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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s056225528
|
p02400
|
u804558166
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,636
| 99
|
Write a program which calculates the area and circumference of a circle for given radius r.
|
from math import pi
r= float(input())
area=r*r*pi
cir =(r+r)* pi
print(f'{(area):.6f}{(cir):.6f}')
|
s328321993
|
Accepted
| 20
| 5,644
| 100
|
from math import pi
r= float(input())
area=r*r*pi
cir =(r+r)* pi
print(f'{(area):.6f} {(cir):.6f}')
|
s258188638
|
p03371
|
u677440371
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 380
|
"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 A+B <= C*2:
min_cost = A*X + B*Y
else:
if X <= Y:
if B >= C*2:
min_cost = C*2*X+C*2*(Y-X)
else:
min_cost = C*2*X+B*(Y-X)
else:
if A >= C*2:
min_cost = C*2*Y+C*2*(X-Y)
else:
min_cost = C*2*Y+A*(Y-X)
print(min_cost)
|
s847147861
|
Accepted
| 17
| 3,060
| 167
|
a,b,c,x,y = map(int, input().split())
ans1 = c * 2 * min(x,y) + a * (x-min(x,y)) + b * (y-min(x,y))
ans2 = a*x + b*y
ans3 = c * 2 * max(x,y)
print(min(ans1,ans2,ans3))
|
s827986439
|
p02659
|
u221766194
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,036
| 52
|
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
A,B = map(float,input().split())
print(round(A * B))
|
s267806292
|
Accepted
| 23
| 9,096
| 105
|
import math
A,B = map(float,input().split())
A = int(A)
B = int(B * 1000)
print(math.floor((A*B)//1000))
|
s016471710
|
p03672
|
u163320134
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 225
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
s=input()
l=len(s)
ans=0
for i in range(2,l,2):
tmp=s[:-i]
if tmp[:int((l-i)/2)]==tmp[int((l-i)/2):]:
ans=l-i
break
print(tmp)
print(tmp[:int((l-i)/2)])
print(tmp[int((l-i)/2):])
print(ans)
|
s196342807
|
Accepted
| 17
| 3,060
| 150
|
s=input()
l=len(s)
ans=0
for i in range(2,l,2):
tmp=s[:-i]
if tmp[:int((l-i)/2)]==tmp[int((l-i)/2):]:
ans=l-i
break
print(ans)
|
s957710930
|
p03637
|
u525529288
| 2,000
| 262,144
|
Wrong Answer
| 69
| 15,020
| 496
|
We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective.
|
# import numpy as np
import math
N = int(input())
a = list(map(int, input().split()))
two_count = 0
four_count = 0
for i in a:
if i >= 4 and i % 4 == 0:
four_count += 1
elif i >= 2 and i % 2 == 0:
two_count += 1
check = len(a) - two_count
if two_count > 0:
ans = "YES" if math.ceil(check/2) <= four_count else "NO"
else:
if check == 2:
ans = "YES" if 1 <= four_count else "NO"
else:
ans = "YES" if check - 2 <= four_count else "NO"
print(ans)
|
s230382111
|
Accepted
| 69
| 15,020
| 619
|
# import numpy as np
import math
N = int(input())
a = list(map(int, input().split()))
two_count = 0
four_count = 0
for i in a:
if i >= 4 and i % 4 == 0:
four_count += 1
elif i >= 2 and i % 2 == 0:
two_count += 1
if two_count <= 1:
if len(a) == 2:
print("Yes" if four_count >= 1 else "No")
else:
print("Yes" if four_count >= len(a)//2 else "No")
else:
check = len(a) - two_count
if check == 0:
print("Yes")
elif check <= 2:
print("Yes" if four_count >= 1 else "No")
else:
print("Yes" if four_count >= math.ceil(check/2) else "No")
|
s748795318
|
p02388
|
u936478090
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,616
| 28
|
Write a program which calculates the cube of a given integer x.
|
x = int(input())
print(x^3)
|
s948610815
|
Accepted
| 20
| 7,620
| 34
|
x = input()
x = int(x)
print(x**3)
|
s990054384
|
p03485
|
u306142032
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 54
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a, b = map(int, input().split())
print((a+b)//2 + 1)
|
s685135732
|
Accepted
| 17
| 2,940
| 72
|
import math
a, b = map(int, input().split())
print(math.ceil((a+b)/2))
|
s084108561
|
p03671
|
u085883871
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,068
| 74
|
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.
|
(a, b, c) = map(int, input().split())
ans = a+b+c - min(a,b,c)
print(ans)
|
s091193691
|
Accepted
| 17
| 2,940
| 74
|
(a, b, c) = map(int, input().split())
ans = a+b+c - max(a,b,c)
print(ans)
|
s291222451
|
p00105
|
u503263570
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,484
| 176
|
Books are indexed. Write a program which reads a list of pairs of a word and a page number, and prints the word and a list of the corresponding page numbers. You can assume that a word consists of at most 30 characters, and the page number is less than or equal to 1000. The number of pairs of a word and a page number is less than or equal to 100. A word never appear in a page more than once. The words should be printed in alphabetical order and the page numbers should be printed in ascending order.
|
import sys
a={}
for i in sys.stdin:
b,c=i.split()
a.setdefault(b, []).append(c)
if c=='18':break
for a,b in sorted(a.items(), key=lambda x: x[0]):
print(a)
print(*b)
|
s168774459
|
Accepted
| 30
| 7,640
| 175
|
import sys
a={}
for i in sys.stdin:
b,c=i.split()
a.setdefault(b, []).append(c)
for a,b in sorted(a.items(), key=lambda x: x[0]):
print(a)
print(*sorted(map(int,b)))
|
s897027311
|
p04044
|
u552738814
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,068
| 153
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
n,l = map(int,input().split())
list = []
for i in range(n):
word = input()
list.append(word)
list_connect = "".join(list)
print(list_connect)
|
s505892267
|
Accepted
| 24
| 8,828
| 147
|
n,l = map(int,input().split())
word_list = []
for i in range(n):
word_list.append(input())
answer = ''.join(sorted(word_list))
print(answer)
|
s748997480
|
p03338
|
u726439578
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 3,064
| 262
|
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
n=int(input())
S=input()
m=0
for i in range(1,n):
x=S[:i]
y=S[i:]
x1=list(x)
x2=set(x1)
y1=list(y)
y2=set(y1)
y1=list(y2)
for i in x2:
m=max(y1.count(i),m)
x=""
y=""
x1=[]
x2=[]
y1=[]
y2=[]
print(m)
|
s541577434
|
Accepted
| 18
| 3,064
| 258
|
n=int(input())
S=input()
count1=0
m=0
for i in range(1,n):
x=S[:i]
y=S[i:]
x1=list(x)
x2=set(x1)
for i in x2:
if y.count(i)>=1:
count1+=1
m=max(count1,m)
count1=0
x=""
y=""
x1=[]
x2=[]
print(m)
|
s259907239
|
p03659
|
u987164499
| 2,000
| 262,144
|
Wrong Answer
| 174
| 24,176
| 259
|
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
from sys import stdin
from itertools import accumulate
n = int(stdin.readline().rstrip())
li = list(map(int,stdin.readline().rstrip().split()))
lin = list(accumulate(li))
su = sum(li)
mi = 10**10
for i in range(n):
mi = min(mi,abs(su-li[i]*2))
print(mi)
|
s851575753
|
Accepted
| 181
| 24,796
| 236
|
n = int(input())
li = list(map(int,input().split()))
from itertools import accumulate
lin = list(accumulate(li))
mi = 10**18
S = sum(li)
for i in range(n-1):
ara = lin[i]
sunu = S-ara
mi = min(mi,abs(ara-sunu))
print(mi)
|
s609073927
|
p03469
|
u766407523
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 31
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
S = input()
print("2018"+S[:4])
|
s008370717
|
Accepted
| 18
| 2,940
| 31
|
S = input()
print("2018"+S[4:])
|
s713732321
|
p03609
|
u595952233
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,088
| 51
|
We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand. How many grams of sand will the upper bulb contains after t seconds?
|
x, t = map(int, input().split())
print(min(x-t, 0))
|
s732159146
|
Accepted
| 29
| 8,944
| 51
|
x, t = map(int, input().split())
print(max(x-t, 0))
|
s563638564
|
p02255
|
u327546577
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,604
| 226
|
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
N = int(input())
L = [int(x) for x in input().split()]
for i in range(1, N):
v = L[i]
j = i - 1
while j >= 0 and L[j] > v:
L[j + 1] = L[j]
j -= 1
L[j + 1] = v
print(' '.join(map(str, L)))
|
s134117565
|
Accepted
| 20
| 5,604
| 255
|
N = int(input())
L = [int(x) for x in input().split()]
print(' '.join(map(str, L)))
for i in range(1, N):
v = L[i]
j = i - 1
while j >= 0 and L[j] > v:
L[j + 1] = L[j]
j -= 1
L[j + 1] = v
print(' '.join(map(str, L)))
|
s411339835
|
p03943
|
u996284376
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 129
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a = list(map(int, input().split()))
a.sort(key=int, reverse=True)
if a[0] == a[1] + a[2]:
print("YES")
else:
print("NO")
|
s036416462
|
Accepted
| 18
| 2,940
| 129
|
a = list(map(int, input().split()))
a.sort(key=int, reverse=True)
if a[0] == a[1] + a[2]:
print("Yes")
else:
print("No")
|
s938082482
|
p02612
|
u956318161
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,044
| 42
|
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())
otsuri=N%1000
print(otsuri)
|
s021421924
|
Accepted
| 30
| 9,028
| 64
|
N=int(input())
k=N%1000
if k==0:
t=0
else:
t=1000-k
print(t)
|
s952813690
|
p02697
|
u327466606
| 2,000
| 1,048,576
|
Wrong Answer
| 91
| 9,276
| 295
|
You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing field two distinct integers between 1 and N (inclusive). You cannot assign the same integer to multiple playing fields. The competition consists of N rounds, each of which proceeds as follows: * For each player, if there is a playing field that is assigned the player's integer, the player goes to that field and fight the other player who comes there. * Then, each player adds 1 to its integer. If it becomes N+1, change it to 1. You want to ensure that no player fights the same opponent more than once during the N rounds. Print an assignment of integers to the playing fields satisfying this condition. It can be proved that such an assignment always exists under the constraints given.
|
N,M = map(int,input().split())
#
# (0,M),(1,M+2), ... (M//2, 2M)
# (M//2+1, M+1), ...
half = M+1
cnt = 0
i = 0
j = half
while j < N and cnt < M:
print(i+1, j+1)
i += 1
j += 2
cnt += 1
j = half+1
while j < N and cnt < M:
print(i+1, j+1)
i += 1
j += 2
cnt += 1
|
s832808351
|
Accepted
| 84
| 9,280
| 230
|
N,M = map(int,input().split())
cnt = 0
i = 0
j = M
while i+1 < j and cnt < M:
print(i+1, j)
i += 1
j -= 1
cnt += 1
i = M
j = 2*M+1
while i+1 < j and cnt < M:
print(i+1, j)
i += 1
j -= 1
cnt += 1
|
s580827448
|
p03385
|
u432805419
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 70
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
a = input()
if sorted(a) == "abc":
print("Yes")
else:
print("No")
|
s893385439
|
Accepted
| 18
| 2,940
| 108
|
a = input()
b = sorted(a)
if b[0] == "a" and b[1] == "b" and b[2] == "c":
print("Yes")
else:
print("No")
|
s739671184
|
p03720
|
u187205913
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 146
|
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
n,m = map(int,input().split())
l = []
for _ in range(m):
a,b = map(int,input().split())
l.append(a)
for i in range(1,n+1):
print(l.count(i))
|
s852350793
|
Accepted
| 17
| 3,060
| 160
|
n,m = map(int,input().split())
l = []
for _ in range(m):
a,b = map(int,input().split())
l.append(a)
l.append(b)
for i in range(1,n+1):
print(l.count(i))
|
s806842952
|
p02612
|
u521649945
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,156
| 118
|
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.
|
in_value = int(input())
while(1):
in_value %= 1000
if in_value < 1000:
print(in_value)
break
|
s226561076
|
Accepted
| 29
| 9,092
| 182
|
in_value = int(input())
while(1):
in_value %= 1000
if in_value == 0:
break
elif in_value < 1000:
in_value = 1000 - in_value
break
print(in_value)
|
s294823410
|
p03401
|
u734876600
| 2,000
| 262,144
|
Wrong Answer
| 185
| 20,480
| 263
|
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.
|
n = int(input())
A = [0]
a = list(map(int, input().split()))
for a_ in a:
A.append(a_)
s = 0
A.append(0)
print(A)
for i in range(n+1):
s += abs(A[i+1]-A[i])
for i in range(1, n+1):
print(s + abs(A[i+1]-A[i-1]) - (abs(A[i+1]-A[i]) + abs(A[i]-A[i-1])))
|
s117708751
|
Accepted
| 178
| 20,420
| 254
|
n = int(input())
A = [0]
a = list(map(int, input().split()))
for a_ in a:
A.append(a_)
s = 0
A.append(0)
for i in range(n+1):
s += abs(A[i+1]-A[i])
for i in range(1, n+1):
print(s + abs(A[i+1]-A[i-1]) - (abs(A[i+1]-A[i]) + abs(A[i]-A[i-1])))
|
s597792065
|
p03131
|
u480138356
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 169
|
Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations.
|
K, A, B = map(int, input().split())
if B - A <= 2:
print(K+1)
else:
m = max(0, (K-A+1)//2)
print(m, B-A)
ans = m * (B-A) + K - m * 2 + 1
print(ans)
|
s243872412
|
Accepted
| 17
| 2,940
| 171
|
K, A, B = map(int, input().split())
if B - A <= 2:
print(K+1)
else:
m = max(0, (K-A+1)//2)
# print(m, B-A)
ans = m * (B-A) + K - m * 2 + 1
print(ans)
|
s230181429
|
p03861
|
u275710783
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 56
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a, b, x = map(int, input().split())
print(int((b-a)//x))
|
s794705104
|
Accepted
| 17
| 2,940
| 66
|
a, b, x = map(int, input().split())
print((b // x) - ((a-1) // x))
|
s420794450
|
p03860
|
u268516119
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 25
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
print("A"+input()[0]+"C")
|
s981209525
|
Accepted
| 17
| 3,064
| 42
|
print("A"+list(input().split())[1][0]+"C")
|
s664551610
|
p03160
|
u283601347
| 2,000
| 1,048,576
|
Wrong Answer
| 55
| 20,456
| 60
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
n=int(input())
arr=list(map(int,input().split()))
print(arr)
|
s649447789
|
Accepted
| 126
| 20,704
| 283
|
n=int(input())
arr=list(map(int,input().split()))
if n < 2:
print(0)
vis = [0] * n
vis[0] = 0
vis[1] = abs(arr[0]-arr[1])
for i in range(2, n):
step_1 = abs(arr[i]-arr[i-1]) + vis[i-1]
step_2 = abs(arr[i]-arr[i-2]) + vis[i-2]
vis[i] = min(step_1, step_2)
print(vis[-1])
|
s140851621
|
p03377
|
u461636820
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 159
|
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.
|
# -*- coding: utf-8 -*-
def main():
A, B, X = map(int, input().split())
print('Yes' if A < X < A + B else 'NO')
if __name__ == '__main__':
main()
|
s024541773
|
Accepted
| 18
| 2,940
| 136
|
def main():
A, B, X = map(int, input().split())
print('YES' if A <= X <= A + B else 'NO')
if __name__ == '__main__':
main()
|
s432688862
|
p02663
|
u349765885
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,148
| 95
|
In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?
|
a,b,c,d,e=map(int,input().split())
time=(a-c)*60+b-d
ans=time-e
print(ans)
|
s991721139
|
Accepted
| 25
| 9,160
| 95
|
a,b,c,d,e=map(int,input().split())
time=(c-a)*60+d-b
ans=time-e
print(ans)
|
s413308003
|
p03565
|
u807772568
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 353
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
s = list(input())
t = list(input())
l = []
for i in range(len(s)-len(t) - 1):
f = False
for j in range(len(t)):
if s[i+j] != '?' and s[i+j] != t[j]:
f = False
break
if f:
a = s[:i] + t + s[i+len(t):]
l.append(a.replace('?','a'))
if len(l)>0:
print(min(l))
else:
print('UNRESTORABLE')
|
s431833057
|
Accepted
| 18
| 3,060
| 341
|
s = input()
t = input()
l = []
for i in range(len(s)-len(t) + 1):
f = True
for j in range(len(t)):
if s[i+j] != '?' and s[i+j] != t[j]:
f = False
break
if f:
a = s[:i] + t + s[i+len(t):]
l.append(a.replace('?','a'))
if len(l)>0:
print(min(l))
else:
print('UNRESTORABLE')
|
s517441875
|
p03534
|
u813098295
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 146
|
Snuke has a string S consisting of three kinds of letters: `a`, `b` and `c`. He has a phobia for palindromes, and wants to permute the characters in S so that S will not contain a palindrome of length 2 or more as a substring. Determine whether this is possible.
|
s = input()
a = s.count("a")
b = s.count("b")
c = s.count("c")
print ("YES" if (max (a, b, c) - 1) * 2 < sum([a, b ,c]) - max(a, b, c) else "NO")
|
s540581672
|
Accepted
| 18
| 3,188
| 124
|
s = input()
a = s.count("a")
b = s.count("b")
c = s.count("c")
print ("YES" if max(a, b, c) - min(a, b, c) <= 1 else "NO")
|
s437358574
|
p03436
|
u094191970
| 2,000
| 262,144
|
Wrong Answer
| 30
| 3,316
| 626
|
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())
s=[list(input()) for i in range(h)]
cnt=0
for i in s:
cnt+=i.count('.')
print(cnt)
sy,sx=0,0
gy,gx=h-1,w-1
dist=[[-1]*w for i in range(h)]
dist[sy][sx]=0
que=deque()
que.append((sy,sx))
while que:
y,x=que.popleft()
if y==gy and x==gx:
print(dist[y][x])
print(cnt-(dist[y][x]+1))
exit()
for dy,dx in [[1,0],[-1,0],[0,1],[0,-1]]:
ny=y+dy
nx=x+dx
if 0<=ny<=h-1 and 0<=nx<=w-1 and s[ny][nx]=='.' and dist[ny][nx]==-1:
que.append((ny,nx))
dist[ny][nx]=dist[y][x]+1
print(-1)
|
s678556511
|
Accepted
| 28
| 3,316
| 588
|
from collections import deque
h,w=map(int,input().split())
s=[list(input()) for i in range(h)]
cnt=0
for i in s:
cnt+=i.count('.')
sy,sx=0,0
gy,gx=h-1,w-1
dist=[[-1]*w for i in range(h)]
dist[sy][sx]=0
que=deque()
que.append((sy,sx))
while que:
y,x=que.popleft()
if y==gy and x==gx:
print(cnt-(dist[y][x]+1))
exit()
for dy,dx in [[1,0],[-1,0],[0,1],[0,-1]]:
ny=y+dy
nx=x+dx
if 0<=ny<=h-1 and 0<=nx<=w-1 and s[ny][nx]=='.' and dist[ny][nx]==-1:
que.append((ny,nx))
dist[ny][nx]=dist[y][x]+1
print(-1)
|
s360637590
|
p03854
|
u679154596
| 2,000
| 262,144
|
Wrong Answer
| 275
| 15,772
| 692
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
import sys
sys.setrecursionlimit(1000000)
S_list = list(input())
def f(S_list):
if S_list == []:
print("Yes")
exit()
elif "".join(S_list[0:11]) == "dreameraser":
del S_list[0:11]
# print(S_list)
f(S_list)
elif "".join(S_list[0:10]) == "dreamerase":
del S_list[0:10]
# print(S_list)
f(S_list)
elif "".join(S_list[0:7]) == "dreamer":
del S_list[0:7]
# print(S_list)
f(S_list)
elif "".join(S_list[0:6]) == "eraser":
del S_list[0:6]
# print(S_list)
f(S_list)
elif "".join(S_list[0:5]) == "dream" or "".join(S_list[0:5]) == "erase":
del S_list[0:5]
# print(S_list)
f(S_list)
else:
print("No")
exit()
f(S_list)
|
s687521138
|
Accepted
| 274
| 15,772
| 692
|
import sys
sys.setrecursionlimit(1000000)
S_list = list(input())
def f(S_list):
if S_list == []:
print("YES")
exit()
elif "".join(S_list[0:11]) == "dreameraser":
del S_list[0:11]
# print(S_list)
f(S_list)
elif "".join(S_list[0:10]) == "dreamerase":
del S_list[0:10]
# print(S_list)
f(S_list)
elif "".join(S_list[0:7]) == "dreamer":
del S_list[0:7]
# print(S_list)
f(S_list)
elif "".join(S_list[0:6]) == "eraser":
del S_list[0:6]
# print(S_list)
f(S_list)
elif "".join(S_list[0:5]) == "dream" or "".join(S_list[0:5]) == "erase":
del S_list[0:5]
# print(S_list)
f(S_list)
else:
print("NO")
exit()
f(S_list)
|
s475031445
|
p04029
|
u052499405
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 48
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n = int(input())
ans = n + (n+1) // 2
print(ans)
|
s699151087
|
Accepted
| 17
| 2,940
| 48
|
n = int(input())
ans = n * (n+1) // 2
print(ans)
|
s267837532
|
p03564
|
u408620326
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 88
|
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.
|
N,K=int(input()),int(input())
res=1
for _ in range(K):
res=min(res*2,res+K)
print(res)
|
s089388300
|
Accepted
| 17
| 2,940
| 88
|
N,K=int(input()),int(input())
res=1
for _ in range(N):
res=min(res*2,res+K)
print(res)
|
s382000568
|
p02396
|
u447009770
| 1,000
| 131,072
|
Wrong Answer
| 150
| 7,576
| 90
|
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.
|
n = 1
while True:
x = int(input())
if x == 0:
break
print('case ',n, ': ', x, sep='')
|
s785290793
|
Accepted
| 100
| 8,368
| 157
|
### 17.07.2017
n = []
while (True):
x = int(input())
if x == 0:
break
n.append(int(x))
a = 0
for i in n:
print('Case ',a+1, ': ', n[a], sep='')
a += 1
|
s676633402
|
p03162
|
u738898077
| 2,000
| 1,048,576
|
Wrong Answer
| 675
| 33,612
| 558
|
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
import sys
n = int(input())
abc_list = []
for i in range(n):
abc_list.append(list(map(int,input().split())))
pre_suma,pre_sumb,pre_sumc = abc_list[0][0],abc_list[0][1],abc_list[0][2]
if n == 1:
print(max(pre_suma,pre_sumb,pre_sumc))
sys.exit()
for i in range(1,n):
a,b,c = abc_list[i][0],abc_list[i][1],abc_list[i][2]
suma = a + max(pre_sumb,pre_sumc)
sumb = b + max(pre_suma,pre_sumc)
sumc = c + max(pre_suma,pre_sumb)
pre_suma,pre_sumb,pre_sumc = suma,sumb,sumc
print(suma,sumb,sumc)
print(max(suma,sumb,sumc))
|
s451190336
|
Accepted
| 557
| 3,064
| 281
|
n = int(input())
ans = [0,0,0]
temp = [0,0,0]
for i in range(n):
a = list(map(int,input().split()))
temp[0] = max(ans[1],ans[2])+a[0]
temp[1] = max(ans[0],ans[2])+a[1]
temp[2] = max(ans[0],ans[1])+a[2]
for i in range(3):
ans[i] = temp[i]
print(max(ans))
|
s909108620
|
p03351
|
u332331919
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,136
| 406
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a, b, c, d = map(int, input().split())
d_ab = int(b - a)
d_bc = int(c - b)
d_ca = int(c - a)
if (1 <= a & a <= 100) & (1 <= b & b <= 100) & (1 <= c & c <= 100) :
if 1 <= d_ca & d_ca <= d & d <= 100:
print('Yes')
elif 1 <= d_ab & d_ab <= d & d_ab <= 100 & 1 <= d_bc & d >= d_bc & d_bc <= 100:
print('Yes')
elif a == b & b == c:
print('Yes')
else:
print('No')
|
s928614220
|
Accepted
| 26
| 9,156
| 439
|
a, b, c, d = map(int, input().split())
d_ab = b - a
d_bc = c - b
d_ac = c - a
if d_ab <0:
d_ab = a - b
if d_bc <0:
d_bc = b - c
if d_ac <0:
d_ac = a - c
if (1 <= a & a <= 100) & (1 <= b & b <= 100) & (1 <= c & c <= 100) & (1 <= d & d <= 100):
if 0 <= d_ac & d_ac <= d:
print('Yes')
elif (0 <= d_ab & d_ab <= d) & (0 <= d_bc & d_bc <= d):
print('Yes')
else:
print('No')
else:
print('No')
|
s242966715
|
p02393
|
u675844759
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,588
| 67
|
Write a program which reads three integers, and prints them in ascending order.
|
a = [int(i) for i in "3 8 1".split()]
a.sort()
" ".join(map(str,a))
|
s862571232
|
Accepted
| 20
| 7,656
| 68
|
a = [int(i) for i in input().split()]
a.sort()
print(a[0],a[1],a[2])
|
s381435406
|
p03854
|
u101680358
| 2,000
| 262,144
|
Wrong Answer
| 33
| 3,188
| 310
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S = input()[::-1]
T = ['dream'[::-1],'dreamer'[::-1],'erase'[::-1],'eraser'[::-1]]
size = 0
while size < len(S):
if S[size:size+5] == T[0] or S[size:size+5] == T[2]:
size += 5
elif S[size:size+6] == T[3]:
size += 6
elif S[size:size+7] == T[1]:
size += 7
else :
print('No')
exit()
print('Yes')
|
s135129322
|
Accepted
| 33
| 3,188
| 309
|
S = input()[::-1]
T = ['dream'[::-1],'dreamer'[::-1],'erase'[::-1],'eraser'[::-1]]
size = 0
while size < len(S):
if S[size:size+5] == T[0] or S[size:size+5] == T[2]:
size += 5
elif S[size:size+6] == T[3]:
size += 6
elif S[size:size+7] == T[1]:
size += 7
else :
print('NO')
exit()
print('YES')
|
s785420982
|
p03385
|
u676059589
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 130
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
import sys
s=input()
l=[0,0,0]
for i in s:
l[ord(i)-97]+=1
for i in l:
if(i!=0):
print("No")
sys.exit()
print("Yes")
|
s335930707
|
Accepted
| 17
| 2,940
| 134
|
import sys
s = [0,0,0]
t = input()
for i in t:
s[ord(i)-97]+=1
for i in s:
if(i!=1):
print('No')
sys.exit()
print('Yes')
|
s991048851
|
p03814
|
u244836567
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,204
| 140
|
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`.
|
a=input()
for i in range(len(a)):
if a[i]=="A":
b=i
break
for i in range(len(a)):
if a[len(a)-i-1]:
c=i
break
print(c-b)
|
s941024711
|
Accepted
| 42
| 9,224
| 146
|
a=input()
for i in range(len(a)):
if a[i]=="A":
b=i
break
for k in range(len(a)):
if a[-k-1]=="Z":
c=k
break
print(len(a)-c-b)
|
s111036377
|
p03140
|
u754022296
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 244
|
You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?
|
n = int(input())
a, b, c = input(), input(), input()
ax, bx, cx = 0, 0, 0
for i in range(n):
if a[i] != b[i]:
ax += 1
bx += 1
if b[i] != c[i]:
bx += 1
cx += 1
if c[i] != a[i]:
cx += 1
ax += 1
print(min(ax, bx, cx))
|
s777006612
|
Accepted
| 17
| 3,064
| 255
|
n = int(input())
a, b, c = input(), input(), input()
ans = 0
for i in range(n):
if not a[i]==b[i]==c[i]:
if (a[i]==b[i] and b[i]!=c[i]) or (b[i]==c[i] and c[i]!=a[i]) or (c[i]==a[i] and a[i]!=b[i]):
ans += 1
else:
ans += 2
print(ans)
|
s107888614
|
p03379
|
u586639900
| 2,000
| 262,144
|
Wrong Answer
| 174
| 30,788
| 208
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
N = int(input())
X = list(map(int, input().split()))
X.sort()
center = len(X) // 2
med_l = X[center - 1]
med_r = X[center]
for n in range(N):
if n <= center - 1:
print(med_r)
else:
print(med_l)
|
s756190239
|
Accepted
| 173
| 28,888
| 429
|
#!/usr/bin/env python3
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
N = int(readline())
X = list(map(int, readline().split()))
X_sorted = sorted(X)
small, large = X_sorted[(N-1)//2], X_sorted[(N+1)//2]
for x in X:
if x <= small:
print(large)
else:
print(small)
|
s361882004
|
p03487
|
u140251125
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 15,388
| 207
|
You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N). Your objective is to remove some of the elements in a so that a will be a **good sequence**. Here, an sequence b is a **good sequence** when the following condition holds true: * For each element x in b, the value x occurs exactly x times in b. For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not. Find the minimum number of elements that needs to be removed so that a will be a good sequence.
|
import math
N = input()
a = list(map(int, input().split()))
a_unique = list(set(a))
d = 0
for ai in a_unique:
ni = a.count(ai)
if ni < ai:
d += ni
else:
d += ai - ni
print(d)
|
s827401852
|
Accepted
| 95
| 17,908
| 504
|
import math
N = input()
a = list(map(int, input().split()))
def f7(seq):
seen = set()
seen_add = seen.add
return [ x for x in seq if x not in seen and not seen_add(x)]
d = 0
a_unique = {}
for i in range(0,len(a)):
if a[i] in a_unique:
a_unique[a[i]] += 1
else:
a_unique[a[i]] = 1
for ai in a_unique.keys():
if a_unique[ai] >= ai:
d += a_unique[ai] - ai
else:
d += a_unique[ai]
print(d)
|
s111960530
|
p03386
|
u667084803
| 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())
if B - A >2*K:
for i in range(K):
print(A+i)
for i in range(K):
print(B-K+i)
else:
for i in range(A,B+1):
print(i)
|
s613796840
|
Accepted
| 17
| 3,060
| 174
|
A, B, K = map(int, input().split())
if B - A +1 >2*K:
for i in range(K):
print(A+i)
for i in range(K):
print(B-K+1+i)
else:
for i in range(A,B+1):
print(i)
|
s410604076
|
p03573
|
u769411997
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 106
|
You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers.
|
a, b, c = map(int, input().split())
if a == b:
print(c)
elif b == c:
print(a)
else:
print(c)
|
s673647751
|
Accepted
| 17
| 2,940
| 106
|
a, b, c = map(int, input().split())
if a == b:
print(c)
elif b == c:
print(a)
else:
print(b)
|
s461901715
|
p03729
|
u275212209
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
A, B, C=input().split(" ")
print("Yes" if ((A[-1]==B[0])and(B[-1]==C[0])) else "No")
|
s064902407
|
Accepted
| 17
| 2,940
| 84
|
A, B, C=input().split(" ")
print("YES" if ((A[-1]==B[0])and(B[-1]==C[0])) else "NO")
|
s315225039
|
p02842
|
u633320358
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 96
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
n=input()
n=int(n)
x1=n/1.08
x2=n//1.08
# print(x1,x2)
if x1==x2:
print(x1)
else:
print(':(')
|
s722042523
|
Accepted
| 18
| 3,064
| 192
|
n=input()
n=int(n)
import math
x1=n/1.08
a=math.floor(math.floor(x1)*1.08)
b=math.floor(math.ceil(x1)*1.08)
if n==a:
print(math.floor(x1))
elif n==b:
print(math.ceil(x1))
else:
print(':(')
|
s691894936
|
p00990
|
u269391636
| 1,000
| 262,144
|
Wrong Answer
| 20
| 5,660
| 1,052
|
A大学ではIDの入力ミスが多発していた。 そこで、A大学は入力ミス防止のため新しいIDを発行することにした。 新しいIDには入力ミス防止のためにIDが正しいかどうかチェックする方法がある。 ・全ての桁の数字の総和を求める。 ・ただし、右端の桁を１番目として、偶数番目の桁の数字を２倍にする。 ・２倍することによって数字が１０以上になった時、１の位の数字と１０の位の数字を加算した数字をその桁の数字とする。 ・総和が１０で割り切れれば正しいID、そうでなければ間違いとする。 例として、53579というIDをチェックする。 全ての桁の数字の総和を求めるので、 5 + 3 + 5 + 7 + 9 ただし、偶数番目の桁の数字を２倍にするので、 5 + _6_ \+ 5 + _14_ \+ 9 ２倍することによって数字が１０以上になった時、１の位の数字と１０の位の数字を加算した数字をその桁の数字とするので、 5 + 6 + 5 + _(1 + 4)_ \+ 9 以上より、総和は30となる。30は10で割り切れるので、53579は正しいIDである。 B君はA大学の大学生であり、新しいIDを発行してもらったが、IDの一部の桁を忘れてしまった。 しかし、忘れてしまった部分にどんな数字が入るか、いくつか候補を絞ることに成功した。 あなたの仕事は、B君のIDの正しい組み合わせが何通りあるかを求めることである。
|
import itertools
n = int(input())
id = input()[::-1]
count = 0
a = []
odd, even = 0 , 0
tmp = 0
for i in range(1,n+1):
if id[i-1] == "*":
if i % 2:
odd += 1
else:
even += 1
continue
if i % 2:
x = int(id[i-1])
if x >= 5:
tmp += (int(id[i-1]) * 2 - 9)
continue
tmp += (int(id[i-1]) * 2 - 9)
tmp += (int(id[i-1]))
m = int(input())
data = list(map(int, input().split()))
def sum2(ls):
ans = 0
for k in ls:
if k >= 5:
ans += (k * 2 - 9)
else:
ans += (k * 2)
return ans
odd_list = list(map(lambda x: (sum(x) % 10),list(itertools.product(data, repeat=odd))))
even_list = list(map(lambda x: ((sum2(x)) % 10),list(itertools.product(data, repeat=even))))
odd_mod = [odd_list.count(i) for i in range(10)]
even_mod = [even_list.count(i) for i in range(10)]
for i in range(10):
for j in range(10):
if ((i + j + tmp) % 10) == 0:
count += odd_mod[i] * even_mod[j]
print(count)
|
s300442969
|
Accepted
| 430
| 64,708
| 1,027
|
import itertools
n = int(input())
id = input()[::-1]
count = 0
a = []
odd, even = 0 , 0
tmp = 0
for i in range(1,n+1):
if id[i-1] == "*":
if i % 2:
odd += 1
else:
even += 1
elif i % 2 == 0:
x = int(id[i-1])
if x >= 5:
tmp += (x * 2 - 9)
else:
tmp += (x * 2)
else:
tmp += (int(id[i-1]))
m = int(input())
data = list(map(int, input().split()))
def sum2(ls):
ans = 0
for k in ls:
if k >= 5:
ans += (k * 2 - 9)
else:
ans += (k * 2)
return ans
odd_list = list(map(lambda x: (sum(x) % 10),list(itertools.product(data, repeat=odd))))
even_list = list(map(lambda x: ((sum2(x)) % 10),list(itertools.product(data, repeat=even))))
odd_mod = [odd_list.count(i) for i in range(10)]
even_mod = [even_list.count(i) for i in range(10)]
for i in range(10):
for j in range(10):
if ((i + j + tmp) % 10) == 0:
count += odd_mod[i] * even_mod[j]
print(count)
|
s550026196
|
p03448
|
u665224938
| 2,000
| 262,144
|
Wrong Answer
| 54
| 3,060
| 253
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
count = 0
for a in range(A):
for b in range(B):
for c in range(C):
s = a * 500 + b * 100 + c * 50
if s == X:
count += 1
print(count)
|
s266389248
|
Accepted
| 53
| 3,060
| 291
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
count = 0
for a in range(A+1):
for b in range(B+1):
for c in range(C+1):
s = a * 500 + b * 100 + c * 50
# print(a, b, c, s)
if s == X:
count += 1
print(count)
|
s314547171
|
p02612
|
u814663076
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,168
| 203
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
import sys
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(rs())
def rs_(): return [_ for _ in rs().split()]
def ri_(): return [int(_) for _ in rs().split()]
N = ri()
print(N % 1000)
|
s243205051
|
Accepted
| 26
| 9,168
| 233
|
import sys
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(rs())
def rs_(): return [_ for _ in rs().split()]
def ri_(): return [int(_) for _ in rs().split()]
N = ri()
print(1000 - N % 1000 if N % 1000 > 0 else 0)
|
s873173300
|
p03434
|
u799479335
| 2,000
| 262,144
|
Wrong Answer
| 154
| 12,508
| 215
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
import numpy as np
N = int(input())
a_inputs = input().split()
a = np.zeros(N)
for i in range(N):
a[i] = int(a_inputs[i])
a = np.sort(a)[::-1]
alice = a[0::2].sum()
bob = a[1::2].sum()
print(alice - bob)
|
s444891754
|
Accepted
| 379
| 18,732
| 220
|
import numpy as np
N = int(input())
a_inputs = input().split()
a = np.zeros(N)
for i in range(N):
a[i] = int(a_inputs[i])
a = np.sort(a)[::-1]
alice = a[0::2].sum()
bob = a[1::2].sum()
print(int(alice - bob))
|
s152720206
|
p03597
|
u934740772
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 85
|
We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black?
|
N = int(input())
A = int(input())
S = N * N
ANS = N - S
ANS = abs(ANS)
print (ANS)
|
s147452401
|
Accepted
| 17
| 2,940
| 97
|
N = int(input())
A = int(input())
S = N * N
ANS = A - S
ANS = abs(ANS)
print (ANS)
|
s258689500
|
p03759
|
u310855433
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 71
|
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a, b, c = map(int,input().split())
print("Yes" if b-a == c-b else "No")
|
s837252752
|
Accepted
| 17
| 2,940
| 71
|
a, b, c = map(int,input().split())
print("YES" if b-a == c-b else "NO")
|
s622491224
|
p04029
|
u873715358
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 73
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
child = int(input())
sum = 0
for i in range(child):
sum += i
print(sum)
|
s249790762
|
Accepted
| 18
| 2,940
| 75
|
child = int(input())
sum = 0
for i in range(child+1):
sum += i
print(sum)
|
s172092178
|
p03140
|
u561231954
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 3,316
| 257
|
You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?
|
n=int(input())
a=input()
b=input()
c=input()
ans=0
for j in range(n):
d=a[j]
e=b[j]
f=c[j]
if d==e and d!=f:
ans+=1
elif d==f and d!=e:
ans+=1
elif e==f and e!=d:
ans+=1
elif d!=e and e!=f and f!=d:
ans+=2
|
s293626871
|
Accepted
| 17
| 3,064
| 285
|
n=int(input())
a=input()
b=input()
c=input()
ans=0
for j in range(n):
d=a[j]
e=b[j]
f=c[j]
if d==e and d!=f:
ans+=1
elif d==f and d!=e:
ans+=1
elif e==f and e!=d:
ans+=1
elif d!=e and e!=f and f!=d:
ans+=2
print(ans)
|
s981758252
|
p03110
|
u584317622
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 147
|
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
N = int(input())
c = 0
for i in range(N):
a = list(input().split())
if a[1] == 'BTC':
c += float(a[0])*38000
else:
c += float(a[0])
print(c)
|
s083736783
|
Accepted
| 17
| 2,940
| 150
|
N = int(input())
c = 0
for i in range(N):
a = list(input().split())
if a[1] == 'BTC':
c += float(a[0])*380000.0
else:
c += float(a[0])
print(c)
|
s025961132
|
p03501
|
u118211443
| 2,000
| 262,144
|
Wrong Answer
| 21
| 2,940
| 62
|
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
N,A,B=map(int,input().split())
print("B" if N+A>B else "N+A" )
|
s209988534
|
Accepted
| 18
| 2,940
| 58
|
N,A,B=map(int,input().split())
print(B if N*A>B else N*A )
|
s599270480
|
p03546
|
u162660367
| 2,000
| 262,144
|
Wrong Answer
| 38
| 4,336
| 653
|
Joisino the magical girl has decided to turn every single digit that exists on this world into 1. Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points). She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive). You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows: * If A_{i,j}≠-1, the square contains a digit A_{i,j}. * If A_{i,j}=-1, the square does not contain a digit. Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end.
|
H,W=map(int,input().split())
C=[]
A=[]
from queue import Queue
def djcstra(frm,to,D):
dist=D[frm]
Q=sorted(range(len(D)) , key=lambda x:-dist[x])
Q.pop(-1)
while len(Q):
u=Q.pop(-1)
if u==to:break
for i,d in enumerate(D[u]):
if dist[i] >dist[u]+d:
dist[i]=dist[u]+d
Q=sorted(sorted(Q , key=lambda x:-dist[x]))
return dist[to]
for i in range(10):
C.append(list(map(int,input().split())))
for i in range(H):
A.append(list(map(int,input().split())))
cst=[float('inf')]*10
for i in range(10):
cst[i]=djcstra(i,1,C)
sum(sum(cst[i] for i in AA if i >= 0) for AA in A)
|
s095918708
|
Accepted
| 29
| 3,316
| 353
|
H,W=map(int,input().split())
C=[]
A=[]
for i in range(10):
C.append(list(map(int,input().split())))
for i in range(H):
A.append(list(map(int,input().split())))
# WF
for i in range(10):
for j in range(10):
for k in range(10):
C[j][k]=min(C[j][k], C[j][i]+C[i][k])
print(sum(sum(C[i][1] for i in AA if i >= 0) for AA in A))
|
s749151631
|
p03338
|
u116670634
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,064
| 264
|
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
def types(st1, st2):
ret = 0
for i in range(26):
j = chr(i + 97)
if j in st1 and j in st2:
ret += 1
return ret
n = int(input())
s = input()
ans = 0
for i in range(n-1):
ans = max(ans, types(s[:i], s[i+1:]))
print(ans)
|
s746673413
|
Accepted
| 18
| 3,060
| 262
|
def types(st1, st2):
ret = 0
for i in range(26):
j = chr(i + 97)
if j in st1 and j in st2:
ret += 1
return ret
n = int(input())
s = input()
ans = 0
for i in range(n-1):
ans = max(ans, types(s[:i], s[i:]))
print(ans)
|
s653978895
|
p03862
|
u480847874
| 2,000
| 262,144
|
Wrong Answer
| 64
| 14,284
| 467
|
There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies. Snuke can perform the following operation any number of times: * Choose a box containing at least one candy, and eat one of the candies in the chosen box. His objective is as follows: * Any two neighboring boxes contain at most x candies in total. Find the minimum number of operations required to achieve the objective.
|
def m():
n, x = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
for i in range(1, n-1, 2):
if a[i+1] > a[i-1]:
if a[i+1] + a[i] < x:
continue
ans += (a[i+1] + a[i]) -x
else:
if a[i-1] + a[i] < x:
continue
ans += (a[i-1] + a[i]) -x
if a[-1] + a[-2] < x:
return ans
ans += (a[-1] + a[-2]) -x
return ans
print(m())
|
s233617547
|
Accepted
| 86
| 14,052
| 384
|
def m():
n, x = map(int, input().split())
a = list(map(int, input().split()))
b = a.copy()
ans = 0
for i in range(n):
if b[i] > x:
b[i] = x
for j in range(1, n):
if b[j-1] + b[j] > x:
b[j] -= (b[j-1] + b[j]) - x
else:
pass
for k in range(n):
ans += a[k] - b[k]
return ans
print(m())
|
s213563406
|
p02396
|
u805716376
| 1,000
| 131,072
|
Wrong Answer
| 130
| 5,616
| 86
|
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.
|
for i in range(10000):
n = int(input())
if n:
print(f'Case {i}: {n}')
|
s413387578
|
Accepted
| 140
| 5,616
| 88
|
for i in range(1,10001):
n = int(input())
if n:
print(f'Case {i}: {n}')
|
s816195707
|
p03543
|
u947327691
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 130
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
n=input()
cnt=0
for i in range(1,len(n)):
if n[i]==n[i-1]:
cnt+=1
if cnt >=3:
print("Yes")
else:
print("No")
|
s081453081
|
Accepted
| 17
| 3,060
| 112
|
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")
|
s623594280
|
p02613
|
u067694718
| 2,000
| 1,048,576
|
Wrong Answer
| 147
| 16,620
| 236
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
from collections import Counter
n = int(input())
a = [input() for i in range(n)]
b = Counter(a)
b.setdefault("AC", 0)
b.setdefault("WA", 0)
b.setdefault("TLE", 0)
b.setdefault("RE", 0)
for i in b.keys():
print(i + " x " + str(b[i]))
|
s599957728
|
Accepted
| 148
| 16,620
| 326
|
from collections import Counter
n = int(input())
a = [input() for i in range(n)]
b = Counter(a)
b.setdefault("AC", 0)
b.setdefault("WA", 0)
b.setdefault("TLE", 0)
b.setdefault("RE", 0)
print("AC" + " x " + str(b["AC"]))
print("WA" + " x " + str(b["WA"]))
print("TLE" + " x " + str(b["TLE"]))
print("RE" + " x " + str(b["RE"]))
|
s322745906
|
p03377
|
u058592821
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 87
|
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 = (i for i in input().split())
if x <= b:
print('Yes')
else:
print('No')
|
s733457228
|
Accepted
| 17
| 2,940
| 102
|
a, b, x = (int(i) for i in input().split())
if x >= a and x-a <= b:
print('YES')
else:
print('NO')
|
s083021785
|
p03645
|
u970197315
| 2,000
| 262,144
|
Wrong Answer
| 977
| 38,320
| 606
|
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
# ABC068 C - Cat Snuke and a Voyage
ni = lambda: int(input())
nm = lambda: map(int, input().split())
nl = lambda: list(map(int, input().split()))
n,m = nm()
G = [[] for _ in range(n)]
for i in range(m):
a,b = map(int,input().split())
G[a-1].append(b-1)
G[b-1].append(a-1)
# print(G)
for g in G:
island_1toi = False
island_iton = False
for gg in g:
if gg == 1:
island_1toi = True
if gg == n-1:
island_iton = True
if island_1toi == True and island_iton == True:
print('POSSIBLE')
exit()
print('IMPOSSIBLE')
|
s595351908
|
Accepted
| 968
| 38,320
| 425
|
n,m=map(int,input().split())
G=[[] for i in range(n)]
for i in range(m):
a,b=map(int,input().split())
a-=1
b-=1
G[a].append(b)
G[b].append(a)
for g in G:
island_1=False
island_n=False
for gg in g:
if gg==0:
island_1=True
if gg==n-1:
island_n=True
if island_1==True and island_n==True:
print('POSSIBLE')
exit()
print('IMPOSSIBLE')
|
s075445951
|
p02399
|
u725841747
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,648
| 95
|
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
i=list(map(int,input().split(" ")))
print("{0} {1} {2}".format(i[0]//i[1],i[0]%i[1],i[0]/i[1]))
|
s948649636
|
Accepted
| 30
| 7,648
| 114
|
i=list(map(int,input().split(" ")))
print("{0} {1} {2:0.5f}".format(i[0]//i[1],i[0]%i[1],float(i[0])/float(i[1])))
|
s375152295
|
p03592
|
u163703551
| 2,000
| 262,144
|
Wrong Answer
| 25
| 3,572
| 649
|
We have a grid with N rows and M columns of squares. Initially, all the squares are white. There is a button attached to each row and each column. When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa. When a button attached to a column is pressed, the colors of all the squares in that column are inverted. Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.
|
import sys
import socket
hostname = socket.gethostname()
if hostname == 'F451C':
sys.stdin = open('b1.in')
def read_int_list():
return list(map(int, input().split()))
def read_str_list():
return input().split()
def read_int():
return int(input())
def read_str():
return input()
def main():
n, m, k = read_int_list()
if n <= m:
for i in range(n + 1):
if k == (n * n - i * n) + (i * (m - n)):
print('Yes')
if n > m:
for i in range(m + 1):
if k == (m * m - i * m) + (i * (n - m)):
print('Yes')
else:
print('No')
main()
|
s738711137
|
Accepted
| 193
| 3,696
| 536
|
import sys
import socket
hostname = socket.gethostname()
if hostname == 'F451C':
sys.stdin = open('b1.in')
def read_int_list():
return list(map(int, input().split()))
def read_str_list():
return input().split()
def read_int():
return int(input())
def read_str():
return input()
def main():
n, m, k = read_int_list()
for i in range(m + 1):
for j in range(n + 1):
if k == (m - i) * j + (n - j) * i:
print('Yes')
return
print('No')
main()
|
s483829211
|
p03671
|
u294385082
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 69
|
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.
|
a = list(sorted(list(map(int,input().split()))))
print(a[-1] + a[-2])
|
s273087888
|
Accepted
| 17
| 2,940
| 69
|
a = list(sorted(list(map(int,input().split()))))
print(a[0] + a[-2])
|
s889552233
|
p03557
|
u103902792
| 2,000
| 262,144
|
Wrong Answer
| 352
| 24,180
| 420
|
The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower. He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i. To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar. How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.
|
n = int(input())
AS = list(map(int,input().split()))
AS.sort()
BS = list(map(int,input().split()))
BS.sort()
CS = list(map(int,input().split()))
CS.sort()
import bisect
AB = [0 for _ in range(n)]
BC = [0 for _ in range(n)]
a = 0
for b in range(n):
AB[b] = bisect.bisect_left(AS,BS[b])
for b in range(n):
BC[b] = n-bisect.bisect_left(CS,BS[b])
ans = 0
for i in range(n):
ans += AB[b] * BC[b]
print(ans)
|
s826418345
|
Accepted
| 381
| 23,248
| 279
|
n = int(input())
AS = list(map(int,input().split()))
AS.sort()
BS = list(map(int,input().split()))
CS = list(map(int,input().split()))
CS.sort()
import bisect
ans = 0
for b in range(n):
ans += bisect.bisect_left(AS,BS[b])*(n-bisect.bisect_right(CS,BS[b]))
print(ans)
|
s393624648
|
p03860
|
u716660050
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 42
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
l=input().split()
print(l[0]+l[1][0]+l[2])
|
s176286709
|
Accepted
| 17
| 2,940
| 48
|
l=input().split()
print(l[0][0]+l[1][0]+l[2][0])
|
s654702980
|
p03729
|
u371467115
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 106
|
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
A,B,C=input().split()
if A[0]==(B[::-1])[0] and B[0]==(C[::-1])[0]:
print("YES")
else:
print("NO")
|
s514923780
|
Accepted
| 18
| 2,940
| 107
|
A,B,C=input().split()
if (A[::-1])[0]==B[0] and (B[::-1])[0]==C[0]:
print("YES")
else:
print("NO")
|
s095315858
|
p03485
|
u565093947
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 47
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b=map(int,input().split())
print(-(-(a+b)/2))
|
s609627088
|
Accepted
| 17
| 2,940
| 48
|
a,b=map(int,input().split())
print(-(-(a+b)//2))
|
s015450154
|
p03998
|
u891847179
| 2,000
| 262,144
|
Wrong Answer
| 117
| 27,136
| 1,109
|
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.
|
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import accumulate, permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from bisect import bisect, bisect_left
from fractions import gcd
from heapq import heappush, heappop
from functools import reduce
import numpy as np
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def ZIP(n): return zip(*(MAP() for _ in range(n)))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7
strings = {}
strings['a'] = input()
strings['b'] = input()
strings['c'] = input()
cnts = {s: 0 for s in ['a', 'b', 'c']}
next = 'a'
while True:
if cnts[next] < len(strings[next]) - 1:
tmp = strings[next][cnts[next] + 1]
cnts[next] += 1
next = tmp
else:
break
print(next.upper())
|
s020764679
|
Accepted
| 30
| 9,144
| 451
|
A =list(str(input()))
B =list(str(input()))
C =list(str(input()))
P=A
for i in range(len(A)+len(B)+len(C)+3):
if P[0] == 'a':
P.pop(0)
P=A
if len(A) ==0:
print('A')
break
elif P[0] =='b':
P.pop(0)
P=B
if len(B) ==0:
print('B')
break
elif P[0] =='c':
P.pop(0)
P=C
if len(C) ==0:
print('C')
break
|
s132152492
|
p03861
|
u667024514
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 53
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a,b,n = map(int,input().split())
print(a//n-(b-1)//n)
|
s730167316
|
Accepted
| 17
| 2,940
| 54
|
a,b,n = map(int,input().split())
print(b//n-(a-1)//n)
|
s772178351
|
p03963
|
u883048396
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 64
|
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.
|
iN,iK = [int(_) for _ in input().split()]
print(iN*(iK**(iN-1)))
|
s441986446
|
Accepted
| 17
| 2,940
| 69
|
iN,iK = [int(_) for _ in input().split()]
print(iK*((iK-1)**(iN-1)))
|
s633902892
|
p02578
|
u135116520
| 2,000
| 1,048,576
|
Wrong Answer
| 122
| 32,008
| 123
|
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()))
s=0
for i in range(N-1):
if A[i]>A[i+1]:
s+=A[i]-A[i+1]
print(s)
|
s807037066
|
Accepted
| 155
| 32,388
| 167
|
N=int(input())
A=list(map(int,input().split()))
s=0
t=0
for i in range(N):
if t>=A[i]:
t=max(t,A[i])
s+=t-A[i]
else:
t=max(t,A[i])
s=s
print(s)
|
s683887547
|
p02394
|
u903708019
| 1,000
| 131,072
|
Wrong Answer
| 30
| 5,592
| 206
|
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
W, H, x, y, r = list(map(int, input().split()))
x_right = x + r
x_left = x - r
y_up = y + r
y_down = y - r
if x_right <= W & x_left >= 0 & y_up <= H & y_down >= 0:
print("Yes")
else:
print("No")
|
s251173411
|
Accepted
| 20
| 5,596
| 212
|
W, H, x, y, r = list(map(int, input().split()))
x_right = x + r
x_left = x - r
y_up = y + r
y_down = y - r
if x_right <= W and x_left >= 0 and y_up <= H and y_down >= 0:
print("Yes")
else:
print("No")
|
s283589879
|
p04043
|
u752898745
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 147
|
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 irohachan(A,B,C):
if (A == 5 or A ==7) and (B == 5 or B ==7) and (C == 5 or C ==7) and (A+B+C ==17):
print("YES")
else:
print("NO")
|
s346043673
|
Accepted
| 17
| 3,060
| 218
|
def irohachan(A,B,C):
if (A == 5 or A ==7) and (B == 5 or B ==7) and (C == 5 or C ==7) and (A+B+C ==17):
print("YES")
else:
print("NO")
A,B,C = list(map(int, input().split()))
irohachan(A,B,C)
|
s261578170
|
p03836
|
u268554510
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 257
|
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx,sy,tx,ty = map(int,input().split())
diff_x = tx-sx
diff_y = ty-sy
ans = ''
ans += 'U'*diff_y
ans += 'R'*diff_x
ans += 'L'
ans += 'U'*(diff_y+1)
ans += 'R'*(diff_x+1)
ans += 'D'
ans += 'R'
ans += 'D'*(diff_y+1)
ans += 'L'*(diff_x+1)
ans += 'U'
print(ans)
|
s994728688
|
Accepted
| 17
| 3,064
| 293
|
sx,sy,tx,ty = map(int,input().split())
diff_x = tx-sx
diff_y = ty-sy
ans = ''
ans += 'U'*diff_y
ans += 'R'*diff_x
ans += 'D'*diff_y
ans += 'L'*diff_x
ans += 'L'
ans += 'U'*(diff_y+1)
ans += 'R'*(diff_x+1)
ans += 'D'
ans += 'R'
ans += 'D'*(diff_y+1)
ans += 'L'*(diff_x+1)
ans += 'U'
print(ans)
|
s438117846
|
p03544
|
u595952233
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,292
| 3,368,908
| 86
|
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
n = int(input())
l = [2, 1]
i = 0
while i < n:
l.append(l[-1] + l[-2])
print(l[n-2])
|
s821664103
|
Accepted
| 28
| 9,092
| 92
|
n = int(input())
l = [2, 1]
i = 0
while i < n:
l.append(l[-1] + l[-2])
i+=1
print(l[n])
|
s215381162
|
p02694
|
u692054751
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 9,160
| 112
|
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())
cnt = 0
yokin = 100
while (X >= yokin):
yokin = int(yokin * 1.01)
cnt += 1
print(cnt)
|
s063249580
|
Accepted
| 23
| 9,160
| 111
|
X = int(input())
cnt = 0
yokin = 100
while (X > yokin):
yokin = int(yokin * 1.01)
cnt += 1
print(cnt)
|
s777676365
|
p02255
|
u576398884
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,692
| 308
|
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.
|
def insertionSort(A, N):
for i in range(1, N):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j+1] = A[j]
j -= 1
A[j+1] = v
print(" ".join(map(str,A)))
return A
n = int(input())
a = list(map(int, input().split()))
insertionSort(a, n)
|
s287007089
|
Accepted
| 20
| 7,792
| 340
|
def pd(A):
print(" ".join(map(str,A)))
def insertionSort(A, N):
pd(A)
for i in range(1, N):
v = A[i]
j = i - 1
while j >= 0 and A[j] > v:
A[j+1] = A[j]
j -= 1
A[j+1] = v
pd(A)
return A
n = int(input())
a = list(map(int, input().split()))
insertionSort(a, n)
|
s424355806
|
p04011
|
u564060397
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 117
|
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.
|
a=list(input())
b=set(a)
for i in b:
c=a.count(i)
if c%2!=0:
print("No")
exit()
print("Yes")
|
s817123135
|
Accepted
| 17
| 2,940
| 110
|
a=int(input())
b=int(input())
c=int(input())
d=int(input())
if a>b:
print(b*c+d*(a-b))
else:
print(a*c)
|
s988340993
|
p03759
|
u076764813
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c=map(int,input().split())
if b-a == c-b:
print("Yes")
else:
print("No")
|
s192943891
|
Accepted
| 17
| 2,940
| 85
|
a,b,c=map(int,input().split())
if b-a == c-b:
print("YES")
else:
print("NO")
|
s360197080
|
p03623
|
u251123951
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 78
|
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|.
|
a,b,c=map(int,input().split())
print("A") if abs(a-b)>abs(a-c) else print("B")
|
s022347443
|
Accepted
| 17
| 2,940
| 78
|
a,b,c=map(int,input().split())
print("A") if abs(a-b)<abs(a-c) else print("B")
|
s537335964
|
p03711
|
u414558682
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 486
|
Based on some criterion, Snuke divided the integers from 1 through 12 into three groups as shown in the figure below. Given two integers x and y (1 ≤ x < y ≤ 12), determine whether they belong to the same group.
|
a = [1, 3, 5, 7, 8, 10, 12]
b = [4, 6, 9, 11]
c = [2]
# print(a)
# print(b)
# print(c)
x, y = map(int, input().split())
# print(a.index(x))
# print(b.index(x))
# print(c.index(x))
# print(a.index(y))
# print(b.index(y))
# print(c.index(y))
# print(x in a)
# print(x in b)
# print(x in c)
# print(b in a)
# print(b in b)
# print(b in c)
if x in a and b in a:
print('Yes')
elif x in b and b in b:
print('Yes')
elif x in c and b in c:
print('Yes')
else:
print('No')
|
s119933127
|
Accepted
| 17
| 3,064
| 486
|
a = [1, 3, 5, 7, 8, 10, 12]
b = [4, 6, 9, 11]
c = [2]
# print(a)
# print(b)
# print(c)
x, y = map(int, input().split())
# print(a.index(x))
# print(b.index(x))
# print(c.index(x))
# print(a.index(y))
# print(b.index(y))
# print(c.index(y))
# print(x in a)
# print(x in b)
# print(x in c)
# print(b in a)
# print(b in b)
# print(b in c)
if x in a and y in a:
print('Yes')
elif x in b and y in b:
print('Yes')
elif x in c and y in c:
print('Yes')
else:
print('No')
|
s146689644
|
p03679
|
u609061751
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 188
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
import sys
input = sys.stdin.readline
X, A, B = [int(x) for x in input().split()]
if (A - B) <= 0:
print("delicious")
elif (B - A) <= X:
print("safe")
else:
print("dangerous")
|
s464249298
|
Accepted
| 17
| 3,060
| 188
|
import sys
input = sys.stdin.readline
X, A, B = [int(x) for x in input().split()]
if (A - B) >= 0:
print("delicious")
elif (B - A) <= X:
print("safe")
else:
print("dangerous")
|
s352974941
|
p03679
|
u588633699
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 123
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
X, A, B = map(int, input().split())
if A>=B:
print('delicious')
elif B-A>=X:
print('safe')
else:
print('dangerous')
|
s020713902
|
Accepted
| 17
| 2,940
| 123
|
X, A, B = map(int, input().split())
if A>=B:
print('delicious')
elif B-A<=X:
print('safe')
else:
print('dangerous')
|
s890456207
|
p03944
|
u277802731
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 239
|
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
#47b
w,h,n = map(int,input().split())
s = w*h
for _ in range(n):
x,y,a = map(int,input().split())
if a==1:
s-=(w-x)*h
elif a==2:
s-=x*h
elif a==3:
s-=(h-y)*w
else:
s-=y*w
print(max(0,s))
|
s753952630
|
Accepted
| 17
| 3,064
| 292
|
#47b
w,h,n = map(int,input().split())
x,y=0,0
for _ in range(n):
xi,yi,a = map(int,input().split())
if a==1:
x = max(x,xi)
elif a==2:
w = min(xi,w)
elif a==3:
y = max(y,yi)
else:
h = min(h,yi)
print(0 if x>=w or y>=h else (w-x)*(h-y))
|
s125092708
|
p02392
|
u874395007
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,584
| 104
|
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
a, b, c = map(int, input().split())
if a < b and b < c:
ans = 'YES'
else:
ans = 'NO'
print(ans)
|
s987071768
|
Accepted
| 20
| 5,584
| 96
|
a, b, c = map(int, input().split())
if a < b and b < c:
print('Yes')
else:
print('No')
|
s567283941
|
p03140
|
u740284863
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 275
|
You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?
|
int(input())
a = str(input())
b = str(input())
c = str(input())
def f(a,b,c):
ans = 0
for i in range(len(a)):
if a[i] != b[i]:
ans += 1
if a[i] != c[i]:
ans += 1
return ans
ans = min(f(a,b,c),f(b,c,a),f(c,a,b))
print(ans)
|
s521724193
|
Accepted
| 17
| 3,064
| 297
|
int(input())
a = str(input())
b = str(input())
c = str(input())
def f(a,b,c):
ans = 0
for i in range(len(a)):
if a[i] != b[i]:
ans += 1
if a[i] != c[i] and c[i] != b[i]:
ans += 1
return ans
ans = min(f(a,b,c),f(b,c,a),f(c,a,b))
print(ans)
|
s203981810
|
p03415
|
u636684559
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 62
|
We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}. Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right.
|
res=''
for i in range(3):
s=input()
res+=s[i-1]
print(res)
|
s693714973
|
Accepted
| 19
| 2,940
| 60
|
res=''
for i in range(3):
s=input()
res+=s[i]
print(res)
|
s941203972
|
p03719
|
u182047166
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 254
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
#N = int(input())
#A = [int(x) for x in input().split()]
#a, b, c = map(int, input().split())
#name1 = str(input())
#alph = {"A": 1, "B": 2, "C": 3, "D": 4, "E": 5}
a, b, c = map(int, input().split())
if a <= b <=c:
print("Yes")
else:
print("No")
|
s422388025
|
Accepted
| 17
| 2,940
| 256
|
#N = int(input())
#A = [int(x) for x in input().split()]
#a, b, c = map(int, input().split())
#name1 = str(input())
#alph = {"A": 1, "B": 2, "C": 3, "D": 4, "E": 5}
a, b, c = map(int, input().split())
if a <= c <= b:
print("Yes")
else:
print("No")
|
s385793827
|
p03493
|
u371350984
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,108
| 53
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
s = list(map(int, input().split()))
print(s.count(1))
|
s043453687
|
Accepted
| 24
| 8,860
| 25
|
print(input().count("1"))
|
s720532327
|
p03473
|
u189487046
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 44
|
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
s = input()
print(s.replace("2017", "2018"))
|
s317398060
|
Accepted
| 17
| 2,940
| 32
|
s = int(input())
print(24-s+24)
|
s401838265
|
p03494
|
u545213416
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,060
| 215
|
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.
|
num = int(input())
listName = list(map(int, input().split()))
counter = 0
while (len(listName) == num):
listName = [i / 2 for i in listName if i%2 == 0]
print(listName)
counter += 1
print(counter - 1)
|
s991191187
|
Accepted
| 20
| 3,060
| 195
|
num = int(input())
listName = list(map(int, input().split()))
counter = 0
while (len(listName) == num):
listName = [i / 2 for i in listName if i%2 == 0]
counter += 1
print(counter - 1)
|
s280675339
|
p00010
|
u546285759
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,652
| 417
|
Write a program which prints the central coordinate $(p_x, p_y)$ and the radius $r$ of a circumscribed circle of a triangle which is constructed by three points $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ on the plane surface.
|
from math import sqrt
for _ in range(int(input())):
x1, y1, x2, y2, x3, y3= map(float, input().split())
a= sqrt((x3-x2)**2 + (y3-y2)**2)
b= sqrt((x3-x1)**2 + (y3-y1)**2)
c= sqrt((x2-x1)**2 + (y2-y1)**2)
numerator, denominator= 2*b*c, b**2 + c**2 - a**2
cosA= denominator / numerator
sinA= sqrt(1 - (cosA)**2)
r= a / (2*sinA)
px= (x2-x1) / 2
py= (y3-y1) / 2
print(px, py, r)
|
s516115277
|
Accepted
| 20
| 5,684
| 598
|
n = int(input())
for _ in range(n):
x1, y1, x2, y2, x3, y3 = map(float, input().split())
a = pow((x3-x2) ** 2 + (y3-y2) ** 2, 0.5)
b = pow((x3-x1) ** 2 + (y3-y1) ** 2, 0.5)
c = pow((x1-x2) ** 2 + (y1-y2) ** 2, 0.5)
cosA = (b**2 + c**2 - a**2) / (2*b*c)
sinA = pow(1 - cosA**2, 0.5)
R = a / sinA / 2
a, b, c = x1-x2, y1-y2, -(x1**2 + y1**2) + (x2**2 + y2**2)
d, e, f = x2-x3, y2-y3, -(x2**2 + y2**2) + (x3**2 + y3**2)
l = (c*e - b*f) / (e*a - b*d)
m = (c*d - a*f) / (b*d - a*e)
l, m = l*-0.5, m*-0.5
print("{:.3f} {:.3f} {:.3f}".format(l, m, R))
|
s569987276
|
p03696
|
u466331465
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 318
|
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.
|
from itertools import groupby
N = int(input())
S = input()
ans = []
#S1 = groupby(S)
a=0
a1=0
b=0
b1=0
cnt = 0
c=0
S1 = groupby(list(S))
for key,group in S1:
group = list(group)
if list(key)==['(']:
a=len(group)
a1+=a
else:
b=len(group)
b1+=b
a1-=b
S = "("*b1+S
S = S+")"*a1
print("".join(S))
|
s105857200
|
Accepted
| 18
| 3,064
| 344
|
from itertools import groupby
N = int(input())
S = input()
ans = []
#S1 = groupby(S)
a=0
a1=0
b=0
b1=0
S1 = groupby(list(S))
for key,group in S1:
group = list(group)
if list(key)==['(']:
a1+=len(group)
else:
if a1<=len(group):
b1+=len(group)-a1
a1 = 0
else:
a1 -=len(group)
S = S+")"*a1
S = "("*b1+S
print(S)
|
s414867676
|
p03478
|
u181012033
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 17,140
| 302
|
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).
|
def find_sum_digits(n):
sum = 0
while (n>0):
sum += (n % 10)
n /= 10
print (n)
return sum
n,a,b = map(int,input().split(" "))
print(n)
total = 0
for i in range(n + 1):
print (i)
ans = find_sum_digits(i)
if a <= ans <= b:
total += i
print(total)
|
s711255920
|
Accepted
| 36
| 3,060
| 144
|
n, a, b = map(int, input().split())
ans = 0
for i in range(n+1):
if a <= sum(list(map(int, list(str(i))))) <= b:
ans += i
print(ans)
|
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