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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s983923449
|
p03698
|
u519939795
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 159
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
import sys
s=input()
for i in range(len(s)+1):
for j in range(len(s)+1):
if s[i]==s[j]:
print('no')
sys.exit()
print('yes')
|
s962151455
|
Accepted
| 17
| 2,940
| 134
|
#ABC063.B
S = input()
cnt = 0
for i in S:
if S.count(i) != 1:
cnt +=1
if cnt == 0:
print('yes')
else:
print('no')
|
s446686574
|
p03836
|
u848173626
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 530
|
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.
|
from sys import stdin
SX, SY, TX, TY = map(int, stdin.readline().rstrip().split())
# S->T
print('U' * (TY - SY), end='')
print('R' * (TX - SX), end='')
# T->S
print('D' * (TY - SY), end='')
print('L' * (TX - SX), end='')
print('L', end='')
print('U' * (TY - SY + 1), end='')
print('R' * (TX - SX + 1), end='')
print('D', end='')
print('R', end='')
print('U' * (TY - SY + 1), end='')
print('L' * (TX - SX + 1), end='')
print('U', end='')
|
s576167548
|
Accepted
| 17
| 3,064
| 522
|
from sys import stdin
SX, SY, TX, TY = map(int, stdin.readline().rstrip().split())
# S->T
print('U' * (TY - SY), end='')
print('R' * (TX - SX), end='')
# T->S
print('D' * (TY - SY), end='')
print('L' * (TX - SX), end='')
print('L', end='')
print('U' * (TY - SY + 1), end='')
print('R' * (TX - SX + 1), end='')
print('D', end='')
print('R', end='')
print('D' * (TY - SY + 1), end='')
print('L' * (TX - SX + 1), end='')
print('U')
|
s017670845
|
p02262
|
u286589639
| 6,000
| 131,072
|
Wrong Answer
| 20
| 5,600
| 454
|
Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$
|
def insertion_sort(A ,N, g):
for i in range(g, N):
v = A[i]
j = i - g
while j>=0 and A[j]>v:
A[j+g] = A[j]
j = j - g
A[j+g] = v
def shell_sort(A, N):
cnt = 0
m = N//2
G = [i for i in reversed(range(1, m, 2))]
for g in G:
insertion_sort(A, N, g)
N = int(input())
A = [int(input()) for i in range(N)]
shell_sort(A, N)
m = N//2
G = [i for i in reversed(range(1, m, 2))]
print(m)
print(' '.join(map(str, G)))
for a in A:
print(a)
|
s128961482
|
Accepted
| 22,860
| 45,516
| 497
|
def insertion_sort(A ,N, g):
global cnt
for i in range(g, N):
v = A[i]
j = i - g
while j>=0 and A[j]>v:
A[j+g] = A[j]
j = j - g
cnt += 1
A[j+g] = v
def shell_sort(A, N):
global cnt
cnt = 0
G = []
h = 1
while h<=len(A):
G.append(h)
h = 3*h+1
G.reverse()
m = len(G)
print(m)
print(' '.join(map(str, G)))
for g in G:
insertion_sort(A, N, g)
return m, G
N = int(input())
A = [int(input()) for i in range(N)]
shell_sort(A, N)
print(cnt)
for a in A:
print(a)
|
s839819775
|
p02390
|
u586792237
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,580
| 138
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
x = int(input())
h = x // 3600
m = (x - (h*3600))//60
s = x-((h*3600)+(m*60))
print(h, ':', m, ':', s)
print(h, ':', m, ':', s, sep='')
|
s985973005
|
Accepted
| 20
| 5,588
| 114
|
x = int(input())
h = x // 3600
m = (x - (h*3600))//60
s = x-((h*3600)+(m*60))
print(h, ':', m, ':', s, sep='')
|
s560826771
|
p02578
|
u625864724
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 32,188
| 284
|
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())
lst = list(map(int,input().split()))
ans = 0
for i in range(n):
tall = 0
for j in range(i):
if (lst[j] > tall):
tall = lst[j]
print (i, tall)
if (lst[i] < tall):
sa = tall - lst[i]
print(i,sa)
ans = ans + sa
lst[i] = tall
print(ans)
|
s883303435
|
Accepted
| 161
| 32,144
| 210
|
n = int(input())
lst = list(map(int,input().split()))
ans = 0
for i in range(n - 1):
if (lst[i] > lst[i + 1]):
sa = lst[i] - lst[i + 1]
ans = ans + sa
lst[i + 1] = lst[i]
print(ans)
|
s946480299
|
p03360
|
u597047658
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 224
|
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?
|
import heapq
A = list(map(int, input().split()))
A = [-a for a in A]
K = int(input())
heapq.heapify(A)
print(A)
for i in range(K):
max_val = heapq.heappop(A)
heapq.heappush(A, 2*max_val)
print(sum([-a for a in A]))
|
s237517291
|
Accepted
| 17
| 3,060
| 215
|
import heapq
A = list(map(int, input().split()))
A = [-a for a in A]
K = int(input())
heapq.heapify(A)
for i in range(K):
max_val = heapq.heappop(A)
heapq.heappush(A, 2*max_val)
print(sum([-a for a in A]))
|
s951949203
|
p01102
|
u196653484
| 8,000
| 262,144
|
Wrong Answer
| 30
| 5,584
| 414
|
The programming contest named _Concours de Programmation Comtemporaine Interuniversitaire_ (CPCI) has a judging system similar to that of ICPC; contestants have to submit correct outputs for two different inputs to be accepted as a correct solution. Each of the submissions should include the program that generated the output. A pair of submissions is judged to be a correct solution when, in addition to the correctness of the outputs, they include an identical program. Many contestants, however, do not stop including a different version of their programs in their second submissions, after modifying a single string literal in their programs representing the input file name, attempting to process different input. The organizers of CPCI are exploring the possibility of showing a special error message for such _close_ submissions, indicating contestants what's wrong with such submissions. Your task is to detect such close submissions.
|
results=[]
while True:
a=input()
if a==".":
break
b=input()
count=0
for i,j in zip(a,b):
if(a!=b):
count+=1
if count==0:
results.append(0)
elif count==1:
results.append(1)
else:
results.append(2)
for i in results:
if i==0:
print("IDENTICAL")
if i==1:
print("CLOSE")
if i==2:
print("DIFFERENT")
|
s622750504
|
Accepted
| 30
| 5,596
| 1,019
|
results=[]
while True:
a=input().split("\"")
if a[0]==".":
break
b=input().split("\"")
count1=0
count2=0
if len(a) != len(b):
results.append(2)
else:
for i in range(len(a)):
if(i%2==1):
if len(a[i]) != len(b[i]):
count1+=1
else:
for k,l in zip(a[i],b[i]):
if(k!=l):
count1+=1
else:
if len(a[i]) != len(b[i]):
count2+=1
else:
for k,l in zip(a[i],b[i]):
if(k!=l):
count2+=1
if count1==0 and count2==0:
results.append(0)
elif count1==1 and count2==0:
results.append(1)
else:
results.append(2)
for i in results:
if i==0:
print("IDENTICAL")
if i==1:
print("CLOSE")
if i==2:
print("DIFFERENT")
|
s787192434
|
p03377
|
u513081876
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 98
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A, B, X = map(int, input().split())
if A <= X and X <= A+B:
print('Yes')
else:
print('No')
|
s962694027
|
Accepted
| 17
| 2,940
| 98
|
A, B, X = map(int, input().split())
if A <= X and X <= A+B:
print('YES')
else:
print('NO')
|
s208362457
|
p03612
|
u955125992
| 2,000
| 262,144
|
Wrong Answer
| 51
| 14,008
| 127
|
You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this.
|
n = int(input())
p = list(map(int, input().split()))
ans = 0
for i in range(n):
if i == p[i]:
ans += 1
print(ans)
|
s342132612
|
Accepted
| 73
| 14,004
| 206
|
n = int(input())
p = list(map(int, input().split()))
ans = 0
for i in range(n-1):
if i+1 == p[i]:
p[i], p[i+1] = p[i+1], p[i]
ans += 1
if p[n-1] == n:
ans += 1
print(ans)
|
s202800417
|
p03457
|
u727551259
| 2,000
| 262,144
|
Wrong Answer
| 450
| 11,816
| 478
|
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())
ti,xi,yi = [0],[0],[0]
ans = 'NO'
for ni in range(n):
t,x,y = [int(x) for x in input().split()]
ti.append(t)
xi.append(x)
yi.append(y)
for i in range(n):
dt = ti[i+1] - ti[i]
dx = xi[i+1] - xi[i]
dy = yi[i+1] - yi[i]
if dt < abs(dx)+abs(dy):
break
odev1 = (xi[i]+yi[i])%2
odev2 = (xi[i+1]+yi[i+1])%2
if dt%2 == 0:
if odev1 != odev2:
break
else:
if odev1 == odev2:
break
if i == n-1:
ans = 'YES'
print(ans)
|
s044920149
|
Accepted
| 441
| 11,728
| 478
|
n = int(input())
ti,xi,yi = [0],[0],[0]
ans = 'No'
for ni in range(n):
t,x,y = [int(x) for x in input().split()]
ti.append(t)
xi.append(x)
yi.append(y)
for i in range(n):
dt = ti[i+1] - ti[i]
dx = xi[i+1] - xi[i]
dy = yi[i+1] - yi[i]
if dt < abs(dx)+abs(dy):
break
odev1 = (xi[i]+yi[i])%2
odev2 = (xi[i+1]+yi[i+1])%2
if dt%2 == 0:
if odev1 != odev2:
break
else:
if odev1 == odev2:
break
if i == n-1:
ans = 'Yes'
print(ans)
|
s060851670
|
p03637
|
u239375815
| 2,000
| 262,144
|
Wrong Answer
| 75
| 14,252
| 233
|
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.
|
n = int(input())
a = list(map(int,input().split()))
a_n4 = a_4 = [x for x in a if x%4!=0]
a_4 = [x for x in a if x%4==0]
a_2 = [x for x in a if x%4!=0 and x%2==0]
print("Yes "if (len(a_n4)-1-(int(len(a_2)/2))<=len(a_4)) else "No")
|
s865007864
|
Accepted
| 60
| 15,020
| 278
|
N = int(input())
a = list(map(int, input().split()))
c = [0,0,0,0]
for x in a:
c[x % 4] += 1
# if (c[1]==0) and (c[3]==0):
# print('Yes')
if c[1] + c[3] < c[0] + 1:
print('Yes')
elif (c[1] + c[3] == c[0] + 1)and(c[2] == 0):
print('Yes')
else:
print('No')
|
s398507147
|
p03449
|
u905203728
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 308
|
We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You will collect all the candies you visit during the travel. The top-left and bottom-right squares also contain candies, and you will also collect them. At most how many candies can you collect when you choose the best way to travel?
|
n=int(input())
box=[list(map(int,input().split())) for i in range(2)]
count=box[0][0]
point=-1
add=0
for j in range(n):
for i in range(n):
if add<=i:
count +=box[1][i]
else:
count +=box[0][i]
point=max(point,count)
count=box[0][0]
add +=1
print(point)
|
s547577492
|
Accepted
| 20
| 3,064
| 310
|
n=int(input())
box=[list(map(int,input().split())) for i in range(2)]
count=box[0][0]
point=-1
add=0
for j in range(n):
for i in range(n):
if add<=i:
count +=box[1][i]
else:
count +=box[0][i+1]
point=max(point,count)
count=box[0][0]
add +=1
print(point)
|
s014761454
|
p02602
|
u038216098
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 31,548
| 234
|
M-kun is a student in Aoki High School, where a year is divided into N terms. There is an exam at the end of each term. According to the scores in those exams, a student is given a grade for each term, as follows: * For the first through (K-1)-th terms: not given. * For each of the K-th through N-th terms: the multiplication of the scores in the last K exams, including the exam in the graded term. M-kun scored A_i in the exam at the end of the i-th term. For each i such that K+1 \leq i \leq N, determine whether his grade for the i-th term is **strictly** greater than the grade for the (i-1)-th term.
|
N,K=map(int,input().split())
A=list(map(int,input().split()))
point=[1]*N
for i in range(K-1,N):
for j in range(K):
point[i]*=A[i-j]
for i in range(N-K-2,N-1):
if(point[i]>=point[i+1]):
print("No")
else:
print("Yes")
|
s066100657
|
Accepted
| 147
| 31,756
| 155
|
N,K=map(int,input().split())
A=list(map(int,input().split()))
point=[1]*N
for i in range(N-K):
if(A[K+i]>A[i]):
print("Yes")
else:
print("No")
|
s348347758
|
p03844
|
u533039576
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 13
|
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.
|
eval(input())
|
s285055337
|
Accepted
| 17
| 2,940
| 21
|
print(eval(input()))
|
s828260923
|
p02795
|
u575101291
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,316
| 104
|
We have a grid with H rows and W columns, where all the squares are initially white. You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions: * Choose one row, then paint all the squares in that row black. * Choose one column, then paint all the squares in that column black. At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations.
|
h = int(input())
w = int(input())
n = int(input())
if h < w:
t = n / w
else:
t = n / h
print(int(t))
|
s090362995
|
Accepted
| 17
| 3,060
| 123
|
import math
h = int(input())
w = int(input())
n = int(input())
if h < w:
t = n / w
else:
t = n / h
print(math.ceil(t))
|
s035435965
|
p03860
|
u331464808
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 40
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
s = input()
print("A" + str(s[0]) + "C")
|
s521662914
|
Accepted
| 17
| 2,940
| 53
|
a,x,c=map(str,input().split())
print('A'+ x[0] + 'C')
|
s891786161
|
p03674
|
u075012704
| 2,000
| 262,144
|
Wrong Answer
| 2,109
| 32,380
| 561
|
You are given an integer sequence of length n+1, a_1,a_2,...,a_{n+1}, which consists of the n integers 1,...,n. It is known that each of the n integers 1,...,n appears at least once in this sequence. For each integer k=1,...,n+1, find the number of the different subsequences (not necessarily contiguous) of the given sequence with length k, modulo 10^9+7.
|
from statistics import mode
from scipy.misc import comb
N = int(input())
A = list(map(int, input().split()))
MOD = 10 ** 9
C = mode(A)
C_indexes = [i for i, x in enumerate(A) if x == C]
outside = C_indexes[0] + len(A) - C_indexes[1] - 1
for k in range(1, len(A)+1):
if k > outside + 1:
print(comb((N+1), k, exact=True) % MOD)
else:
print(comb((N+1), k, exact=True) % MOD - comb(outside, (k-1), exact=True) % MOD)
|
s744882268
|
Accepted
| 305
| 31,324
| 796
|
N = int(input())
A = list(map(int, input().split()))
MOD = 10 ** 9 + 7
duplicate_x = None
checked = set()
for a in A:
if a in checked:
duplicate_x = a
checked.add(a)
x_l_index = A.index(duplicate_x)
x_r_index = N + 1 - A[::-1].index(duplicate_x) - 1
factorial = [1, 1]
inverse = [1, 1]
invere_base = [0, 1]
for i in range(2, N + 2):
factorial.append((factorial[-1] * i) % MOD)
invere_base.append((-invere_base[MOD % i] * (MOD // i)) % MOD)
inverse.append((inverse[-1] * invere_base[-1]) % MOD)
def nCr(n, r):
if not 0 <= r <= n:
return 0
return factorial[n] * inverse[r] * inverse[n - r] % MOD
for k in range(1, N + 1 + 1):
print((nCr(N + 1, k) - nCr(max(0, x_l_index) + max(N + 1 - x_r_index - 1, 0), k - 1)) % MOD)
|
s460658553
|
p02261
|
u294922877
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,612
| 1,199
|
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def bubble_sort(n, key=0):
c = 0
r = [x if isinstance(x, list) else [x] for x in n[:]]
l = len(r) - 1
for i in range(0, l):
for j in range(l, i, -1):
if r[j][key] < r[j-1][key]:
r[j], r[j-1] = r[j-1], r[j]
c += 1
r = [x[0] if len(x) == 1 else x for x in r]
return {'list': r, 'count': c}
def selection_sort(n, key=0):
c = 0
r = [x if isinstance(x, list) else [x] for x in n[:]]
l = len(r)
for i in range(0, l):
mini = i
for j in range(i, l):
if (r[j][key] < r[mini][key]):
mini = j
if mini != i:
r[i], r[mini] = r[mini], r[i]
c += 1
r = [x[0] if len(x) == 1 else x for x in r]
return {'list': r, 'count': c}
def card_format(cards):
return ' '.join([x[0] + str(x[1]) for x in cards])
if __name__ == '__main__':
n = int(input())
c = list(map(list, input().split(' ')))
c = [[x[0], int(x[1])] for x in c]
bsc = card_format(bubble_sort(c, 1)['list'])
ssc = card_format(selection_sort(c, 1)['list'])
print(bsc)
print('Stable')
print(ssc)
print('Stable' if bsc == ssc else 'Unstable')
|
s960394481
|
Accepted
| 20
| 5,616
| 1,201
|
def bubble_sort(n, key=0):
c = 0
r = [x if isinstance(x, list) else [x] for x in n[:]]
l = len(r) - 1
for i in range(0, l):
for j in range(l, i, -1):
if r[j][key] < r[j-1][key]:
r[j], r[j-1] = r[j-1], r[j]
c += 1
r = [x[0] if len(x) == 1 else x for x in r]
return {'list': r, 'count': c}
def selection_sort(n, key=0):
c = 0
r = [x if isinstance(x, list) else [x] for x in n[:]]
l = len(r)
for i in range(0, l):
mini = i
for j in range(i, l):
if (r[j][key] < r[mini][key]):
mini = j
if mini != i:
r[i], r[mini] = r[mini], r[i]
c += 1
r = [x[0] if len(x) == 1 else x for x in r]
return {'list': r, 'count': c}
def card_format(cards):
return ' '.join([x[0] + str(x[1]) for x in cards])
if __name__ == '__main__':
n = int(input())
c = list(map(list, input().split(' ')))
c = [[x[0], int(x[1])] for x in c]
bsc = card_format(bubble_sort(c, 1)['list'])
ssc = card_format(selection_sort(c, 1)['list'])
print(bsc)
print('Stable')
print(ssc)
print('Stable' if bsc == ssc else 'Not stable')
|
s941733337
|
p03548
|
u756988562
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 197
|
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?
|
input_lines = input().split(" ")
X = int(input_lines[0])
Y = int(input_lines[1])
Z = int(input_lines[2])
judge = X//(Y)
if judge*(Y+Z)>=Z:
print(str(X//(Y+Z)-1))
else:
print(str(X//(Y+Z)))
|
s047206028
|
Accepted
| 17
| 2,940
| 202
|
input_lines = input().split(" ")
X = int(input_lines[0])
Y = int(input_lines[1])
Z = int(input_lines[2])
judge = X//(Y+Z)
if judge*(Y+Z)+Z>X:
print(str(X//(Y+Z)-1))
else:
print(str(X//(Y+Z)))
|
s925754360
|
p03369
|
u948911484
| 2,000
| 262,144
|
Wrong Answer
| 25
| 8,884
| 29
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
print(700+input().count("o"))
|
s870511309
|
Accepted
| 24
| 9,028
| 41
|
print([7,8,9,10][input().count("o")]*100)
|
s426114435
|
p02397
|
u042882066
| 1,000
| 131,072
|
Wrong Answer
| 60
| 7,652
| 109
|
Write a program which reads two integers x and y, and prints them in ascending order.
|
while True:
x, y = map(int, input().split(" "))
if x == 0 and y == 0:
break
print(x, y)
|
s986515807
|
Accepted
| 60
| 7,612
| 181
|
# -*- coding: utf-8 -*-
while True:
x, y = map(int, input().split(" "))
if x == 0 and y == 0:
break
elif x <= y:
print(x, y)
elif x >= y:
print(y, x)
|
s844267679
|
p03007
|
u606045429
| 2,000
| 1,048,576
|
Wrong Answer
| 232
| 13,964
| 617
|
There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.
|
N = int(input())
A = [int(i) for i in input().split()]
A.sort()
L = [a for a in A if a <= 0]
R = [a for a in A if a > 0]
if L != [] and R != []:
low = L[0]
for r in R[:-1]:
print(low, r)
low -= r
up = R[-1]
for l in L[1:]:
print(up, l)
up -= l
print(up, low)
print(up - low)
elif L == []:
low, up = R[0], R[-1]
for r in R[1:-1]:
print(low, r)
low -= r
print(up, low)
print(up - low)
elif R == []:
low, up = L[0], L[-1]
for l in L[1:-1]:
print(up, l)
up -= l
print(up, low)
print(up - low)
|
s106425612
|
Accepted
| 307
| 24,016
| 746
|
N = int(input())
A = [int(i) for i in input().split()]
A.sort()
L = [a for a in A if a <= 0]
R = [a for a in A if a > 0]
ans = []
if L != [] and R != []:
low = L[0]
for r in R[:-1]:
ans.append([low, r])
low -= r
up = R[-1]
for l in L[1:]:
ans.append([up, l])
up -= l
ans.append([up, low])
ans.append([up - low])
elif L == []:
low, up = R[0], R[-1]
for r in R[1:-1]:
ans.append([low, r])
low -= r
ans.append([up, low])
ans.append([up - low])
elif R == []:
low, up = L[0], L[-1]
for l in L[1:-1]:
ans.append([up, l])
up -= l
ans.append([up, low])
ans.append([up - low])
print(*ans[-1])
for a in ans[:-1]:
print(*a)
|
s707729913
|
p02407
|
u853619096
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,524
| 81
|
Write a program which reads a sequence and prints it in the reverse order.
|
n=input()
n=int(n)
a=input().split()
a=a[::-1]
for i in a:
print(" ".join(a))
|
s541308233
|
Accepted
| 20
| 7,636
| 76
|
n=int(input())
a=list(map(str,input().split()))
a=a[::-1]
print(" ".join(a))
|
s201514092
|
p02612
|
u878473776
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,164
| 95
|
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(""))
if N % 1000 == 0:
N1 = N // 1000
else:
N1 = N // 1000 + 1
print(N1)
|
s384324305
|
Accepted
| 31
| 9,124
| 106
|
N = int(input(""))
if N % 1000 == 0:
N1 = N // 1000
else:
N1 = N // 1000 + 1
print(N1 * 1000 - N)
|
s180342043
|
p04044
|
u022871813
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 132
|
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):
a = input()
list.append(a)
for i in sorted(list):
print(i)
|
s377386699
|
Accepted
| 17
| 3,060
| 125
|
n, l = map(int,input().split())
list = []
for i in range(n):
a = input()
list.append(a)
print(''.join(sorted(list)))
|
s881254886
|
p03574
|
u637551956
| 2,000
| 262,144
|
Wrong Answer
| 24
| 3,188
| 1,667
|
You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process.
|
H,W = map(int, input().split())
S = [list(input()) for i in range(H)]
S=[[0 if j == "." else "#" for j in i] for i in S]
for i in range(H):
for j in range(W):
if S[i][j]=='#':
try:S[i-1][j-1]
except IndexError:pass
else:
if i-1>=0 and j-1>0 and S[i-1][j-1]!='#':
S[i-1][j-1]+=1
try:S[i-1][j]
except IndexError:pass
else:
if i-1>=0 and S[i-1][j]!='#':
S[i-1][j]+=1
try:S[i-1][j+1]
except IndexError:pass
else:
if i-1>=0 and S[i-1][j+1]!='#':
S[i-1][j+1]+=1
try:S[i][j-1]
except IndexError:pass
else:
if j-1>=0 and S[i][j-1]!='#':
S[i][j-1]+=1
try:S[i][j+1]
except IndexError:pass
else:
if S[i][j+1]!='#':
S[i][j+1]+=1
try:S[i+1][j-1]
except IndexError:pass
else:
if j-1>=0 and S[i+1][j-1]!='#':
S[i+1][j-1]+=1
try:S[i+1][j]
except IndexError:pass
else:
if S[i+1][j]!='#':
S[i+1][j]+=1
try:S[i+1][j+1]
except IndexError:pass
else:
if S[i+1][j+1]!='#':
S[i+1][j+1]+=1
print(S)
|
s863312092
|
Accepted
| 25
| 3,572
| 1,692
|
H,W = map(int, input().split())
S = [list(input()) for i in range(H)]
S=[[0 if j == "." else "#" for j in i] for i in S]
for i in range(H):
for j in range(W):
if S[i][j]=='#':
try:S[i-1][j-1]
except IndexError:pass
else:
if i-1>=0 and j-1>=0 and S[i-1][j-1]!='#':
S[i-1][j-1]+=1
try:S[i-1][j]
except IndexError:pass
else:
if i-1>=0 and S[i-1][j]!='#':
S[i-1][j]+=1
try:S[i-1][j+1]
except IndexError:pass
else:
if i-1>=0 and S[i-1][j+1]!='#':
S[i-1][j+1]+=1
try:S[i][j-1]
except IndexError:pass
else:
if j-1>=0 and S[i][j-1]!='#':
S[i][j-1]+=1
try:S[i][j+1]
except IndexError:pass
else:
if S[i][j+1]!='#':
S[i][j+1]+=1
try:S[i+1][j-1]
except IndexError:pass
else:
if j-1>=0 and S[i+1][j-1]!='#':
S[i+1][j-1]+=1
try:S[i+1][j]
except IndexError:pass
else:
if S[i+1][j]!='#':
S[i+1][j]+=1
try:S[i+1][j+1]
except IndexError:pass
else:
if S[i+1][j+1]!='#':
S[i+1][j+1]+=1
for i in S:
print(*i,sep="")
|
s873417098
|
p02409
|
u092736322
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,608
| 409
|
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
n=int(input())
S=[]
for i in range(1200):
S.append(0)
for i in range(n):
k=input().split()
b=int(k[0])
f=int(k[1])
r=int(k[2])
v=int(k[3])
S[(b-1)*30+(f-1)*10+r-1]+=v
for i in range(4):
for j in range(3):
msg=""
for t in range(9):
msg+=str(S[i*30+j*10+t])+" "
msg+=str(S[i*30+j*10+9])
print(msg)
print("####################")
|
s965441481
|
Accepted
| 20
| 5,604
| 393
|
n=int(input())
S=[]
for i in range(1200):
S.append(0)
for i in range(n):
k=input().split()
b=int(k[0])
f=int(k[1])
r=int(k[2])
v=int(k[3])
S[(b-1)*30+(f-1)*10+r-1]+=v
for i in range(4):
for j in range(3):
msg=""
for t in range(10):
msg+=" "+str(S[i*30+j*10+t])
print(msg)
if i<3:
print("####################")
|
s397470814
|
p03644
|
u865413330
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 210
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n = int(input())
count = 0
currCount = 0
for i in range(n):
currCount = 0
while i % 2 != 0:
i = int(i / 2)
currCount += 1
if count < currCount:
count = currCount
print(count)
|
s956216876
|
Accepted
| 17
| 3,060
| 296
|
n = int(input())
count = 0
currCount = 0
ans = 0
for i in range(n+1):
currCount = 0
currNum = i
while (i % 2) == 0:
if i == 0:
break
i = int(i / 2)
currCount += 1
if count <= currCount:
count = currCount
ans = currNum
print(ans)
|
s177911106
|
p03597
|
u717626627
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 49
|
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?
|
a = int(input())
b = int(input())
print(a^2 - b)
|
s734445527
|
Accepted
| 17
| 2,940
| 49
|
a = int(input())
b = int(input())
print(a*a - b)
|
s749504771
|
p03474
|
u056486147
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 326
|
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
a, b = map(int, input().split())
s = list(input())
print(s)
print(len(s), a+b+1)
flag = True
if len(s) == a+b+1:
if s[a] == "-":
s.pop(a)
for i in s:
if i == "-":
flag = False
else:
flag = False
else:
flag = False
if flag:
print("Yes")
else:
print("No")
|
s068371991
|
Accepted
| 18
| 2,940
| 319
|
import sys
a, b = map(int, input().split())
s = list(input())
if len(s) == a+b+1:
if s[a] == "-":
s.pop(a)
for i in s:
if i == "-":
print("No")
sys.exit()
else:
print("No")
sys.exit()
else:
print("No")
sys.exit()
print("Yes")
|
s927786355
|
p03370
|
u253952966
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 141
|
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
n, x = map(int, input().split())
mn = []
for _ in range(n):
mi = int(input())
mn.append(mi)
mn.sort()
x -= sum(mn)
print(n + (x % mn[0]))
|
s725284889
|
Accepted
| 18
| 2,940
| 142
|
n, x = map(int, input().split())
mn = []
for _ in range(n):
mi = int(input())
mn.append(mi)
mn.sort()
x -= sum(mn)
print(n + (x // mn[0]))
|
s445656750
|
p03485
|
u374146618
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 66
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a, b = [int(x) for x in input().split()]
x = (a+b)/2
print(x//1+1)
|
s456911962
|
Accepted
| 17
| 2,940
| 110
|
a, b = [int(x) for x in input().split()]
x = (a+b)/2
if x%1==0:
print(int(x))
else:
print(int(x//1+1))
|
s034344095
|
p03456
|
u762474234
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 172
|
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.
|
import math
a, b = input().split()
c = int(a + b)
square_root = math.sqrt(c)
# print(c/square_root, square_root)
print('YES') if square_root.is_integer() else print('NO')
|
s669032642
|
Accepted
| 17
| 2,940
| 172
|
import math
a, b = input().split()
c = int(a + b)
square_root = math.sqrt(c)
# print(c/square_root, square_root)
print('Yes') if square_root.is_integer() else print('No')
|
s337216561
|
p04043
|
u518958552
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 221
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a, b, c = map(int,input().split())
if a == 5 and b == 7 and c == 5:
print("Yes")
elif a == 5 and b == 5 and c == 7:
print("Yes")
elif a == 7 and b == 5 and c == 5:
print("Yes")
else:
print("No")
|
s240913636
|
Accepted
| 17
| 2,940
| 132
|
a, b, c = map(int,input().split())
d = [a, b, c]
if d.count(5) == 2 and d.count(7) == 1:
print("YES")
else:
print("NO")
|
s880785767
|
p03943
|
u842696304
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 126
|
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.
|
x = list(input().split())
x.sort()
if int(x[0]) + int(x[1]) == int(x[2]):
print('YES')
else:
print('FALSE')
print(x)
|
s341185995
|
Accepted
| 17
| 2,940
| 119
|
a, b, c = map(int, input().split())
if a + b == c or b + c == a or c + a == b:
print("Yes")
else:
print("No")
|
s466541235
|
p02288
|
u357267874
| 2,000
| 131,072
|
Wrong Answer
| 30
| 5,604
| 773
|
A binary heap which satisfies max-heap property is called max-heap. In a max- heap, for every node $i$ other than the root, $A Write a program which reads an array and constructs a max-heap from the array based on the following pseudo code. $maxHeapify(A, i)$ move the value of $A[i]$ down to leaves to make a sub-tree of node $i$ a max-heap. Here, $H$ is the size of the heap. 1 maxHeapify(A, i) 2 l = left(i) 3 r = right(i) 4 // select the node which has the maximum value 5 if l ≤ H and A[l] > A[i] 6 largest = l 7 else 8 largest = i 9 if r ≤ H and A[r] > A[largest] 10 largest = r 11 12 if largest ≠ i // value of children is larger than that of i 13 swap A[i] and A[largest] 14 maxHeapify(A, largest) // call recursively The following procedure buildMaxHeap(A) makes $A$ a max-heap by performing maxHeapify in a bottom-up manner. 1 buildMaxHeap(A) 2 for i = H/2 downto 1 3 maxHeapify(A, i)
|
def get_left_index(A, i):
if len(A) > 2 * i:
return 2 * i
else:
return None
def get_right_index(A, i):
if len(A) > 2 * i + 1:
return 2 * i + 1
else:
return None
def max_heapfy(A, i):
left = get_left_index(A, i)
right = get_right_index(A, i)
largest = i
if left and A[left] > A[i]:
largest = left
if right and A[right] > A[largest]:
largest = right
if largest != i:
A[i], A[largest] = A[largest], A[i]
max_heapfy(A, largest)
H = int(input())
A = [0]
for elem in map(int, input().split()):
A.append(elem)
def build_max_heap(A):
for i in range(H//2, 0, -1):
print(i)
max_heapfy(A, i)
build_max_heap(A)
print(' ', end='')
print(*A[1:])
|
s635196351
|
Accepted
| 1,200
| 64,160
| 756
|
def get_left_index(A, i):
if len(A) > 2 * i:
return 2 * i
else:
return None
def get_right_index(A, i):
if len(A) > 2 * i + 1:
return 2 * i + 1
else:
return None
def max_heapfy(A, i):
left = get_left_index(A, i)
right = get_right_index(A, i)
largest = i
if left and A[left] > A[i]:
largest = left
if right and A[right] > A[largest]:
largest = right
if largest != i:
A[i], A[largest] = A[largest], A[i]
max_heapfy(A, largest)
H = int(input())
A = [0]
for elem in map(int, input().split()):
A.append(elem)
def build_max_heap(A):
for i in range(H//2, 0, -1):
max_heapfy(A, i)
build_max_heap(A)
print(' ', end='')
print(*A[1:])
|
s772676751
|
p02865
|
u448751328
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 3,316
| 85
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
N = int(input())
if N % 2 == 0:
ans = N / 2 - 1
else:
ans = (N -1) / 2
print(ans)
|
s020301845
|
Accepted
| 18
| 3,064
| 71
|
N=int(input())
if N%2==0:
print(N//2-1)
else:
print((N+1)//2-1)
|
s141688080
|
p03543
|
u565380863
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 4
|
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**?
|
1234
|
s663789335
|
Accepted
| 17
| 2,940
| 106
|
#-*- coding: utf-8 -*-
N = input()
print('Yes' if N[0] == N[1] == N[2] or N[1] == N[2] == N[3] else 'No')
|
s848553784
|
p03401
|
u143492911
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 19,808
| 281
|
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.
|
n=int(input())
a=list(map(int,input().split()))
ans=[]
a.sort()
cnt=0
for i in range(n):
v=a[i]
a[i]=0
a.sort()
for j in range(1,n):
cnt+=a[j]-a[j-1]
print(cnt)
ans.append(cnt+(max(a)-min(a)))
cnt=0
a[i]=v
for i in ans:
print(i)
|
s859810469
|
Accepted
| 216
| 14,048
| 223
|
n=int(input())
a=list(map(int,input().split()))
a.insert(0,0)
a.insert(n+1,0)
total=0
for i in range(n+1):
total+=abs(a[i+1]-a[i])
for i in range(n):
print(total-abs(a[i+1]-a[i])-abs(a[i+2]-a[i+1])+abs(a[i+2]-a[i]))
|
s700654759
|
p03759
|
u629709614
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 86
|
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-a:
print("YES")
else:
print("NO")
|
s812293212
|
Accepted
| 17
| 2,940
| 86
|
a, b, c = map(int, input().split())
if b-a == c-b:
print("YES")
else:
print("NO")
|
s335110353
|
p03854
|
u899395423
| 2,000
| 262,144
|
Wrong Answer
| 80
| 3,188
| 345
|
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()
s = s[::-1]
words = ["dream", "dreamer", "erase", "eraser"]
words = [w[::-1] for w in words]
lens = [len(w) for w in words]
while True:
for l, w in zip(lens, words):
if s[:l] == w:
s = s[l:]
break
else:
print("No")
break
if len(s) == 0:
print("Yes")
break
|
s496662948
|
Accepted
| 83
| 3,316
| 345
|
s = input()
s = s[::-1]
words = ["dream", "dreamer", "erase", "eraser"]
words = [w[::-1] for w in words]
lens = [len(w) for w in words]
while True:
for l, w in zip(lens, words):
if s[:l] == w:
s = s[l:]
break
else:
print("NO")
break
if len(s) == 0:
print("YES")
break
|
s346474431
|
p03657
|
u717626627
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 128
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a, b = map(int, input().split())
if a % 3 == 0 or b % 3 == 0 or a + b % 3 == 0:
print('Possible')
else:
print('Impossible')
|
s677877937
|
Accepted
| 17
| 2,940
| 130
|
a, b = map(int, input().split())
if a % 3 == 0 or b % 3 == 0 or (a + b) % 3 == 0:
print('Possible')
else:
print('Impossible')
|
s193860091
|
p03637
|
u219607170
| 2,000
| 262,144
|
Wrong Answer
| 69
| 15,020
| 208
|
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.
|
N = int(input())
A = list(map(int, input().split()))
four, odd = 0, 0
for a in A:
if a%4 == 0:
four += 1
elif a%2 == 1:
odd += 1
result = 'No' if four > odd else 'Yes'
print('result')
|
s653363903
|
Accepted
| 65
| 15,020
| 257
|
N = int(input())
A = list(map(int, input().split()))
four, two, odd = 0, 0, 0
for a in A:
if a%4 == 0:
four += 1
elif a%2 == 0:
two += 1
else:
odd += 1
result = 'Yes' if four + 1 >= odd + bool(two) else 'No'
print(result)
|
s132657025
|
p03737
|
u702786238
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 83
|
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
s = input()
s.upper().split(" ")
print("{}{}{}".format(s[0][0], s[1][0], s[2][0]))
|
s129069150
|
Accepted
| 17
| 2,940
| 88
|
s = input()
s = s.upper().split(" ")
print("{}{}{}".format(s[0][0], s[1][0], s[2][0]))
|
s272715292
|
p03737
|
u363836311
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 95
|
You are given three words s_1, s_2 and s_3, each composed of lowercase English letters, with spaces in between. Print the acronym formed from the uppercased initial letters of the words.
|
s1,s2,s3=map(str, input().split())
S1=list(s1)
S2=list(s2)
S3=list(s3)
print(S1[0]+S2[0]+S3[0])
|
s918920961
|
Accepted
| 17
| 2,940
| 135
|
s1,s2,s3=map(str, input().split())
S1=list(s1)
S2=list(s2)
S3=list(s3)
print(chr(ord(S1[0])-32)+chr(ord(S2[0])-32)+chr(ord(S3[0])-32))
|
s557740880
|
p03657
|
u798260206
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,164
| 168
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a,b = map(int,input().split())
Flag = "Impossible"
if a%3==0:
Flag = "Possible"
elif b%3 ==0:
Flag="Possible"
elif (a+b)%3:
Flag = "Possible"
print(Flag)
|
s695192039
|
Accepted
| 22
| 9,132
| 172
|
a,b = map(int,input().split())
Flag = "Impossible"
if a%3==0:
Flag = "Possible"
elif b%3 ==0:
Flag="Possible"
elif (a+b)%3==0:
Flag = "Possible"
print(Flag)
|
s509203331
|
p03475
|
u581187895
| 3,000
| 262,144
|
Wrong Answer
| 113
| 3,188
| 548
|
A railroad running from west to east in Atcoder Kingdom is now complete. There are N stations on the railroad, numbered 1 through N from west to east. Tomorrow, the opening ceremony of the railroad will take place. On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated. The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds. Here, it is guaranteed that F_i divides S_i. That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains. For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains.
|
import math
N = int(input())
CSF = [list(map(int, input().split())) for _ in range(N-1)]
for i in range(N):
t = 0
for j in range(i, N-1):
c, s, f = CSF[j]
if t <= s:
t = s+c
else:
t += math.ceil(t / f) * f + c
print(t)
|
s846190009
|
Accepted
| 88
| 3,188
| 547
|
import math
N = int(input())
CSF = [list(map(int, input().split())) for _ in range(N-1)]
for i in range(N):
t = 0
for j in range(i, N-1):
c, s, f = CSF[j]
if t <= s:
t = s+c
else:
t = math.ceil(t / f) * f + c
print(t)
|
s504075519
|
p03645
|
u257974487
| 2,000
| 262,144
|
Wrong Answer
| 2,106
| 52,620
| 374
|
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.
|
N, M = map(int,input().split())
P = [list(map(int,input().split())) for _ in range(M)]
P.sort()
Q = []
s = 0
ans = ("IMPOSSIBLE")
for i in range (M):
if P[i][0] == 1:
Q.append(P[i][1])
else:
s = i - 1
break
if s > 0:
for j in range(s, M):
if P[j][1] == N:
if P[j][0] in Q:
ans = ("POSSIBLE")
print(ans)
|
s301061127
|
Accepted
| 1,174
| 52,616
| 639
|
N, M = map(int,input().split())
P = [list(map(int,input().split())) for _ in range(M)]
P.sort()
q = -1
ans = ("IMPOSSIBLE")
for i in range (M):
if P[i][0] == 1:
p = P[i][1]
if q < 0:
if i > 0:
q = i - 1
else:
q = 0
for j in range(q, M):
if P[j][0] == p:
if P[j][1] == N:
ans = ("POSSIBLE")
break
elif P[j][0] > p:
if q >= 0:
q = j - 1
else:
q = 0
break
else:
break
print(ans)
|
s552954041
|
p03416
|
u201234972
| 2,000
| 262,144
|
Wrong Answer
| 111
| 3,740
| 143
|
Find the number of _palindromic numbers_ among the integers between A and B (inclusive). Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.
|
A,B = map(int,input().split())
ans = 0
for x in range(A,B+1):
x = str(x)
print(x)
if x[:1] == x[:2:-1]:
ans += 1
print(ans)
|
s508432199
|
Accepted
| 65
| 2,940
| 130
|
A,B = map(int,input().split())
ans = 0
for x in range(A,B+1):
x = str(x)
if x[:2] == x[:2:-1]:
ans += 1
print(ans)
|
s372490203
|
p02608
|
u111652094
| 2,000
| 1,048,576
|
Wrong Answer
| 2,031
| 9,216
| 548
|
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
import math
N=int(input())
for i in range(1,N+1):
ans=0
if i<=5:
print(0)
else:
ma=int(math.sqrt(i)/6)
for x in range(1,ma):
for y in range(x,ma):
for z in range(y,ma):
if x**2+y**2+z**2+x*y+y*z+z*x==i:
if x==y and y==z:
ans+=1
elif x==y or y==z or z==x:
ans+=3
else:
ans+=6
print(ans)
|
s022472906
|
Accepted
| 198
| 9,348
| 474
|
import math
N=int(input())
anslist=[0]*(N+1)
ma=int(math.sqrt(N))+1
for x in range(1,ma):
for y in range(x,ma):
for z in range(y,ma):
n=x**2+y**2+z**2+x*y+y*z+z*x
if n <=N:
if x==y and y==z:
anslist[n]+=1
elif x==y or y==z or z==x:
anslist[n]+=3
else:
anslist[n]+=6
for i in range(1,N+1):
print(anslist[i])
|
s933019076
|
p03599
|
u136090046
| 3,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 413
|
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.
|
ABCDEF = input()
split_ABCDEF = ABCDEF.split()
int_split_ABCDEF = [int(i) for i in split_ABCDEF]
A = int_split_ABCDEF[0]*100
B = int_split_ABCDEF[1]*100
C = int_split_ABCDEF[2]
D = int_split_ABCDEF[3]
E = int_split_ABCDEF[4]
F = int_split_ABCDEF[5]
if A > B:
sugar = int((B / 100) * E)
print("{} {}".format(sugar+B, sugar))
else:
sugar = int((A / 100) * E)
print("{} {}".format(sugar+A, sugar))
|
s864454293
|
Accepted
| 340
| 39,648
| 840
|
def solve():
A, B, C, D, E, F = map(int, input().split())
memo = {}
for a in range(31):
for b in range(31):
for c in range(101):
for d in range(101):
w = (100 * A) * a + (100 * B) * b
s = C * c + D * d
if w == 0:
break
if w + s > F:
break
else:
if w / 100 * E >= s:
density = 100 * s / (w + s)
memo[a, b, c, d] = density
max_v = max(memo.values())
for k, v in memo.items():
if v == max_v:
print((100 * A) * k[0] + (100 * B) * k[1] + C * k[2] + D * k[3], C * k[2] + D * k[3])
break
if __name__ == '__main__':
solve()
|
s354467169
|
p03080
|
u317587743
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 120
|
There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat.
|
N = int(input())
s = list(input())
r = s.count('R')
b = s.count('B')
if r > b:
print('YES')
else:
print('NO')
|
s635286887
|
Accepted
| 17
| 2,940
| 120
|
N = int(input())
s = list(input())
r = s.count('R')
b = s.count('B')
if r > b:
print('Yes')
else:
print('No')
|
s054437622
|
p02659
|
u297651868
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,164
| 81
|
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
a,b=map(float,input().split())
left=a//100*b
right=a%100*b
print(int(left+right))
|
s944073948
|
Accepted
| 27
| 9,752
| 74
|
from decimal import Decimal
a,b=map(Decimal,input().split())
print(a*b//1)
|
s983429924
|
p02613
|
u165436807
| 2,000
| 1,048,576
|
Wrong Answer
| 150
| 16,024
| 214
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
s = [0]*n
for i in range(n):
s[i] = input()
ac = s.count('AC')
wa = s.count('WA')
tle = s.count('TLE')
re = s.count('RE')
print('AC x',ac)
print('WA x',wa)
print('TLE x',tle)
print('Re x',re)
|
s569081540
|
Accepted
| 147
| 16,088
| 214
|
n = int(input())
s = [0]*n
for i in range(n):
s[i] = input()
ac = s.count('AC')
wa = s.count('WA')
tle = s.count('TLE')
re = s.count('RE')
print('AC x',ac)
print('WA x',wa)
print('TLE x',tle)
print('RE x',re)
|
s685373556
|
p03471
|
u616522759
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 3,060
| 328
|
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
N, Y = map(int, input().split())
ans = '-1 -1 -1'
for i in range(N + 1):
if ans != '-1 -1 -1':
break
for j in range(N + 1):
for k in range(N + 1):
if Y == 10000 * i + 5000 * j + 1000 * k \
and i + j + k <= N:
ans = str(i) + ' ' + str(j) + ' ' + str(k)
print(ans)
|
s967599352
|
Accepted
| 1,365
| 3,060
| 299
|
N, Y = map(int, input().split())
ans = '-1 -1 -1'
for i in range(N + 1):
for j in range(N + 1):
if Y == 10000 * i + 5000 * j + 1000 * (N - i - j)\
and i + j + (N - i - j) <= N\
and N - i - j >= 0:
ans = str(i) + ' ' + str(j) + ' ' + str(N - i - j)
print(ans)
|
s425499607
|
p02389
|
u435300817
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,592
| 111
|
Write a program which calculates the area and perimeter of a given rectangle.
|
input_height_width = input()
height, width = [int(x) for x in input_height_width.split()]
print(height * width)
|
s098961731
|
Accepted
| 20
| 7,644
| 112
|
values = input()
height, width = [int(x) for x in values.split()]
print(height * width, height * 2 + width * 2)
|
s790941918
|
p02392
|
u192877548
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,636
| 146
|
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
# coding: utf-8
a,b,c = map(int,input().split())
if a < b:
if b < c:
print("YES")
else:
print("NO")
else:
print("NO")
|
s233321890
|
Accepted
| 20
| 7,640
| 156
|
# coding: utf-8
a,b,c = map(int,input().split())
if a < b and a < c:
if b < c:
print("Yes")
else:
print("No")
else:
print("No")
|
s696765142
|
p02388
|
u630518143
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 322
|
Write a program which calculates the cube of a given integer x.
|
x = int(input("整数xを入力してください。（0<=x<=100）："))
while not(0<=x and x<=100):
print("0<=x<=100を満たす整数を入力してください。")
x = int(input("整数xを入力してください。（0<=x<=100）："))
x = x ** 3
print(str(x) + "の３乗は" + str(x) + "です。")
|
s925347446
|
Accepted
| 20
| 5,576
| 35
|
x = int(input())
x = x**3
print(x)
|
s216104514
|
p02578
|
u724088399
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,176
| 155
|
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.
|
A = list(map(int,input().split()))
sm = 0
now = A[0]
for num in A:
if num < now:
sm += now-num
elif num > now:
now = num
print(sm)
|
s079866888
|
Accepted
| 104
| 32,220
| 172
|
N = int(input())
A = list(map(int,input().split()))
sm = 0
now = A[0]
for num in A:
if num < now:
sm += now-num
elif num > now:
now = num
print(sm)
|
s508245986
|
p03478
|
u840310460
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 262
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
N, A, B = [int(i) for i in input().split()]
def func_10(x):
ans = 0
while x/10 != 0:
ans += x % 10
x = x//10
return ans
ans = 0
for i in (1, N+1):
if A <= func_10(i) <= B:
ans += i
print(ans)
|
s995689448
|
Accepted
| 29
| 2,940
| 235
|
N, A, B = [int(i) for i in input().split()]
def func_de(n):
ans = 0
while n/10 != 0:
ans += n % 10
n //= 10
return ans
ANS = 0
for i in range(N+1):
if A <= func_de(i) <= B:
ANS += i
print(ANS)
|
s447342192
|
p02390
|
u822442916
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,680
| 73
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
S=int(input())
h=S/3600
k=S//3600
m=k/60
s=k-m
print("%d:%d:%d" %(h,m,s))
|
s704594958
|
Accepted
| 40
| 7,684
| 73
|
S=int(input())
h=S/3600
k=S%3600
m=k/60
s=k%60
print("%d:%d:%d" %(h,m,s))
|
s373533578
|
p03455
|
u747602774
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 119
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
import math
s=list(input())
X=int(s[0]+s[1])
root=math.sqrt(X)
a=root//1
if a**2!=X:
print('No')
else:
print('Yes')
|
s381465032
|
Accepted
| 17
| 2,940
| 81
|
x,y = map(int,input().split())
if x*y%2==0:
print('Even')
else:
print('Odd')
|
s145951453
|
p03861
|
u633450100
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 115
|
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 = [int(i) for i in input().split()]
if (b-a) % x == 0:
print((b-a) // x)
else:
print((b-a) // x + 1)
|
s312378485
|
Accepted
| 17
| 2,940
| 127
|
a,b,x = [int(i) for i in input().split()]
if a % x == 0:
A = a // x
else:
A = a // x + 1
B = b // x
print(B - A + 1)
|
s644017821
|
p03457
|
u871867619
| 2,000
| 262,144
|
Wrong Answer
| 700
| 29,584
| 523
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
import numpy as np
N = int(input())
travel_list = [np.array([0, 0, 0])]
add_list = [np.array([int(i) for i in input().split()]) for i in range(N)]
travel_list.extend(add_list)
def is_travel(arr1, arr2):
dt, dx, dy = abs(arr2 - arr1)
if dt <= dx + dy:
return False
elif dt % 2 != (dx + dy) % 2:
return False
else:
return True
for i in range(N):
if is_travel(travel_list[i], travel_list[i+1]):
continue
else:
print('NO')
exit()
print('Yes')
|
s045849113
|
Accepted
| 1,406
| 29,564
| 539
|
import numpy as np
N = int(input())
travel_list = [np.array([0, 0, 0])]
add_list = [np.array([int(i) for i in input().split()]) for i in range(N)]
travel_list.extend(add_list)
def is_travel(arr1, arr2):
dt, dx, dy = abs(arr2 - arr1)
if dt < dx + dy:
return False
elif dt % 2 != (dx + dy) % 2:
return False
else:
return True
travel = 'Yes'
for i in range(N):
if is_travel(travel_list[i], travel_list[i+1]):
continue
else:
travel = 'No'
break
print(travel)
|
s729307755
|
p03612
|
u513434790
| 2,000
| 262,144
|
Wrong Answer
| 168
| 14,008
| 428
|
You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this.
|
N = int(input())
p = tuple(map(int, input().split()))
a = list(p)
b = list(p)
c = 0
for i in range(N):
if i+1 == a[i] and i == N-1:
c += 1
elif i+1 == a[i]:
a[i], a[i+1] = a[i+1], a[i]
c += 1
cc = 0
for i in range(N)[::-1]:
if i+1 == b[i] and i == 0:
cc += 1
elif i+1 == b[i]:
b[i], b[i-1] = b[i-1], b[i]
cc += 1
print(i)
print(min(c,cc))
|
s372502019
|
Accepted
| 130
| 14,132
| 411
|
N = int(input())
p = tuple(map(int, input().split()))
a = list(p)
b = list(p)
c = 0
for i in range(N):
if i+1 == a[i] and i == N-1:
c += 1
elif i+1 == a[i]:
a[i], a[i+1] = a[i+1], a[i]
c += 1
cc = 0
for i in range(N)[::-1]:
if i+1 == b[i] and i == 0:
cc += 1
elif i+1 == b[i]:
b[i], b[i-1] = b[i-1], b[i]
cc += 1
print(min(c,cc))
|
s083368030
|
p02865
|
u929585607
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 22
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
print(int(input())//2)
|
s902358224
|
Accepted
| 17
| 2,940
| 26
|
print((int(input())-1)//2)
|
s560147228
|
p03049
|
u228223940
| 2,000
| 1,048,576
|
Wrong Answer
| 38
| 3,828
| 561
|
Snuke has N strings. The i-th string is s_i. Let us concatenate these strings into one string after arranging them in some order. Find the maximum possible number of occurrences of `AB` in the resulting string.
|
import re
N = int(input())
s = [input() for _ in range(N)]
tmp = 0
print(s[1],s[1][-1:],s[1][0:1])
B_A = 0
B_any = 0
any_A = 0
for i in range(N):
tmp = tmp + s[i].count("AB")
if s[i][0:1] == "B":
if s[i][-1:] == "A":
B_A = B_A + 1
else:
B_any = B_any + 1
elif s[i][-1:] == "A":
any_A = any_A + 1
add = 0
add1 = min(B_any,any_A)
if B_A % 2 == 0:
add = int(B_A / 2)
else:
if B_any == any_A:
add = int((B_A-1) / 2)
else:
add = int((B_A-1) / 2) + 1
print(tmp+add+add1)
|
s467543949
|
Accepted
| 38
| 3,956
| 469
|
import re
N = int(input())
s = [input() for _ in range(N)]
tmp = 0
B_A = 0
B_any = 0
any_A = 0
for i in range(N):
tmp = tmp + s[i].count("AB")
if s[i][0:1] == "B":
if s[i][-1:] == "A":
B_A = B_A + 1
else:
B_any = B_any + 1
elif s[i][-1:] == "A":
any_A = any_A + 1
add = 0
add1 = min(B_any,any_A)
add = max(B_A - 1,0)
if B_any + any_A > 0:
if B_A >= 1:
add1 = add1 + 1
print(tmp+add+add1)
|
s577738630
|
p03693
|
u896741788
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 75
|
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
a,s,d=map(int,input().split())
print("YES") if 10*s+d%4==0 else print("NO")
|
s069314628
|
Accepted
| 17
| 2,940
| 77
|
a,s,d=map(int,input().split())
print("YES") if (10*s+d)%4==0 else print("NO")
|
s803813019
|
p02845
|
u535236942
| 2,000
| 1,048,576
|
Wrong Answer
| 2,108
| 14,396
| 201
|
N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green. The person numbered i says: * "In front of me, exactly A_i people are wearing hats with the same color as mine." Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats. Since the count can be enormous, compute it modulo 1000000007.
|
n = int(input())
a = list(map(int,input().split()))
ans = 1
for i in range(n):
if a[:i].count(a[i]) == 2:
ans = ans*3%1000000007
if a[:i].count(a[i]) == 1:
ans = ans*2%1000000007
print(ans)
|
s089389363
|
Accepted
| 194
| 14,396
| 412
|
n = int(input())
a = list(map(int,input().split()))
mod = 1000000007
ans = 1
l = []
for i in range(n):
if a[i] == 0:
l.append(0)
ans = ans*(4-len(l))%mod
else:
id = -1
count = 0
for j in range(len(l)):
if l[j] == a[i] - 1:
count += 1
id = j
if id == -1:
ans = 0
break
ans = (ans*count)%mod
l[id] = a[i]
if ans == 0:
break
print(ans%mod)
|
s578341497
|
p03457
|
u335973735
| 2,000
| 262,144
|
Wrong Answer
| 453
| 39,456
| 482
|
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())
z_t = []*N
t0 = 0
x0 = 0
y0 = 0
ans = 'NO'
for i in range(N):
s = input().rstrip().split()
z_t.append(s)
print(z_t)
for i in range(N):
dt = int(z_t[i][0]) - t0
dx = int(z_t[i][1]) - x0
dy = int(z_t[i][2]) - y0
extra_t = dt - (dx+dy)
if extra_t >= 0 and extra_t%2 == 0:
t0 = int(z_t[i][0])
x0 = int(z_t[i][1])
y0 = int(z_t[i][2])
if i == N-1:
ans = 'YES'
else:
break
print(ans)
|
s168546270
|
Accepted
| 412
| 32,144
| 478
|
N = int(input())
z_t = []*N
t0 = 0
x0 = 0
y0 = 0
ans = 'No'
for i in range(N):
s = input().rstrip().split()
z_t.append(s)
for i in range(N):
dt = int(z_t[i][0]) - t0
dx = abs(int(z_t[i][1]) - x0)
dy = abs(int(z_t[i][2]) - y0)
extra_t = dt - (dx+dy)
if extra_t >= 0 and extra_t%2 == 0:
t0 = int(z_t[i][0])
x0 = int(z_t[i][1])
y0 = int(z_t[i][2])
if i == N-1:
ans = 'Yes'
else:
break
print(ans)
|
s498857940
|
p03696
|
u580225095
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 250
|
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.
|
# -*- coding: utf-8 -*-
x = int(input())
y = input()
z = y
while z.find("()") != -1:
z = z.replace("()","")
a = z.count("(")
b = z.count(")")
ans = z
for i in range(a):
ans = ans + ")"
for i in range(b):
ans = "(" + ans
print(ans)
|
s543302317
|
Accepted
| 17
| 3,060
| 250
|
# -*- coding: utf-8 -*-
x = int(input())
y = input()
z = y
while z.find("()") != -1:
z = z.replace("()","")
a = z.count("(")
b = z.count(")")
ans = y
for i in range(a):
ans = ans + ")"
for i in range(b):
ans = "(" + ans
print(ans)
|
s098490102
|
p03547
|
u202877219
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 104
|
In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger?
|
a,b = map(str, input().split())
if a>b:
print (">")
elif b<a:
print ("<")
else:
print ("=")
|
s085563504
|
Accepted
| 17
| 2,940
| 108
|
x,y = map(str, input().split())
if x>y:
print (">")
elif x < y:
print ("<")
else:
print ("=")
|
s692279587
|
p03471
|
u350049649
| 2,000
| 262,144
|
Wrong Answer
| 765
| 3,060
| 202
|
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
N,Y=map(int,input().split())
a1=-1
a2=-1
a3=-1
for a in range(N+1):
for b in range(N-a,N+1):
c=N-a-b
if a*10000+b*5000+c*1000==Y:
a1=a
a2=b
a3=c
break
print(a1,a2,a3)
|
s520515793
|
Accepted
| 759
| 3,060
| 200
|
N,Y=map(int,input().split())
a1=-1
a2=-1
a3=-1
for a in range(N+1):
for b in range(N-a+1):
c=N-a-b
if a*10000+b*5000+c*1000==Y:
a1=a
a2=b
a3=c
break
print(a1,a2,a3)
|
s844743716
|
p03168
|
u589303510
| 2,000
| 1,048,576
|
Wrong Answer
| 1,627
| 222,760
| 463
|
Let N be a positive odd number. There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i. Taro has tossed all the N coins. Find the probability of having more heads than tails.
|
def coins(n,prob):
dp = [[0]*(n+1) for _ in range(n+1)]
dp[0][0] = 1
for i in range(1,n+1):
for j in range(i+1):
if j == 0:
dp[i][j] = dp[i-1][j]*prob[i-1]
else:
dp[i][j] = dp[i-1][j]*(1-prob[i-1])+dp[i-1][j-1]*prob[i-1]
total = 0
for i in range(n,n//2,-1):
total+= dp[n][i]
return total
n = int(input())
prob = list(map(float,input().split()))
print(coins(n,prob))
|
s254566404
|
Accepted
| 1,607
| 222,560
| 467
|
def coins(n,prob):
dp = [[0]*(n+1) for _ in range(n+1)]
dp[0][0] = 1
for i in range(1,n+1):
for j in range(i+1):
if j == 0:
dp[i][j] = dp[i-1][j]*(1-prob[i-1])
else:
dp[i][j] = dp[i-1][j]*(1-prob[i-1])+dp[i-1][j-1]*prob[i-1]
total = 0
for i in range(n,n//2,-1):
total+= dp[n][i]
return total
n = int(input())
prob = list(map(float,input().split()))
print(coins(n,prob))
|
s195933623
|
p03847
|
u786020649
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,188
| 441
|
You are given a positive integer N. Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.
|
import sys
p=10**9+7
dthbit=lambda d,n: (n>>d)&1
def main(n):
dp=[[0 for _ in range(3)] for _ in range(64)]
dp[63][0]=1
for d in range(62,-1,-1):
b=dthbit(d,n)
s=dp[d+1][:]
dp[d][0]=dp[d+1][0]+(1^b)*dp[d+1][1] % p
dp[d][1]=b*dp[d+1][0]+dp[d+1][1] % p
dp[d][2]=(1+b)*dp[d+1][1]+3*dp[d+1][2] % p
print(s,dp[d+1][:],dp[d][:])
return sum(dp[0][:])
n=int(input())
print(main(n))
|
s843596843
|
Accepted
| 29
| 9,176
| 408
|
import sys
p=10**9+7
dthbit=lambda d,n: (n>>d)&1
def main(n):
dp=[[0 for _ in range(3)] for _ in range(64)]
dp[63][0]=1
for d in range(62,-1,-1):
b=dthbit(d,n)
s=dp[d+1][:]
dp[d][0]=dp[d+1][0]+(1^b)*dp[d+1][1] % p
dp[d][1]=b*dp[d+1][0]+dp[d+1][1] % p
dp[d][2]=(1+b)*dp[d+1][1]+3*dp[d+1][2] % p
return sum(dp[0][:]) % p
n=int(input())
print(main(n))
|
s261826424
|
p03352
|
u802772880
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 229
|
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())
l2=[1,2,4,8,16,32,64,128,256,512]
l3=[3,9,27,81,243,729]
l5=[5,25,125,625]
l6=[6,36,216]
l7=[7,49,343]
l10=[10,100,1000]
le=[n*n for n in range(11,32)]
l=l2+l3+l5+l6+l7+l10+le
m=max([i for i in l if i<=X])
print(m)
|
s148607886
|
Accepted
| 18
| 2,940
| 151
|
X=int(input())
Y=1
for b in range(1,X):
for p in range(2,X):
if b**p<=X:
Y=max(Y,b**p)
else:
break
print(Y)
|
s981577920
|
p03069
|
u528005130
| 2,000
| 1,048,576
|
Wrong Answer
| 78
| 12,836
| 455
|
There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored.
|
# coding: utf-8
# Your code here!
N = int(input())
S = [x for x in input()]
cum_white = []
count = 0
for i, s in enumerate(S):
if s == '.':
count += 1
cum_white.append(count)
if len(cum_white) == 0 or len(cum_white) == N:
print(0)
else:
min_num = int(1e10)
for i, n_white in enumerate(cum_white):
num = (i - n_white) + (N - cum_white[-1] - n_white)
min_num = min(min_num, num)
print(min_num)
|
s158393411
|
Accepted
| 180
| 12,396
| 385
|
# coding: utf-8
# Your code here!
N = int(input())
S = ['.'] + [x for x in input()] + ['#']
cum_white = []
count = 0
for i, s in enumerate(S):
if s == '#':
count += 1
cum_white.append(count)
min_num = int(1e10)
for i, n_white in enumerate(cum_white):
num = n_white + (N + 1 - i - (cum_white[-1] - n_white))
min_num = min(min_num, num)
print(min_num)
|
s816832679
|
p02393
|
u731896389
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,304
| 36
|
Write a program which reads three integers, and prints them in ascending order.
|
n= input().split()
n.sort()
print(n)
|
s281758227
|
Accepted
| 30
| 7,644
| 63
|
a=list(map(int,input().split()))
a.sort()
print(a[0],a[1],a[2])
|
s919249252
|
p03644
|
u475675023
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 75
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n=int(input())
for i in range(9):
if 2**i >= n:
print(2**i)
break
|
s048486483
|
Accepted
| 17
| 2,940
| 64
|
n=int(input())
m=1
while 2**m<=n:
m+=1
else:
print(2**(m-1))
|
s946659324
|
p03369
|
u023229441
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 37
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
print(700+(input().count("○"))*100)
|
s342247335
|
Accepted
| 17
| 2,940
| 36
|
print(700+(input().count("o"))*100)
|
s792245876
|
p03493
|
u284363684
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,020
| 52
|
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.
|
# input
S = list(input())
print(S.count(1))
|
s127624880
|
Accepted
| 25
| 8,996
| 54
|
# input
S = list(input())
print(S.count("1"))
|
s880213233
|
p03474
|
u187516587
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 323
|
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
p=input().split()
A=int(p[0])
B=int(p[1])
S=input()
if S[A]=="-":
t=S.split('-')
if len(t)==2:
a,b=int(t[0]),int(t[1])
print(a)
if 10**(A-1)<=a<10**A and 10**(B-1)<=b<10**B:
print("Yes")
else:
print("No")
else:
print("No")
else:
print("No")
|
s891752581
|
Accepted
| 17
| 3,060
| 175
|
p=input().split()
A=int(p[0])
B=int(p[1])
S=input()
if S[A]=="-":
t=S.split('-')
if len(t)==2:
print("Yes")
else:
print("No")
else:
print("No")
|
s356867701
|
p03407
|
u077075933
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 68
|
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 "NO")
|
s986739614
|
Accepted
| 17
| 2,940
| 68
|
A, B, C = map(int, input().split())
print("Yes" if C<=A+B else "No")
|
s577240503
|
p02601
|
u005569385
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,136
| 293
|
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 = map(int,input().split())
K = int(input())
for i in range(0,K+1):
A = A*(2*i)
for j in range(0,K+1):
B = B*(2*j)
for h in range(0,K+1):
C = C*(2*h)
if (i+j+h) <= K and C>B>A:
print("Yes")
exit()
print("No")
|
s961739469
|
Accepted
| 31
| 9,192
| 252
|
A,B,C = map(int,input().split())
K = int(input())
for i in range(0,K+1):
for j in range(0,K+1):
for h in range(0,K+1):
if A*(2**i)<B*(2**j)<C*(2**h) and (i+j+h)<=K:
print("Yes")
exit()
print("No")
|
s360254824
|
p03759
|
u394482932
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 43
|
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.
|
for s in input().split():print(s[0],end='')
|
s822734946
|
Accepted
| 17
| 2,940
| 58
|
a,b,c=map(int,input().split());print('YNEOS'[b-a!=c-b::2])
|
s862093787
|
p03658
|
u492030100
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 135
|
Snuke has N sticks. The length of the i-th stick is l_i. Snuke is making a snake toy by joining K of the sticks together. The length of the toy is represented by the sum of the individual sticks that compose it. Find the maximum possible length of the toy.
|
N, K = map(int, input().split())
L = [l for l in range(N)]
L.sort(reverse=True)
ans = 0
for i in range(K):
ans += L[i]
print(ans)
|
s041152239
|
Accepted
| 18
| 2,940
| 147
|
N, K = map(int, input().split())
L = [int(l) for l in input().split()]
L.sort(reverse=True)
ans = 0
for i in range(K):
ans += L[i]
print(ans)
|
s575734686
|
p02842
|
u671211357
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 266
|
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.
|
import math
N = int(input())
kari = N*100/108
hako = math.ceil(kari)
ans = math.floor(kari)
if hako*108/100==N or ans*108/100==N:
print(kari)
elif math.floor(hako*108/100)==N:
print(hako)
elif math.floor(ans*108/100)==N:
print(ans)
else:
print(":(")
|
s102884812
|
Accepted
| 17
| 3,060
| 378
|
import math
N = int(input())
kari = N*100/108
hako = math.ceil(kari)
ans = math.floor(kari)
if hako*108/100==N or ans*108/100==N:
print(math.floor(kari))
elif math.floor(hako*108/100)==N:
print(hako)
elif math.floor(ans*108/100)==N:
print(ans)
# print(hako)
# print(ans)
else:
print(":(")
|
s193631754
|
p03469
|
u498575211
| 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[5:])
|
s820371018
|
Accepted
| 17
| 2,940
| 31
|
S = input()
print("2018"+S[4:])
|
s101841550
|
p02613
|
u377989038
| 2,000
| 1,048,576
|
Wrong Answer
| 143
| 16,612
| 213
|
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 as ct
n = int(input())
s = ct([input() for _ in range(n)])
print("AC x " + str(s["AC"]))
print("WA x " + str(s["wA"]))
print("TLE x " + str(s["TLE"]))
print("RE x " + str(s["RE"]))
|
s139659295
|
Accepted
| 143
| 16,508
| 213
|
from collections import Counter as ct
n = int(input())
s = ct([input() for _ in range(n)])
print("AC x " + str(s["AC"]))
print("WA x " + str(s["WA"]))
print("TLE x " + str(s["TLE"]))
print("RE x " + str(s["RE"]))
|
s147743229
|
p03486
|
u239528020
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,092
| 208
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
#!/usr/bin/env python3
import sys
sys.setrecursionlimit(10**6)
s = list(str(input()))
t = str(input())
s.sort()
s = "".join(s)
ans = sorted([s, t])
if ans[0] == t:
print("No")
else:
print("Yes")
|
s058066570
|
Accepted
| 30
| 9,072
| 263
|
#!/usr/bin/env python3
import sys
sys.setrecursionlimit(10**6)
s = list(str(input()))
t = list(str(input()))
s.sort()
t.sort(reverse=True)
s = "".join(s)
t = "".join(t)
ans = sorted([s, t])
# print(ans)
if ans[0] == t:
print("No")
else:
print("Yes")
|
s606374258
|
p04029
|
u824147251
| 2,000
| 262,144
|
Wrong Answer
| 37
| 3,064
| 33
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
n = int(input())
print((n+1)*n/2)
|
s378277987
|
Accepted
| 42
| 3,444
| 34
|
n = int(input())
print((n+1)*n//2)
|
s204546982
|
p02255
|
u734765925
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,628
| 229
|
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())
A = list(map(int,input().split()))
temp = 0
for i in range(1,n):
temp = A[i]
j = i -1
while j >=0 and A[j] > temp :
A[j+1]=A[j]
j-=1
A[j+1] = temp
print(" ".join(map(str,A)))
|
s331689025
|
Accepted
| 30
| 7,660
| 256
|
n = int(input())
A = list(map(int,input().split()))
temp = 0
print(" ".join(map(str,A)))
for i in range(1,n):
temp = A[i]
j = i -1
while j >=0 and A[j] > temp :
A[j+1]=A[j]
j-=1
A[j+1] = temp
print(" ".join(map(str,A)))
|
s070782881
|
p03048
|
u902468164
| 2,000
| 1,048,576
|
Wrong Answer
| 890
| 3,060
| 163
|
Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this?
|
r,g,b,n=[int(i) for i in input().split(" ")]
res = 0
for i in range(0,n+1,r):
for j in range(i,n+1,g):
if j % b == 0:
res += 1
print(res)
|
s141532350
|
Accepted
| 958
| 2,940
| 169
|
r,g,b,n=[int(i) for i in input().split(" ")]
res = 0
for i in range(0,n+1,r):
for j in range(i,n+1,g):
if (n-j) % b == 0:
res += 1
print(res)
|
s648067977
|
p03457
|
u780675733
| 2,000
| 262,144
|
Wrong Answer
| 435
| 27,380
| 422
|
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())
Input = []
for i in range(N):
Input.append(list(map(int,input().split())))
x = 0
y = 0
t = 0
for i in Input:
dif_x = abs(x - i[1])
dif_y = abs(y - i[2])
dif_t = abs(t - i[0])
if dif_x + dif_y <= dif_t and (dif_x + dif_y)%2 == dif_t%2:
x = i[1]
y = i[2]
t = i[0]
else:
print("NO")
break
else:
print("YES")
|
s227703883
|
Accepted
| 442
| 27,380
| 394
|
N = int(input())
Input = []
for i in range(N):
Input.append(list(map(int,input().split())))
x = 0
y = 0
t = 0
for i in Input:
dif_x = abs(x - i[1])
dif_y = abs(y - i[2])
dif_t = abs(t - i[0])
if dif_x + dif_y <= dif_t and (dif_x + dif_y)%2 == dif_t%2:
x = i[1]
y = i[2]
t = i[0]
else:
print("No")
break
else:
print("Yes")
|
s365086804
|
p03643
|
u301823349
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 118
|
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
N = int(input())
p = 0
while True:
if N < 2**p:
print(2**(p-1))
break
else:
p=p+1
|
s688066500
|
Accepted
| 17
| 2,940
| 28
|
N = input()
print("ABC" + N)
|
s936772673
|
p03162
|
u469953228
| 2,000
| 1,048,576
|
Wrong Answer
| 959
| 50,332
| 304
|
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.
|
N = int(input())
dp = [[0] * 3 for _ in range(N+1)]
A = [list(map(int,input().split())) for _ in range(N)]
for i in range(N):
for j in range(3):
for k in range(3):
if j == k:
continue
dp[i+1][k] = max(dp[i+1][k], dp[i][k] + A[i][j])
print(max(dp[N]))
|
s996135608
|
Accepted
| 980
| 50,336
| 304
|
N = int(input())
dp = [[0] * 3 for _ in range(N+1)]
A = [list(map(int,input().split())) for _ in range(N)]
for i in range(N):
for j in range(3):
for k in range(3):
if j == k:
continue
dp[i+1][k] = max(dp[i+1][k], dp[i][j] + A[i][k])
print(max(dp[N]))
|
s050432332
|
p02608
|
u388971072
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,244
| 243
|
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
N = int(input())
N_list = [0]*(N+1)
for i in range(1,11):
for j in range(1,11):
for k in range(1,11):
if(i**2+j**2+k**2+i*j+j*k+k*i<=N):
N_list[i**2+j**2+k**2+i*j+j*k+k*i]+=1
for i in N_list:
print(i)
|
s116003770
|
Accepted
| 1,027
| 9,292
| 250
|
N = int(input())
N_list = [0]*(N+1)
for i in range(1,101):
for j in range(1,101):
for k in range(1,101):
if(i**2+j**2+k**2+i*j+j*k+k*i<=N):
N_list[i**2+j**2+k**2+i*j+j*k+k*i]+=1
for i in N_list[1:]:
print(i)
|
s037132831
|
p02393
|
u255164080
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,720
| 46
|
Write a program which reads three integers, and prints them in ascending order.
|
nums = input().split()
nums.sort()
print(nums)
|
s419016480
|
Accepted
| 30
| 6,724
| 224
|
nums = input().split()
a = int(nums[0])
b = int(nums[1])
c = int(nums[2])
if a > b:
tmp = a
a = b
b = tmp
if b > c:
tmp = b
b = c
c = tmp
if a > b:
tmp = a
a = b
b = tmp
print(a, b, c)
|
s739158398
|
p02613
|
u068844030
| 2,000
| 1,048,576
|
Wrong Answer
| 136
| 16,160
| 234
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
s = [input() for i in range(n)]
AC = s.count("AC")
WA = s.count("WA")
TLE = s.count("TLE")
RE = s.count("RE")
print("AC × " + str(AC))
print("WA × " + str(WA))
print("TLE × " + str(TLE))
print("RE × " + str(RE))
|
s983307635
|
Accepted
| 146
| 16,052
| 293
|
n = int(input())
s = [input() for i in range(n)]
AC = s.count("AC")
WA = s.count("WA")
TLE = s.count("TLE")
RE = s.count("RE")
AC2 = "AC x " + str(AC)
WA2 = "WA x " + str(WA)
TLE2 = "TLE x " + str(TLE)
RE2 = "RE x " + str(RE)
list = [AC2,WA2,TLE2,RE2]
list_n = "\n".join(list)
print(list_n)
|
s152640402
|
p03160
|
u830054172
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 109,972
| 243
|
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())
a = list(map(int, input().split()))
dp = [0]*n
for i in range(n):
if i >= 2:
dp[i] = min(dp[i-1]+abs(a[i]-a[i-1]), dp[i-2]+abs(a[i]-a[i-2]))
elif i == 1:
dp[i] = a[i]-a[i-1]
print(dp)
print(dp[-1])
|
s105303366
|
Accepted
| 181
| 13,928
| 264
|
N = int(input())
h = list(map(int, input().split())) + [0]*100010
dp = [float("inf")]*100010
dp[0] = 0
for i in range(N):
dp[i+1] = min(dp[i+1], dp[i] + abs(h[i+1] - h[i]))
dp[i+2] = min(dp[i+2], dp[i] + abs(h[i+2] - h[i]))
# print(dp)
print(dp[N-1])
|
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