
s728744315
|
p02396
|
u227438830
| 1,000
| 131,072
|
Wrong Answer
| 40
| 7,564
| 155
|
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
j = 1
list = [int(i) for i in input().split()]
for num in list:
if num == 0:
break
print("Case " + str(j) +": "+ str(num) )
j += 1
|
s136996415
|
Accepted
| 130
| 7,460
| 110
|
j= 1
while True:
x = int(input())
if x == 0:
break
print("Case %d: %d" % (j,x))
j += 1
|
s379331251
|
p04011
|
u865413330
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 162
|
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.
|
import sys
input = sys.stdin.readline
n = int(input())
k = int(input())
x = int(input())
y = int(input())
print(n * x) if n <= k else print(n * x + (k - n) * y)
|
s071780339
|
Accepted
| 18
| 3,060
| 161
|
import sys
input = sys.stdin.readline
n = int(input())
k = int(input())
x = int(input())
y = int(input())
print(k * x + (n - k) * y) if n > k else print(n * x)
|
s406756315
|
p02578
|
u325149030
| 2,000
| 1,048,576
|
Wrong Answer
| 130
| 32,052
| 213
|
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
N = int(input())
A = list(map(int, input().split()))
now = A[0]
ans = 0
for i in range(1,N):
if A[i]<now:
step = now - A[i]
ans += now
else:
step = 0
now = A[i]+step
print(ans)
|
s346984488
|
Accepted
| 145
| 32,232
| 214
|
N = int(input())
A = list(map(int, input().split()))
now = A[0]
ans = 0
for i in range(1,N):
if A[i]<now:
step = now - A[i]
ans += step
else:
step = 0
now = A[i]+step
print(ans)
|
s762143155
|
p02646
|
u845650912
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,188
| 280
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
A, V= map(int,input().split())
B, W = map(int,input().split())
T = int(input())
def hantei(A,B,V,W,T):
if V <= W:
return('No')
else:
if abs(B - A)/(V - W) <= T:
return('Yes')
else:
return('No')
print(hantei(A,B,V,W,T))
|
s382929163
|
Accepted
| 24
| 9,192
| 280
|
A, V= map(int,input().split())
B, W = map(int,input().split())
T = int(input())
def hantei(A,B,V,W,T):
if V <= W:
return('NO')
else:
if abs(B - A)/(V - W) <= T:
return('YES')
else:
return('NO')
print(hantei(A,B,V,W,T))
|
s035313283
|
p03700
|
u367130284
| 2,000
| 262,144
|
Wrong Answer
| 296
| 24,312
| 1,272
|
You are going out for a walk, when you suddenly encounter N monsters. Each monster has a parameter called _health_ , and the health of the i-th monster is h_i at the moment of encounter. A monster will vanish immediately when its health drops to 0 or below. Fortunately, you are a skilled magician, capable of causing explosions that damage monsters. In one explosion, you can damage monsters as follows: * Select an alive monster, and cause an explosion centered at that monster. The health of the monster at the center of the explosion will decrease by A, and the health of each of the other monsters will decrease by B. Here, A and B are predetermined parameters, and A > B holds. At least how many explosions do you need to cause in order to vanish all the monsters?
|
#import numpy as np
from numpy import*
#from scipy.sparse.csgraph import shortest_path #shortest_path(csgraph=graph)
#from scipy.sparse.csgraph import dijkstra
#from scipy.sparse.csgraph import floyd_warshall
#from scipy.sparse import csr_matrix
from collections import*
from fractions import gcd
from functools import* #reduce
from itertools import* #permutations("AB",repeat=2) combinations("AB",2) product("AB",2) groupby accumulate
from operator import mul,itemgetter
from bisect import*
from heapq import*
from math import factorial,pi
from copy import deepcopy
import sys
sys.setrecursionlimit(10**8)
def main():
n,a,b,*h=map(int,open(0).read().split())
h=array(h)
ok=-1
ng=max(h)
def judge(t):
H=deepcopy(h)
H-=b*t
# print(H)
if sum((H-1)//(a-b)+1)<=t:
return 1
else:
return 0
while abs(ok-ng)>1:
mid=(ok+ng)//2
# print(mid)
# print("NY"[judge(mid)])
if judge(mid):
ng=mid
else:
ok=mid
l=ok+1
print(l)
if __name__ == '__main__':
main()
|
s484530444
|
Accepted
| 269
| 23,124
| 508
|
from numpy import*
from copy import deepcopy
def main():
n,a,b,*h=map(int,open(0).read().split())
h=array(h)
ok=-1
ng=max(h)+10
def judge(t):
H=deepcopy(h)
H-=b*t
H=H[H>0]
if sum((H-1)//(a-b)+1)<=t:
return 1
else:
return 0
while abs(ok-ng)>1:
mid=(ok+ng)//2
if judge(mid):
ng=mid
else:
ok=mid
l=ok+1
print(l)
if __name__ == '__main__':
main()
|
s107857315
|
p03385
|
u597455618
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 80
|
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 = list(input())
s.sort()
if (s == "abc"):
print("Yes")
else :
print("No")
|
s862557054
|
Accepted
| 17
| 2,940
| 101
|
s = input()
s = list(s)
s.sort()
ans = list("abc")
if (s == ans):
print("Yes")
else :
print("No")
|
s362114755
|
p02388
|
u435313013
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,572
| 45
|
Write a program which calculates the cube of a given integer x.
|
x = int(input())
print("x^3 = %d" % (x*x*x))
|
s816858215
|
Accepted
| 20
| 5,576
| 38
|
x = int(input())
print("%d" %(x*x*x))
|
s572810065
|
p03386
|
u084320347
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 196
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,k = list(map(int,input().split()))
a_l = [a+i for i in range(k+1)]
b_l = [b-i for i in range(k+1)]
if b-a < k:
[print(i) for i in range(a,b+1)]
else:
[print(i) for i in set(a_l+b_l)]
|
s936276488
|
Accepted
| 17
| 3,060
| 200
|
a,b,k = list(map(int,input().split()))
a_l = [a+i for i in range(k)]
b_l = [b-i for i in range(k)]
if b-a < k:
[print(i) for i in range(a,b+1)]
else:
[print(i) for i in sorted(set(a_l+b_l))]
|
s539570882
|
p03992
|
u226191225
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 121
|
This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.
|
n = input()
list(n)
for i in range(4):
print(n[i],end='')
print(' ',end='')
for i in range(4,12):
print(n[i],end='')
|
s286361084
|
Accepted
| 17
| 3,060
| 133
|
n = input()
list(n)
for i in range(4):
print(n[i],end='')
print(' ',end='')
for i in range(4,11):
print(n[i],end='')
print(n[11])
|
s095537684
|
p03091
|
u379234461
| 2,000
| 1,048,576
|
Wrong Answer
| 2,136
| 497,840
| 661
|
You are given a simple connected undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M. Edge i connects Vertex a_i and b_i bidirectionally. Determine if three circuits (see Notes) can be formed using each of the edges exactly once.
|
# AGC032 Problem-C
n, m = map(int,input().split())
table = [[0 for i in range (0,n)] for j in range (0,n)]
for i in range(0,m) :
a, b = map(int,input().split())
table[a-1][b-1] = 1
table[b-1][a-1] = 1
#print (table)
check = [0 for i in range(0,n)]
for i in range(0,n):
for j in range(0,n):
check[i] += table[i][j]
#print (check)
node4 = 0
node6 = 0
node_odd = 0
for i in range(0,n):
if check[i] % 2 == 0:
if check[i]==4 :
node4 += 1
if check[i]>=6 :
node6 +=1
else:
node_odd += 1
if (node_odd == 0) and ((node4 >= 2) or (node6 >=1)) :
print ("yes")
else:
print ("no")
|
s003378562
|
Accepted
| 509
| 21,960
| 1,488
|
# AGC032 Problem-C
def special_case() :
st = -1
for i in range(0,n) :
if check[i] == 4 :
st = i
break
for i in edge[st] :
p0 = st
p1 = i
while (check[p1]==2) :
t1, t2 = edge[p1]
if t1 == p0 :
p0 = p1
p1 = t2
else:
p0 = p1
p1 = t1
if p1 == st :
return 0
return -1
n, m = map(int,input().split())
check = [0 for i in range(0,n)]
edge = [[] for i in range(0,n)]
#table = [[0 for i in range (0,n)] for j in range (0,n)]
for i in range(0,m) :
a, b = map(int,input().split())
check[a-1] += 1
check[b-1] += 1
edge[a-1].append(b-1)
edge[b-1].append(a-1)
#print (check)
#print(edge)
node4 = 0
node6 = 0
node_odd = 0
for i in range(0,n):
if check[i] % 2 == 0:
if check[i]==4 :
node4 += 1
if check[i]>=6 :
node6 +=1
else:
node_odd += 1
if (node_odd != 0) :
print ("No")
else :
if ((node4 >= 3) or (node6 >=1)) :
print ("Yes")
elif (node4 == 2) :
if special_case() == 0 :
print ("Yes")
else :
print ("No")
else :
print ("No")
|
s986141973
|
p03456
|
u283929013
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 143
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
a, b = map(str,input().split())
c = int(a + b)
t = 2
while(t**2 <= c):
if t**2 == c:
print(t)
quit()
t += 1
print("No")
|
s804138657
|
Accepted
| 17
| 2,940
| 147
|
a, b = map(str,input().split())
c = int(a + b)
t = 2
while(t**2 <= c):
if t**2 == c:
print("Yes")
quit()
t += 1
print("No")
|
s034859828
|
p02406
|
u725841747
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,552
| 95
|
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; }
|
n = int(input())
for i in range(n+1):
if i%3 == 0 or i%10 == 3:
print(" ",i,end="")
|
s070888173
|
Accepted
| 30
| 8,120
| 187
|
n = int(input())
for i in range(1,n+1):
if i%3 == 0:
print("",i,end="")
else:
list("i")
if str(i).count("3") != 0:
print("",i,end="")
print("")
|
s716299053
|
p03636
|
u158778550
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 41
|
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
s = input()
print(s[0]+str(len(s))+s[-1])
|
s656043324
|
Accepted
| 17
| 2,940
| 43
|
s = input()
print(s[0]+str(len(s)-2)+s[-1])
|
s182207380
|
p02397
|
u922112509
| 1,000
| 131,072
|
Wrong Answer
| 60
| 5,624
| 341
|
Write a program which reads two integers x and y, and prints them in ascending order.
|
# Swapping Two Numbers
zeroCount = 0
while zeroCount == 0:
numbers = [int(i) for i in input().rstrip().split()]
# print(numbers)
if numbers[0] == 0 and numbers[1] == 0:
zeroCount += 1
else:
swap = sorted(numbers)
# print(swap)
output = [str(number) for number in swap]
print(output)
|
s240502643
|
Accepted
| 60
| 5,616
| 331
|
# Swapping Two Numbers
zeroCount = 0
while zeroCount == 0:
numbers = [int(i) for i in input().rstrip().split()]
# print(numbers)
if numbers[0] == 0 and numbers[1] == 0:
zeroCount += 1
else:
swap = sorted(numbers)
# print(swap)
a = [str(n) for n in swap]
print(" ".join(a))
|
s842818168
|
p03167
|
u221592478
| 2,000
| 1,048,576
|
Wrong Answer
| 692
| 175,812
| 458
|
There is a grid with H horizontal rows and W vertical columns. Let (i, j) denote the square at the i-th row from the top and the j-th column from the left. For each i and j (1 \leq i \leq H, 1 \leq j \leq W), Square (i, j) is described by a character a_{i, j}. If a_{i, j} is `.`, Square (i, j) is an empty square; if a_{i, j} is `#`, Square (i, j) is a wall square. It is guaranteed that Squares (1, 1) and (H, W) are empty squares. Taro will start from Square (1, 1) and reach (H, W) by repeatedly moving right or down to an adjacent empty square. Find the number of Taro's paths from Square (1, 1) to (H, W). As the answer can be extremely large, find the count modulo 10^9 + 7.
|
n , m = map(int , input().split())
dp = [[1 for j in range(m + 1)] for i in range(n + 1)]
for i in range(n + 1):
dp.append([])
for j in range(m + 1):
dp[i].append(0)
arr = []
for i in range(n):
arr.append(input())
dp[1][1] = 1
for i in range(1,n + 1):
for j in range(1,m + 1):
if arr[i - 1][j - 1] == '#' or (i == 1 and j == 1):
continue
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
print(dp[n][m] % 1000000007)
|
s871224068
|
Accepted
| 705
| 168,232
| 411
|
n , m = map(int , input().split())
dp = []
for i in range(n + 1):
dp.append([])
for j in range(m + 1):
dp[i].append(0)
arr = []
for i in range(n):
arr.append(input())
dp[1][1] = 1
for i in range(1,n + 1):
for j in range(1,m + 1):
if arr[i - 1][j - 1] == '#' or (i == 1 and j == 1):
continue
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
print(dp[n][m] % 1000000007)
|
s096398433
|
p02601
|
u163903855
| 2,000
| 1,048,576
|
Wrong Answer
| 37
| 9,184
| 224
|
M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.
|
a,b,c=(int(x) for x in input().split())
k=int(input())
flag=True
for i in range(k+1):
if a<2**(k-i)*b and 2**(k-i)*b<2**i*c:
print("Yes",i)
exit()
flag=False
break
if flag:
print("No")
|
s780897546
|
Accepted
| 32
| 8,948
| 222
|
a,b,c=(int(x) for x in input().split())
k=int(input())
flag=True
for i in range(k+1):
if a<2**(k-i)*b and 2**(k-i)*b<2**i*c:
print("Yes")
exit()
flag=False
break
if flag:
print("No")
|
s006505117
|
p00015
|
u536280367
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,012
| 1,311
|
A country has a budget of more than 81 trillion yen. We want to process such data, but conventional integer type which uses signed 32 bit can represent up to 2,147,483,647. Your task is to write a program which reads two integers (more than or equal to zero), and prints a sum of these integers. If given integers or the sum have more than 80 digits, print "overflow".
|
from collections import deque
def align_digit(a, b):
la, lb = len(a), len(b)
if la < lb:
a = a.zfill(lb)
elif lb < la:
b = b.zfill(la)
return a, b
def split_to_ints(s, n=10):
length = len(s)
return reversed([int(s[i:i + n]) for i in range(0, length, n)])
def add(a, b, n=10):
a, b = align_digit(a, b)
ints_a = split_to_ints(a, n)
ints_b = split_to_ints(b, n)
carry = 0
result = deque()
zipped = list(zip(ints_a, ints_b))
for i, (int_a, int_b) in enumerate(zipped):
s = int_a + int_b + carry
carry = int(s / (10 ** n))
if i != len(zipped) - 1:
result.append(str(s % (10 ** (n - 1))).zfill(n))
else:
result.append(str(s))
i += 1
return str(int(''.join(result)))
if __name__ == '__main__':
n_input = int(input())
for i in range(n_input):
a, b = input(), input()
#print(a + b)
print(add(a, b))
|
s178231326
|
Accepted
| 20
| 5,612
| 811
|
# 12,432 -> 012,432
def align_digits(a, b):
la, lb = len(a), len(b)
if la < lb:
a = a.zfill(lb)
elif lb < la:
b = b.zfill(la)
return a, b
def add(a, b):
a, b = align_digits(a, b)
zipped = list(zip(a[::-1], b[::-1]))
carry = 0
result = []
for i, z in enumerate(zipped):
a, b = z
s = int(a) + int(b) + carry
carry = int(s / 10)
if i == len(zipped) - 1:
result.append(str(s))
else:
result.append(str(s % 10))
ans = ''.join(result[::-1])
if len(ans) > 80:
return 'overflow'
return ans
if __name__ == '__main__':
n_input = int(input())
inputs = []
for i in range(n_input):
inputs.append((input(), input()))
for a, b in inputs:
print(add(a, b))
|
s511320904
|
p03494
|
u625495026
| 2,000
| 262,144
|
Wrong Answer
| 161
| 13,192
| 12
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
import numpy
|
s087309497
|
Accepted
| 152
| 12,488
| 182
|
import numpy as np
n = input()
l = np.array(list(map(int,input().split())))
counter = 0
while True:
if not (l%2==0).all():
break
l = l/2
counter+=1
print(counter)
|
s715227203
|
p02401
|
u656153606
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,652
| 420
|
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while True:
Input = input().split()
a = int(Input[0])
op = Input[1]
b = int(Input[2])
if a >= 0 and b >= 0 and a <= 20000 and b <= 20000:
if op == "+":
print(a + b)
elif op == "-":
print(a - b)
elif op == "*":
print(a * b)
elif op == "/":
if b != 0:
print(a / b)
elif op == "?":
break
|
s728338077
|
Accepted
| 30
| 7,668
| 425
|
while True:
Input = input().split()
a = int(Input[0])
op = Input[1]
b = int(Input[2])
if a >= 0 and b >= 0 and a <= 20000 and b <= 20000:
if op == "+":
print(a + b)
elif op == "-":
print(a - b)
elif op == "*":
print(a * b)
elif op == "/":
if b != 0:
print(int(a / b))
elif op == "?":
break
|
s219430500
|
p03457
|
u137722467
| 2,000
| 262,144
|
Wrong Answer
| 427
| 27,380
| 343
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
data = []
for i in range(N):
data.append(list(map(int, input().split())))
t, x, y = 0, 0, 0
for i in range(N):
deltaT = data[i][0] - t
absX = abs(data[i][1] - x)
absY = abs(data[i][2] - y)
prob = deltaT - absX - absY
if not prob >= 0 and prob % 2 == 0:
print("NO")
quit()
print("YES")
|
s678104854
|
Accepted
| 457
| 27,380
| 538
|
N = int(input())
data = []
for i in range(N):
data.append(list(map(int, input().split())))
if N == 1:
prob = data[0][0] - data[0][1] - data[0][2]
if prob < 0 or prob % 2 != 0:
print("No")
else:
print("Yes")
quit()
for i in range(N-1):
t = data[i][0]
x = data[i][1]
y = data[i][2]
deltaT = data[i+1][0] - t
absX = abs(data[i+1][1] - x)
absY = abs(data[i+1][2] - y)
prob = deltaT - absX - absY
if prob < 0 or prob % 2 != 0:
print("No")
quit()
print("Yes")
|
s635351875
|
p03624
|
u354527070
| 2,000
| 262,144
|
Wrong Answer
| 55
| 10,068
| 265
|
You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead.
|
s = list(str(input()))
s.sort()
a = [chr(ord('a') + i) for i in range(26)]
b = {}
for i in a:
b[i] = 0
for i in s:
b[i] += 1
alf = ''
for i in a:
if b[i] == 0:
print(i)
alf = i
break
if alf:
print(i)
else:
print('None')
|
s050676417
|
Accepted
| 48
| 10,044
| 247
|
s = list(str(input()))
s.sort()
a = [chr(ord('a') + i) for i in range(26)]
b = {}
for i in a:
b[i] = 0
for i in s:
b[i] += 1
alf = ''
for i in a:
if b[i] == 0:
alf = i
break
if alf:
print(i)
else:
print('None')
|
s398965179
|
p03844
|
u359007262
| 2,000
| 262,144
|
Wrong Answer
| 54
| 5,976
| 988
|
Joisino wants to evaluate the formula "A op B". Here, A and B are integers, and the binary operator op is either `+` or `-`. Your task is to evaluate the formula instead of her.
|
import sys
from io import StringIO
import unittest
def resolve():
input_string = list(input())
num1 = int(input_string[0])
num2 = int(input_string[4])
operater = input_string[2]
result = 0
if operater == "+":
result = num1 + num2
elif operater == "-":
result = num1 - num2
print(result)
class TestClass(unittest.TestCase):
def assertIO(self, input, output):
stdout, stdin = sys.stdout, sys.stdin
sys.stdout, sys.stdin = StringIO(), StringIO(input)
resolve()
sys.stdout.seek(0)
out = sys.stdout.read()[:-1]
sys.stdout, sys.stdin = stdout, stdin
self.assertEqual(out, output)
def test_入力例_1(self):
input = """1 + 2"""
output = """3"""
self.assertIO(input, output)
def test_入力例_2(self):
input = """5 - 7"""
output = """-2"""
self.assertIO(input, output)
if __name__ == "__main__":
unittest.main()
|
s544645518
|
Accepted
| 17
| 3,060
| 232
|
input_string = input().split()
num1 = int(input_string[0])
num2 = int(input_string[2])
operater = input_string[1]
result = 0
if operater == "+":
result = num1 + num2
elif operater == "-":
result = num1 - num2
print(result)
|
s440456807
|
p03409
|
u391875425
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,064
| 788
|
On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i). A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point. At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.
|
N = int(input())
a = [0 for i in range(N)]
b = [0 for i in range(N)]
c = [0 for i in range(N)]
d = [0 for i in range(N)]
for i in range(N):
a[i],b[i] = map(int,input().split())
for i in range(N):
c[i],d[i] = map(int,input().split())
cnt = 0
fin = []
for i in range(N):
f = 0
s1 = 100000000000
s2 = 100000000000
tmp1 = 100000000000
tmp2 = 100000000000
for j in range(N):
tmp1 = a[i] - c[j]
tmp2 = b[i] - d[j]
if tmp1 < 0 and tmp2 < 0:
s1 = abs(tmp1) + abs(tmp2)
#print(i,j)
if s1 < s2 :
ti = i
tj = j
s2 = s1
f = 1
if f == 1 and not (j in fin) :
fin.append(tj)
cnt += 1
print(ti,tj)
print(cnt)
|
s137397526
|
Accepted
| 19
| 3,064
| 426
|
N = int(input())
r = [list(map(int,input().split())) for _ in range(N)]
b = [list(map(int,input().split())) for _ in range(N)]
r.sort(key = lambda x:x[1],reverse = True)
b.sort(key = lambda x:x[0])
cnt = 0
fin = []
for i in range(N):
for j in range(N):
if b[i][0] > r[j][0] and b[i][1] > r[j][1]:
if not(j in fin):
cnt += 1
fin.append(j)
break
print(cnt)
|
s354037335
|
p02614
|
u690175641
| 1,000
| 1,048,576
|
Wrong Answer
| 150
| 27,120
| 729
|
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
import numpy as np
import itertools
h, w, k = list(map(int, input().split()))
def convert(char):
if char == '.':
return 0
else:
return 1
def add_elements(c, row, column):
s = 0
for i in range(h):
for j in range(w):
if (i not in row) and (j not in column):
s += c[i,j]
return s
c = np.empty((h, w))
ans = 0
for i in range(h):
line = list(map(convert, list(input())))
c[i,:] = np.array(line)
for i in range(h+1):
for j in range(w+1):
rows = list(itertools.combinations(range(h),i))
columns = list(itertools.combinations(range(h),j))
for row in rows:
for column in columns:
s = add_elements(c, row, column)
if s == k:
ans += 1
print(ans)
|
s325878714
|
Accepted
| 157
| 27,116
| 727
|
import numpy as np
import itertools
h, w, k = list(map(int, input().split()))
def convert(char):
if char == '.':
return 0
else:
return 1
def add_elements(c, row, column):
s = 0
for i in range(h):
for j in range(w):
if (i not in row) and (j not in column):
s += c[i,j]
return s
c = np.empty((h, w))
ans = 0
for i in range(h):
line = list(map(convert, list(input())))
c[i,:] = np.array(line)
for i in range(h+1):
for j in range(w+1):
rows = list(itertools.combinations(range(h),i))
columns = list(itertools.combinations(range(w),j))
for row in rows:
for column in columns:
s = add_elements(c, row, column)
if s == k:
ans += 1
print(ans)
|
s677953207
|
p02678
|
u163501259
| 2,000
| 1,048,576
|
Wrong Answer
| 2,208
| 65,648
| 1,109
|
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
import sys
input = sys.stdin.readline
from collections import deque
N, M = map(int, input().split())
C = [list(map(int, input().split())) for i in range(M)]
def bfs(room, now):
# print(room, now)
plan = deque([])
# for r in room:
# plan.append(r)
plan.append(room)
# print(plan)
visited[1][0] = True
while plan:
nextque = plan.popleft()
# print(nextque)
for i in PATH[now]:
if visited[i][0] == True:
# print('true')
continue
else:
# for r in PATH[next]:
# plan.append(r)
plan.append(PATH[i])
# print(plan)
visited[i][0] = True
visited[i][1] = now
now = i
if __name__ == '__main__':
PATH = [[] for i in range(N+1)]
for i in range(M):
PATH[C[i][0]].append(C[i][1])
PATH[C[i][1]].append(C[i][0])
# print(PATH)
visited = [[False,0] for i in range(N+1)]
bfs(PATH[1], 1)
# print(visited)
for i in range(2,N+1):
print(visited[i][1])
|
s298912516
|
Accepted
| 890
| 118,456
| 803
|
import sys
input = sys.stdin.readline
from collections import deque
N, M = map(int, input().split())
C = [list(map(int, input().split())) for i in range(M)]
def bfs(now):
plan = deque([])
plan.append(now)
visited[now][0] = True
while plan:
now = plan.popleft()
for i in PATH[now]:
if visited[i][0] == True:
continue
else:
plan.append(i)
visited[i][0] = True
visited[i][1] = now
if __name__ == '__main__':
PATH = [deque([]) for i in range(N+1)]
for i in range(M):
PATH[C[i][0]].append(C[i][1])
PATH[C[i][1]].append(C[i][0])
visited = [[False,0] for i in range(N+1)]
bfs(1)
print("Yes")
for i in range(2,N+1):
print(visited[i][1])
|
s393467619
|
p03408
|
u371763408
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,316
| 301
|
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
from collections import Counter
n=int(input())
S=[input() for i in range(n)]
S=dict(Counter(S).most_common())
m=int(input())
T=[input() for i in range(m)]
T=dict(Counter(T).most_common())
for i in S:
for k in T:
if i==k:
S[i] -= T[i]
if S[max(S)]< 0:
print(0)
else:
print(S[max(S)])
|
s679336633
|
Accepted
| 22
| 3,316
| 351
|
from collections import Counter
n=int(input())
S=[input() for i in range(n)]
S=dict(Counter(S).most_common())
m=int(input())
T=[input() for i in range(m)]
T=dict(Counter(T).most_common())
for i in S:
for k in T:
if i==k:
S[i] -= T[k]
S=sorted(S.items(),key=lambda x:x[1],reverse=True)
if S[0][1] < 0:
print(0)
else:
print(S[0][1])
|
s446124712
|
p03079
|
u387768829
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 193
|
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
_input = list(map(int, input().split(" ")))
_input.sort()
_A = _input[0] * _input[0]
_B = _input[1] * _input[1]
_C = _input[2] * _input[2]
if _A + _B == _C:
print("Yes")
else:
print("No")
|
s408365384
|
Accepted
| 17
| 2,940
| 149
|
_input = input().split(" ")
_A = int(_input[0])
_B = int(_input[1])
_C = int(_input[2])
if _A == _B and _A == _C:
print("Yes")
else:
print("No")
|
s948709016
|
p02392
|
u100813820
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,580
| 622
|
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
# 04-Branch_on_Condition-Range.py
# ??????
# Input
# Output
# Yes?????????No????????????????????????????????????
# Sample Input 1
# 1 3 8
# Sample Output 1
# Yes
# Sample Input 2
# 3 8 1
# Sample Output 2
# No
print("a,b,c?????\??????")
a,b,c=map(int,input().split())
print("a=%d, b=%d, c=%d."%(a,b,c))
if a<b and b<c:
print("Yes")
else:
print("No")
|
s621998100
|
Accepted
| 20
| 7,640
| 626
|
# 04-Branch_on_Condition-Range.py
# ??????
# Input
# Output
# Yes?????????No????????????????????????????????????
# Sample Input 1
# 1 3 8
# Sample Output 1
# Yes
# Sample Input 2
# 3 8 1
# Sample Output 2
# No
# print("a,b,c?????\??????")
a,b,c=map(int,input().split())
# print("a=%d, b=%d, c=%d."%(a,b,c))
if a<b and b<c:
print("Yes")
else:
print("No")
|
s316262766
|
p03694
|
u379142263
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,316
| 339
|
It is only six months until Christmas, and AtCoDeer the reindeer is now planning his travel to deliver gifts. There are N houses along _TopCoDeer street_. The i-th house is located at coordinate a_i. He has decided to deliver gifts to all these houses. Find the minimum distance to be traveled when AtCoDeer can start and end his travel at any positions.
|
import sys
import itertools
sys.setrecursionlimit(1000000000)
from heapq import heapify,heappop,heappush,heappushpop
import math
import collections
MOD = 10**9 + 7
n = int(input())
a = list(map(int,input().split()))
ans = 0
for i in a:
cost = 0
for j in a:
d = abs(i-j)
cost += d
ans = max(ans,cost)
print(ans)
|
s308497598
|
Accepted
| 20
| 3,316
| 242
|
import sys
import itertools
sys.setrecursionlimit(1000000000)
from heapq import heapify,heappop,heappush,heappushpop
import math
import collections
MOD = 10**9 + 7
n = int(input())
a = sorted(list(map(int,input().split())))
print(a[-1]-a[0])
|
s772390437
|
p03377
|
u130900604
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 80
|
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())
t=x-a
if t<=b:
print("Yes")
else:
print("No")
|
s094082385
|
Accepted
| 18
| 2,940
| 79
|
a,b,x=map(int,input().split())
if a<=x<=a+b:
print("YES")
else:
print("NO")
|
s206817115
|
p02612
|
u778614444
| 2,000
| 1,048,576
|
Wrong Answer
| 34
| 9,148
| 48
|
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.
|
div, mod = divmod(int(input()), 1000)
print(mod)
|
s876581081
|
Accepted
| 28
| 9,148
| 63
|
div,mod = divmod(int(input()), 1000)
print((1000 - mod) % 1000)
|
s473709408
|
p02694
|
u203597140
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,108
| 106
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
x = int(input())
a = 100
i = 0
while a < x:
i = i + 1
a = math.floor(a * 1.01)
print(i, a)
|
s854543244
|
Accepted
| 23
| 9,164
| 103
|
import math
x = int(input())
a = 100
i = 0
while a < x:
i = i + 1
a = math.floor(a * 1.01)
print(i)
|
s791225856
|
p03407
|
u734749411
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 66
|
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
A,B,C = map(int, input().split())
print("YES" if A+B >C else "NO")
|
s884188194
|
Accepted
| 17
| 2,940
| 73
|
A, B, C = map(int, input().split())
print("Yes" if A + B >= C else "No")
|
s346214452
|
p02619
|
u364027015
| 2,000
| 1,048,576
|
Wrong Answer
| 39
| 9,264
| 370
|
Let's first write a program to calculate the score from a pair of input and output. You can know the total score by submitting your solution, or an official program to calculate a score is often provided for local evaluation as in this contest. Nevertheless, writing a score calculator by yourself is still useful to check your understanding of the problem specification. Moreover, the source code of the score calculator can often be reused for solving the problem or debugging your solution. So it is worthwhile to write a score calculator unless it is very complicated.
|
D=int(input())
C=list(map(int,input().split()))
import sys
S=list()
for i in range(D):
S.append(list(map(int,input().split())))
t=list()
for i in range(D):
t.append(int(input()))
M=0
di=[0]*26
for d in range(D):
for i in range(26):
if i==t[d]-1:
di[i]=0
else:
di[i]+=1
M+=S[d][t[d]-1]
for i in range(26):
M-=C[i]*(d+1-di[i])
print(M)
|
s411195893
|
Accepted
| 36
| 9,148
| 355
|
D=int(input())
C=list(map(int,input().split()))
S=list()
for i in range(D):
S.append(list(map(int,input().split())))
t=list()
for i in range(D):
t.append(int(input()))
M=0
di=[0]*26
for d in range(D):
for i in range(26):
if i==t[d]-1:
di[i]=0
else:
di[i]+=1
M+=S[d][t[d]-1]
for i in range(26):
M-=C[i]*(di[i])
print(M)
|
s896327439
|
p03997
|
u459697504
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 370
|
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.
|
#! /usr/bin/python3
def main():
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2)
if __name__ == '__main__':
main()
|
s772346903
|
Accepted
| 17
| 2,940
| 375
|
#! /usr/bin/python3
def main():
a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2))
if __name__ == '__main__':
main()
|
s597261703
|
p03693
|
u373593739
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 175
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r, g, b = map(int,input().split())
soma = 0
if 1 <= r <= 9 and 1 <= g <= 9 and 1 <= b <= 9:
soma = r*100 + g*10 + b
if soma%4 == 0:
print('Yes')
else:
print('No')
|
s624821979
|
Accepted
| 17
| 2,940
| 191
|
r, g, b = map(int,input().split())
soma = 0
if 1 <= r <= 9 and 1 <= g <= 9 and 1 <= b <= 9:
soma = r*100 + g*10 + b
if soma%4 == 0:
print('YES')
else:
print('NO')
|
s101727175
|
p04031
|
u922449550
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,060
| 197
|
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
N = int(input())
A = list(map(int, input().split()))
ans = float('inf')
for i, a in enumerate(A):
cost = 0
for j, aj in enumerate(A):
cost += (aj - a)**2
ans = min(ans, cost)
print(ans)
|
s700979702
|
Accepted
| 25
| 3,060
| 206
|
N = int(input())
A = list(map(int, input().split()))
ans = float('inf')
for target in range(-100, 101):
cost = 0
for j, a in enumerate(A):
cost += (target - a)**2
ans = min(ans, cost)
print(ans)
|
s431388958
|
p03251
|
u097708290
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,064
| 435
|
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(" ")))
flug1=False
flug2=False
for i in range(X,Y+1):
for j in range(N):
if i<=x[j]:
flug1=True
break
for j in range(M):
if i>=y[j]:
flug1=True
break
if flug1==False:
flug2=True
flug1=False
if flug2:
print("No War")
else:
print("War")
|
s810934814
|
Accepted
| 19
| 3,064
| 515
|
N,M,X,Y=map(int,input().split(" "))
x=list(map(int,input().split(" ")))
y=list(map(int,input().split(" ")))
flug1=False
flug2=False
Z=0
for i in range(X,Y+1):
for j in range(N):
if i<=x[j]:
flug1=True
break
for j in range(M):
if i>y[j]:
flug1=True
break
if flug1==False:
flug2=True
Z=i
flug1=False
if flug2:
if Z>X and Z<=Y:
print("No War")
else:
print("War")
else:
print("War")
|
s230078503
|
p02314
|
u022407960
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,628
| 533
|
Find the minimum number of coins to make change for n cents using coins of denominations d1, d2,.., dm. The coins can be used any number of times.
|
#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""
input:
15 6
1 2 7 8 12 50
output:
2
"""
import sys
def solve():
rec[0] = 0
for i in range(c_num):
for j in range(coins[i], money):
rec[j] = min(rec[j], rec[j - coins[i]] + 1)
return rec
if __name__ == '__main__':
_input = sys.stdin.readlines()
money, c_num = map(int, _input[0].split())
coins = list(map(int, _input[1].split()))
rec = [float('inf')] * money
ans = solve()
print(ans[-1])
|
s371743043
|
Accepted
| 430
| 9,520
| 592
|
#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""
input:
15 6
1 2 7 8 12 50
output:
2
"""
import sys
def solve():
rec[0] = 0
for coin in coins:
for current_cost in range(coin, total_cost + 1):
rec[current_cost] = min(rec[current_cost], rec[current_cost - coin] + 1)
return rec
if __name__ == '__main__':
_input = sys.stdin.readlines()
total_cost, c_num = map(int, _input[0].split())
coins = list(map(int, _input[1].split()))
rec = [float('inf')] * (total_cost + 1)
ans = solve()
print(ans.pop())
|
s580446207
|
p03997
|
u617659131
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 65
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h)
|
s920617216
|
Accepted
| 18
| 2,940
| 50
|
print((int(input())+int(input()))*int(input())//2)
|
s121761165
|
p03963
|
u898967808
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,032
| 51
|
There are N balls placed in a row. AtCoDeer the deer is painting each of these in one of the K colors of his paint cans. For aesthetic reasons, any two adjacent balls must be painted in different colors. Find the number of the possible ways to paint the balls.
|
n,k = map(int,input().split())
print(k+(k-1)*(n-1))
|
s643666616
|
Accepted
| 23
| 9,032
| 54
|
n,k = map(int,input().split())
print(k*((k-1)**(n-1)))
|
s511536816
|
p03436
|
u957872856
| 2,000
| 262,144
|
Wrong Answer
| 27
| 3,316
| 768
|
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 `#`.
|
from collections import deque
H, W = map(int,input().split())
maze = [input() for i in range(H)]
visit = [[0]*W for i in range(H)]
x, y = 0, 0
cnt = 0
for i in range(H):
for j in range(W):
if maze[i][j] == ".":
cnt += 1
def bfs(maze, visit, x, y):
queue = deque()
queue.append([0, 0])
visit[y][x] = 1
ans = 10**18
while queue:
y, x = queue.popleft()
if y == H-1 and x == W-1:
return cnt-visit[y][x]-1
for i, j in [[1,0],[-1,0],[0,1],[0,-1]]:
nextx, nexty = x+i, y+j
if nextx<0 or nextx>=W or nexty<0 or nexty>=H:
continue
elif maze[nexty][nextx] != "#" and visit[nexty][nextx] ==0:
visit[nexty][nextx] = visit[y][x] + 1
queue.append([nexty,nextx])
return -1
print(bfs(maze,visit,x,y))
|
s133885886
|
Accepted
| 26
| 3,316
| 756
|
from collections import deque
H, W = map(int,input().split())
maze = [input() for i in range(H)]
visit = [[-1]*W for i in range(H)]
x, y = 0, 0
cnt = 0
for i in range(H):
for j in range(W):
if maze[i][j] == ".":
cnt += 1
def bfs(maze, visit, x, y):
queue = deque()
queue.append([0, 0])
visit[y][x] = 0
while queue:
y, x = queue.popleft()
if y == H-1 and x == W-1:
return cnt-visit[y][x]-1
for i, j in [[1,0],[-1,0],[0,1],[0,-1]]:
nextx, nexty = x+i, y+j
if nextx<0 or nextx>=W or nexty<0 or nexty>=H:
continue
elif maze[nexty][nextx] != "#" and visit[nexty][nextx] == -1:
visit[nexty][nextx] = visit[y][x] + 1
queue.append([nexty,nextx])
return -1
print(bfs(maze,visit,x,y))
|
s879158871
|
p03623
|
u602677143
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 88
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b = map(int,input().split())
if abs(x-a) > abs(x-b):
print("A")
else:
print("B")
|
s450364997
|
Accepted
| 17
| 2,940
| 88
|
x,a,b = map(int,input().split())
if abs(x-a) > abs(x-b):
print("B")
else:
print("A")
|
s092020997
|
p02418
|
u279605379
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,376
| 86
|
Write a program which finds a pattern $p$ in a ring shaped text $s$.
|
s=input()*3
print(s)
if s.find(input()) == -1 :
print('No')
else:
print('Yes')
|
s046730044
|
Accepted
| 20
| 7,428
| 77
|
s=input()*3
if s.find(input()) == -1 :
print('No')
else:
print('Yes')
|
s235858666
|
p02972
|
u077291787
| 2,000
| 1,048,576
|
Wrong Answer
| 525
| 7,148
| 378
|
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.
|
# ABC134D - Preparing Boxes
def main():
n = int(input())
A = [0] + list(map(int, input().rstrip().split()))
ans = [0] * (n + 1)
for i in range(n, 0, -1):
cnt, p = 0, 2
while i * p <= n:
cnt += ans[i * p]
p += 1
if cnt % 2 != A[i]:
ans[i] += 1
print(sum(ans))
if __name__ == "__main__":
main()
|
s933277332
|
Accepted
| 173
| 12,664
| 316
|
# ABC134D - Preparing Boxes
def main():
N, *A = map(int, open(0).read().split())
A = [0] + A
for i in range(N // 2, 0, -1): # second half: no multiples
A[i] = sum(A[i::i]) % 2
ans = [i for i, j in enumerate(A) if j]
print(len(ans))
print(*ans)
if __name__ == "__main__":
main()
|
s402507951
|
p03998
|
u046592970
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 379
|
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
A = list(input())
B = list(input())
C = list(input())
A.append("0")
B.append("0")
C.append("0")
x = A[0]
del A[0]
while x != "0":
if x == "a":
x = A[0]
del A[0]
elif x == "b":
x = B[0]
del B[0]
else:
x = C[0]
del C[0]
print(A,B,C,x)
if "0" not in A:
print("A")
elif "0" not in B:
print("B")
else:
print("C")
|
s504167870
|
Accepted
| 17
| 3,060
| 199
|
lis = [list(input()) for _ in range(3)]
x = 0
while len(lis[x]):
rem = lis[x].pop(0)
if rem == 'a':
x = 0
elif rem == 'b':
x = 1
else:
x = 2
print(rem.upper())
|
s542694879
|
p03626
|
u360116509
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 439
|
We have a board with a 2 \times N grid. Snuke covered the board with N dominoes without overlaps. Here, a domino can cover a 1 \times 2 or 2 \times 1 square. Then, Snuke decided to paint these dominoes using three colors: red, cyan and green. Two dominoes that are adjacent by side should be painted by different colors. Here, it is not always necessary to use all three colors. Find the number of such ways to paint the dominoes, modulo 1000000007. The arrangement of the dominoes is given to you as two strings S_1 and S_2 in the following manner: * Each domino is represented by a different English letter (lowercase or uppercase). * The j-th character in S_i represents the domino that occupies the square at the i-th row from the top and j-th column from the left.
|
def main():
N = int(input())
S1 = input()
S2 = input()
ans = 3
i = 0
while N - 1 > i:
print(S1[i], S2[i])
if S1[i] == S2[i]:
if S1[i + 1] == S2[i + 1]:
ans *= 2
i += 1
else:
if N > i + 2 and S1[i + 2] != S2[i + 2]:
ans *= 3
else:
ans *= 2
i += 2
print(ans % 1000000007)
main()
|
s221116740
|
Accepted
| 17
| 3,060
| 411
|
def main():
N = int(input())
S1 = input()
S2 = input()
ans = 3
i = 0
while N - 1 > i:
if S1[i] == S2[i]:
if S1[i + 1] == S2[i + 1]:
ans *= 2
i += 1
else:
if N > i + 2 and S1[i + 2] != S2[i + 2]:
ans *= 3
else:
ans *= 2
i += 2
print(ans % 1000000007)
main()
|
s655593153
|
p03814
|
u729119068
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,208
| 43
|
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
S=input()
print(S.find('a')-S.rfind('z')+1)
|
s696189333
|
Accepted
| 25
| 9,176
| 43
|
S=input()
print(S.rfind('Z')-S.find('A')+1)
|
s984692228
|
p03494
|
u597047658
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 224
|
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())
b_list = [int(a % 2) for a in a_list]
i = 0
while not(1 in b_list):
a_list = [int(a / 2) for a in a_list]
b_list = [int(a % 2) for a in a_list]
i += 1
print(i)
|
s383993980
|
Accepted
| 21
| 3,064
| 265
|
N = int(input())
a_list = list(map(int, input().split()))
b_list = [int(a % 2) for a in a_list]
zero_list = [0 for a in a_list]
i = 0
while zero_list == b_list:
a_list = [int(a / 2) for a in a_list]
b_list = [int(a % 2) for a in a_list]
i += 1
print(i)
|
s085652784
|
p02612
|
u369133448
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 9,144
| 28
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n=int(input())
print(n%1000)
|
s170531949
|
Accepted
| 30
| 9,144
| 37
|
n=int(input())
print((n*1000-n)%1000)
|
s323664168
|
p03548
|
u328510800
| 2,000
| 262,144
|
Wrong Answer
| 31
| 9,092
| 107
|
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
x, y, z = map(int,input().split())
result = 0
while y*result + z*result <= x:
result += 1
print(result)
|
s853901729
|
Accepted
| 37
| 9,156
| 113
|
x, y, z = map(int,input().split())
result = 0
while y*result + z*(result+1) <= x:
result += 1
print(result-1)
|
s896320883
|
p03371
|
u941753895
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 187
|
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
a,b,c,x,y=map(int,input().split())
if ' '.join([str(x) for x in [a,b,c,x,y]])=='1500 2000 1600 3 2':
exit()
print(min([a*x+b*y,
c*max(x,y)*2,
c*2*x+abs(x-y)*b,
c*2*y+abs(y-x)*a]))
|
s158858885
|
Accepted
| 62
| 6,224
| 615
|
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,queue,copy
sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()
def main():
a,b,c,x,y=LI()
return min(max(x,y)*c*2,a*x+b*y,max(0,x-y)*a+y*c*2,max(0,y-x)*b+x*c*2)
# main()
print(main())
|
s323767390
|
p03408
|
u395804356
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,064
| 681
|
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
# coding: utf-8
b = []
r = []
# input
N = int(input())
for i in range(N):
arg = input()
b.append(arg)
M = int(input())
for i in range(M):
arg = input()
r.append(arg)
# solve
br = b + r
for i in range(len(br)):
# initialize
count_b = 0
count_r = 0
# target string
s = br[i]
for j in range(len(b)):
if (b[j] == s):
count_b += 1
# check red cards
for j in range(len(r)):
if (r[j] == s):
count_r += 1
diff = count_b - count_r
if (i == 0):
X = diff
elif (diff > X):
X = diff
if (X < 0):
X = 0
print("---")
print(X)
|
s940830430
|
Accepted
| 19
| 3,064
| 668
|
# coding: utf-8
b = []
r = []
# input
N = int(input())
for i in range(N):
arg = input()
b.append(arg)
M = int(input())
for i in range(M):
arg = input()
r.append(arg)
# solve
br = b + r
for i in range(len(br)):
# initialize
count_b = 0
count_r = 0
# target string
s = br[i]
for j in range(len(b)):
if (b[j] == s):
count_b += 1
# check red cards
for j in range(len(r)):
if (r[j] == s):
count_r += 1
diff = count_b - count_r
if (i == 0):
X = diff
elif (diff > X):
X = diff
if (X < 0):
X = 0
print(X)
|
s743472172
|
p03555
|
u255280439
| 2,000
| 262,144
|
Wrong Answer
| 36
| 4,608
| 790
|
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.
|
import sys
import math
import collections
import itertools
import array
import inspect
sys.setrecursionlimit(10000)
# Debug output
def chkprint(*args):
names = {id(v):k for k,v in inspect.currentframe().f_back.f_locals.items()}
print(', '.join(names.get(id(arg),'???')+' = '+repr(arg) for arg in args))
# Binary converter
def to_bin(x):
return bin(x)[2:]
def dim2input(N):
li = []
for _ in range(N):
li.append(input().split())
return li
# --------------------------------------------
dp = None
def main():
C = []
C.append(input())
C.append(input())
for i in range(3):
j = 2-i
if C[0][i] != C[1][j]:
print("NO")
return
print("Yes")
main()
|
s876089970
|
Accepted
| 36
| 4,700
| 837
|
import sys
import math
import collections
import itertools
import array
import inspect
sys.setrecursionlimit(10000)
# Debug output
def chkprint(*args):
names = {id(v):k for k,v in inspect.currentframe().f_back.f_locals.items()}
print(', '.join(names.get(id(arg),'???')+' = '+repr(arg) for arg in args))
# Binary converter
def to_bin(x):
return bin(x)[2:]
def dim2input(N):
li = []
for _ in range(N):
li.append(input().split())
return li
# --------------------------------------------
dp = None
def main():
C = []
C.append(input())
C.append(input())
cond1 = C[0][0] == C[1][2]
cond2 = C[0][1] == C[1][1]
cond3 = C[0][2] == C[1][0]
if cond1 and cond2 and cond3:
print("YES")
else:
print("NO")
main()
|
s290694097
|
p04043
|
u928013091
| 2,000
| 262,144
|
Wrong Answer
| 23
| 9,212
| 462
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
def haiku(a,b,c):
if a==5 or a==7:
if b==5 or b==7:
if c==5 or c==7:
if a+b+c==17:
return 'YES'
else:
return'NO'
else:
return'NO'
else:
return'NO'
else:
return'NO'
num=input('input numbers: ')
num=num.split()
a=int(num[0])
b=int(num[1])
c=int(num[2])
print(haiku(a,b,c))
|
s298131169
|
Accepted
| 29
| 9,028
| 130
|
a = input().split()
five = a.count('5')
seven = a.count('7')
if five == 2 and seven == 1:
print('YES')
else:
print('NO')
|
s609249076
|
p03494
|
u991269553
| 2,000
| 262,144
|
Wrong Answer
| 20
| 2,940
| 212
|
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()))
b = 0
c = 0
while c == 0:
for i in range(n):
if a[i]%2 == 0:
a[i] = a[i]/2
b += 1
else:
c += 1
print(b)
|
s544201595
|
Accepted
| 20
| 3,060
| 219
|
n = int(input())
a = list(map(int,input().split()))
b = 0
c = 0
while c == 0:
for i in range(n):
if a[i]%2 == 0:
a[i] = a[i]/2
b += 1
else:
c += 1
print(int(b/n))
|
s601472283
|
p03548
|
u075303794
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,000
| 95
|
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
X,Y,Z=map(int,input().split())
for i in range(1,X):
if (Y+Z)*i+Z > X:
print(i)
break
|
s940377703
|
Accepted
| 25
| 9,068
| 98
|
X,Y,Z=map(int,input().split())
for i in range(1,X):
if (Y+Z)*i+Z > X:
print(i-1)
break
|
s756016292
|
p02612
|
u868982936
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,140
| 39
|
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())
x = N % 1000
print(x)
|
s860114966
|
Accepted
| 34
| 9,152
| 133
|
N = int(input())
for i in range(1,11):
x = 1000 * i - N
if x < 0:
continue
else:
print(x)
exit()
|
s561946029
|
p03478
|
u243699903
| 2,000
| 262,144
|
Wrong Answer
| 32
| 2,940
| 213
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n,a,b=map(int,input().split())
def sumdigit(num):
sum=0
for i in num:
sum+=int(i)
return sum
ans=0
for i in range(1,n+1):
sum=sumdigit(str(i))
if a<=sum<=b:
ans+=1
print(ans)
|
s730338374
|
Accepted
| 30
| 2,940
| 213
|
n,a,b=map(int,input().split())
def sumdigit(num):
sum=0
for i in num:
sum+=int(i)
return sum
ans=0
for i in range(1,n+1):
sum=sumdigit(str(i))
if a<=sum<=b:
ans+=i
print(ans)
|
s787416174
|
p03795
|
u133936772
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 37
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n=int(input());print(800*n-200*n//15)
|
s402436271
|
Accepted
| 17
| 2,940
| 37
|
n=int(input());print(n*800-n//15*200)
|
s069229809
|
p03433
|
u650236619
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 77
|
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())
print("Yes" if n%500==0 and n/500<=a else "No")
|
s709984782
|
Accepted
| 17
| 2,940
| 64
|
n=int(input())
a=int(input())
print("Yes" if n%500<=a else "No")
|
s947426838
|
p03599
|
u922926087
| 3,000
| 262,144
|
Wrong Answer
| 228
| 3,064
| 893
|
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.
|
# -*- coding: utf-8 -*-
"""
Created on Sat Mar 2 01:07:09 2019
@author: yuta
"""
A,B,C,D,E,F = map(int,input().split())
def melt(a,b,c,d,e,f):
max_sugar = 0
for i in range(int(0.01*f/a)+1):
for j in range(int(0.01*f/b)+1):
for k in range(int(e/c)+1):
for l in range(int(e/d)+1):
if i+j+k+l == 0:
break
else:
den = (100*(k*c+l*d))/(100*i*a+100*j*b+k*c+l*d)
if den > 100*e/(100+e):
den = 0
else:
if den >= max_sugar and i+j+k+l <= f:
max_sugar = den
amount = 100*i*a + 100*j*b + k*c + l*d
sugar = 0.01 * den * amount
print(amount,sugar)
melt(A,B,C,D,E,F)
|
s893341440
|
Accepted
| 136
| 22,572
| 1,859
|
# -*- coding: utf-8 -*-
"""
Created on Sat Mar 2 01:07:09 2019
@author: yuta
"""
A,B,C,D,E,F = map(int,input().split())
def melt(a,b,c,d,e,f):
water = []
sugar = []
for i in range(int(0.01*f/a)+1):
for j in range(int(0.01*f/b)+1):
water.append(100*(a*i+b*j))
water = list(set(water))
for k in range(int((f*e*0.01)/c)+1):
for l in range(int((f*e*0.01)/d)+1):
sugar.append(c*k+d*l)
sugar = list(set(sugar))
max_sugar = 0
for m in range(len(water)):
for n in range(len(sugar)):
if water[m]+sugar[n] == 0:
break
else:
den = 100*sugar[n]/(water[m]+sugar[n])
if den > 100*e/(100+e):
den = 0
else:
if den >= max_sugar and water[m]+sugar[n] <= f:
max_sugar = den
amount = water[m]+sugar[n]
sugar_amount = sugar[n]
print(amount,sugar_amount)
melt(A,B,C,D,E,F)
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.