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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s770240643
|
p02645
|
u089786098
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 9,008
| 44
|
When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.
|
a = input()
a = str(a)
print(a[0],a[1],a[2])
|
s443425220
|
Accepted
| 24
| 8,960
| 24
|
a = input()
print(a[:3])
|
s173709562
|
p03378
|
u186838327
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 134
|
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
n, m, x = map(int, input().split())
a = list(map(int, input().split()))
min(len([b for b in a if b < x]), len([c for c in a if c >x]))
|
s523414579
|
Accepted
| 18
| 2,940
| 141
|
n, m, x = map(int, input().split())
a = list(map(int, input().split()))
print(min(len([b for b in a if b < x]), len([c for c in a if c >x])))
|
s742275139
|
p03437
|
u810356688
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 229
|
You are given positive integers X and Y. If there exists a positive integer not greater than 10^{18} that is a multiple of X but not a multiple of Y, choose one such integer and print it. If it does not exist, print -1.
|
import sys
def input(): return sys.stdin.readline().rstrip()
def main():
x,y=map(int,input().split())
if x==y or y%x==0:
print(-1)
sys.exit()
else:
print(x)
if __name__=='__main__':
main()
|
s246291578
|
Accepted
| 17
| 3,060
| 325
|
import sys
def input(): return sys.stdin.readline().rstrip()
def main():
x,y=map(int,input().split())
if x==y or x%y==0:
print(-1)
sys.exit()
else:
for i in range(2,10*6):
if x*i % y!=0:
print(x*i)
sys.exit()
if __name__=='__main__':
main()
|
s593488433
|
p03369
|
u448655578
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 96
|
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.
|
toppings = list(input())
ans = 700
for i in toppings:
if i == "○":
ans += 100
print(ans)
|
s524133923
|
Accepted
| 17
| 2,940
| 94
|
toppings = list(input())
ans = 700
for i in toppings:
if i == "o":
ans += 100
print(ans)
|
s824438488
|
p02694
|
u366836739
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,232
| 219
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
# coding: utf-8
def main():
x = int(input())
money = 100
year = 0
while money <= x:
year += 1
money *= 1.01
money = money//1
print(year)
if __name__ == '__main__':
main()
|
s967714402
|
Accepted
| 23
| 9,232
| 218
|
# coding: utf-8
def main():
x = int(input())
money = 100
year = 0
while money < x:
year += 1
money *= 1.01
money = money//1
print(year)
if __name__ == '__main__':
main()
|
s738771509
|
p03543
|
u950567701
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 188
|
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**?
|
a = input()
li = []
for s in a:
li.append(s)
count = 0
for i in range(1,4):
if li[i] == li[i-1]:
count += 1
else:
count =0
if count >= 3:
print("Yes")
else:
print("No")
|
s134285046
|
Accepted
| 17
| 2,940
| 221
|
a = input()
li = []
for s in a:
li.append(s)
count = 0
for i in range(1,4):
if li[i] == li[i-1]:
count += 1
if count == 2:
break
else:
count = 0
if count >= 2:
print("Yes")
else:
print("No")
|
s395151183
|
p02613
|
u977961492
| 2,000
| 1,048,576
|
Wrong Answer
| 137
| 16,276
| 186
|
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)]
print("AC x" + str(S.count("AC")))
print("WA x" + str(S.count("WA")))
print("TLE x" + str(S.count("TLE")))
print("RE x" + str(S.count("RE")))
|
s194586360
|
Accepted
| 140
| 16,224
| 183
|
n=int(input())
s=[input() for i in range(n)]
print("AC x "+str(s.count("AC")))
print("WA x "+str(s.count("WA")))
print("TLE x "+str(s.count("TLE")))
print("RE x "+str(s.count("RE")))
|
s584743960
|
p00002
|
u661290476
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,536
| 49
|
Write a program which computes the digit number of sum of two integers a and b.
|
a,b=map(int,input().split())
print(len(str(a+b)))
|
s190651706
|
Accepted
| 20
| 5,592
| 119
|
while True:
try:
a, b = map(int, input().split())
print(len(str(a + b)))
except:
break
|
s257403288
|
p02399
|
u596870465
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,600
| 89
|
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
a, b = map(int,input().split())
d = a // b
r = a%b
f = float(a/b)
print(d,r,f)
|
s740284863
|
Accepted
| 20
| 5,600
| 71
|
a, b = map(int,input().split())
print("%d %d %.5f" % (a/b, a%b, a/b))
|
s084324993
|
p03945
|
u594956556
| 2,000
| 262,144
|
Wrong Answer
| 48
| 3,188
| 139
|
Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa. In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white. Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place.
|
s=input()
n=len(s)
count=0
if n==1:
print(0)
exit()
iro=s[0]
for i in range(1,n):
if s[i]==iro:
count+=1
iro=s[i]
print(count)
|
s088495780
|
Accepted
| 44
| 3,188
| 148
|
s=input()
n=len(s)
count=0
if n==1:
print(0)
exit()
iro=s[0]
for i in range(1,n):
now=s[i]
if now!=iro:
count+=1
iro=now
print(count)
|
s233298065
|
p03964
|
u503901534
| 2,000
| 262,144
|
Wrong Answer
| 20
| 2,940
| 251
|
AtCoDeer the deer is seeing a quick report of election results on TV. Two candidates are standing for the election: Takahashi and Aoki. The report shows the ratio of the current numbers of votes the two candidates have obtained, but not the actual numbers of votes. AtCoDeer has checked the report N times, and when he checked it for the i-th (1≦i≦N) time, the ratio was T_i:A_i. It is known that each candidate had at least one vote when he checked the report for the first time. Find the minimum possible total number of votes obtained by the two candidates when he checked the report for the N-th time. It can be assumed that the number of votes obtained by each candidate never decreases.
|
n = int(input())
kouho = 0
for i in range(n):
t,a = map(int,input().split())
s = t + a
if kouho <= s:
kouho = s
else:
j = 1
while kouho > s:
s = s * j
j = j + 1
kouho = s
print(s)
|
s347653924
|
Accepted
| 21
| 3,064
| 265
|
n = int(input())
import math as m
t1 , a1 = map(int,input().split())
tt = t1
aa = a1
for i in range(n-1):
t, a = map(int,input().split())
n = ((max((tt//t),(aa//a))))
if tt > t*n or aa> a*n:
n = n + 1
tt = t * n
aa = a * n
print(tt + aa)
|
s702147101
|
p03486
|
u412563426
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 237
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s= input()
t= input()
ss = sorted(s)
tt = sorted(t)
sss = "".join(ss)
ttt = "".join(tt)
A = [sss, ttt]
A.sort()
if A[0] == sss:
print("Yes")
else:
print("No")
|
s096127913
|
Accepted
| 18
| 3,064
| 214
|
s= input()
t= input()
if s == t:
print("No")
exit()
ss = sorted(s)
tt = sorted(t, reverse=True)
sss = "".join(ss)
ttt = "".join(tt)
A = [sss, ttt]
A.sort()
if A[0] == sss:
print("Yes")
else:
print("No")
|
s417361080
|
p03436
|
u935558307
| 2,000
| 262,144
|
Wrong Answer
| 65
| 4,204
| 872
|
We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j). Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares. Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game. When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares. The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is `.`; if square (i, j) is initially painted by black, s_{i, j} is `#`.
|
from collections import deque
H,W = map(int,input().split())
maze = [list(input()) for i in range(H)]
blackCount = 0
for row in maze:
for tile in row:
if tile=="#":
blackCount += 1
dist = [[-1 for i in range(W)]for j in range(H)]
dist[0][0] = 1
que = [[0,0]]
patterns = [[1,0],[-1,0],[0,1],[0,-1]]
shortestStep = 0
while que:
print(que)
now = que.pop(0)
nowH = now[0]
nowW = now[1]
nowStep = dist[nowH][nowW]
for pattern in patterns:
nextH = nowH + pattern[0]
nextW = nowW + pattern[1]
if nextH == H-1 and nextW == W-1:
shortestStep = nowStep + 1
break
if 0 <= nextH <= H-1 and 0 <= nextW <= W-1 and maze[nextH][nextW]!="#" and dist[nextH][nextW]==-1:
dist[nextH][nextW] = nowStep + 1
que.append([nextH,nextW])
print(H*W-(shortestStep + blackCount))
|
s293779150
|
Accepted
| 27
| 3,316
| 843
|
from collections import deque
H,W = map(int,input().split())
maze = [list(input()) for _ in range(H)]
default = 0
for i in range(H):
for j in range(W):
if maze[i][j] == ".":
default += 1
que = deque()
que.append((0,0,1))
patterns = [(1,0),(-1,0),(0,1),(0,-1)]
count = 0
while que:
y,x,cost = que.popleft()
if y == H-1 and x == W-1:
count = cost
break
for a,b in patterns:
if 0<=y+a<H and 0<=x+b<W and maze[y+a][x+b]==".":
maze[y+a][x+b] = "o"
que.append((y+a,x+b,cost+1))
if count == 0:
print(-1)
else:
print(default-count)
|
s346495106
|
p02796
|
u934788990
| 2,000
| 1,048,576
|
Wrong Answer
| 477
| 20,104
| 235
|
In a factory, there are N robots placed on a number line. Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative. We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect. Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints. Find the maximum number of robots that we can keep.
|
n = int(input())
xl = []
for i in range(n):
x,l = map(int,input().split())
xl.append([x-l,x+l])
xl.sort()
ans = -10**10
count = 0
for j in range(n):
if xl[i][1] > ans:
count +=1
ans = xl[i][0]
print(count)
|
s333111822
|
Accepted
| 508
| 20,104
| 236
|
n = int(input())
xl = []
for i in range(n):
x,l = map(int,input().split())
xl.append([x+l,x-l])
xl.sort()
ans = -10**10
count = 0
for j in range(n):
if xl[j][1] >= ans:
count +=1
ans = xl[j][0]
print(count)
|
s552573520
|
p02615
|
u440161695
| 2,000
| 1,048,576
|
Wrong Answer
| 113
| 31,504
| 76
|
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?
|
N=int(input())
A=list(map(int,input().split()))
A.sort()
print(sum(A[:N-1]))
|
s759047794
|
Accepted
| 151
| 31,484
| 232
|
N=int(input())
A=list(map(int,input().split()))
A=sorted(A,reverse=True)
ans=0
for i in range((N//2)+1):
for j in range(2):
ans+=A[i]
if N%2==1:
ans-=A[i]
else:
ans=ans-(A[i]*2)
if N//2!=0:
ans-=A[0]
print(max(ans,0))
|
s906712433
|
p03486
|
u187205913
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 173
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
s = input()
t = input()
s = list(s)
t = list(t)
s_ = sorted(s)
t_ = sorted(t)
for i in range(len(s_)):
if not s_[i]<t[i]:
print('No')
exit()
print('Yes')
|
s031095991
|
Accepted
| 17
| 3,060
| 157
|
s = input()
t = input()
s = list(s)
t = list(t)
s_ = sorted(s)
t_ = sorted(t)
t_.reverse()
if ''.join(s_)<''.join(t_):
print('Yes')
else:
print('No')
|
s411366376
|
p00093
|
u150984829
| 1,000
| 131,072
|
Wrong Answer
| 30
| 5,752
| 171
|
西暦 a 年から b 年までの間にあるすべてのうるう年を出力するプログラムを作成してください。 うるう年の条件は、次のとおりとします。ただし、0 < a ≤ b < 3,000 とします。与えられた期間にうるう年がない場合には "NA"と出力してください。 * 西暦年が 4 で割り切れる年であること。 * ただし、100 で割り切れる年はうるう年としない。 * しかし、400 で割り切れる年はうるう年である。
|
for e in iter(input,'0 0'):
s,t=map(int,e.split())
u=[y for y in range(s+1,t)if y%4==0 and y%100!=0 or y%400==0]
if u:
for y in u:print(y)
else:print('NA')
print()
|
s851623828
|
Accepted
| 30
| 5,724
| 159
|
p=print
b=1
for e in iter(input,'0 0'):
b or p()
b=f=0;s,t=map(int,e.split())
for y in range(s,t+1):
if(y%4==0)*(y%100)or y%400==0:p(y);f=1
f or p('NA')
|
s795120359
|
p02831
|
u270467412
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 206
|
Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests.
|
A, B = map(int, input().split())
lcm = 1
cont = 1
while cont == 1:
if A % B == 0 or B % A == 0:
lcm = min(A, B)
break
if A < B:
B = B % A
elif A > B:
A = A % B
print(A * B / lcm)
|
s736569503
|
Accepted
| 17
| 3,060
| 237
|
A, B = map(int, input().split())
AA = A
BB = B
lcm = 1
cont = 1
while cont == 1:
if A % B == 0 or B % A == 0:
lcm = min(A, B)
break
if A < B:
B = B % A
elif A > B:
A = A % B
ans = int(AA * BB / lcm)
print(ans)
|
s736669889
|
p03853
|
u609061751
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 187
|
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
import sys
input = sys.stdin.readline
H, W = [int(x) for x in input().split()]
C = []
for _ in range(H):
C.append(input().rstrip())
for _ in range(2):
for i in C:
print(i)
|
s009257020
|
Accepted
| 18
| 3,060
| 187
|
import sys
input = sys.stdin.readline
H, W = [int(x) for x in input().split()]
C = []
for _ in range(H):
C.append(input().rstrip())
for i in C:
for _ in range(2):
print(i)
|
s112127357
|
p03555
|
u911153222
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 175
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
a = input()
b = input()
a_reversed = sorted(a, reverse=True)
b_reversed = sorted(b, reverse=True)
if a == b_reversed and b == a_reversed:
print('Yes')
else:
print('No')
|
s425741404
|
Accepted
| 17
| 2,940
| 184
|
a = input()
b = input()
a_reversed = ''.join(list(reversed(a)))
b_reversed = ''.join(list(reversed(b)))
if a == b_reversed and b == a_reversed:
print('YES')
else:
print('NO')
|
s308870252
|
p02270
|
u150984829
| 1,000
| 131,072
|
Time Limit Exceeded
| 9,990
| 5,588
| 266
|
In the first line, two integers $n$ and $k$ are given separated by a space character. In the following $n$ lines, $w_i$ are given respectively.
|
import sys
n,k=map(int,input().split())
w=list(map(int,sys.stdin.readlines()))
def check(P):
i=0
for _ in[0]*k:
s=0
while s+w[i]<=P:
s+=w[i];i+=1
if i==n: return n
return i
l,r=max(w),sum(w)
while l<r:
m=(l+r)//2
if check(m)>=n:r=m
else:l=m
print(r)
|
s831860999
|
Accepted
| 230
| 15,824
| 297
|
import sys
def s():
n,k=map(int,input().split())
w=list(map(int,sys.stdin.readlines()))
def f():
c=t=0
for j in w:
t+=j
if t>m:
t=j
c+=1
if c>=k:return 0
return 1
l,r=max(w),sum(w)
while l<r:
m=(l+r)//2
if f():r=m
else:l=m+1
print(r)
if'__main__'==__name__:s()
|
s485864595
|
p03377
|
u756988562
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 95
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
A,B,X = map(int,input().split(" "))
if X-A <B and A <X:
print("Yes")
else:
print("No")
|
s299702084
|
Accepted
| 17
| 2,940
| 168
|
A,B,X = map(int,input().split(" "))
judge = True
for i in range(B):
temp = A+i
if temp == X:
print("YES")
judge =False
if judge:
print("NO")
|
s522528243
|
p03478
|
u024890678
| 2,000
| 262,144
|
Wrong Answer
| 31
| 9,244
| 650
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
n,a,b = map(int, input().split())
ans_list = []
for i in range(1, n+1):
if i <= 9:
if a <= i <= b:
ans_list.append(i)
elif 10 <= i <= 99:
i10 = i // 10
i1 = i % 10
if a <= i10+i1 <= b:
ans_list.append(i)
elif 100 <= i <= 999:
i100 = i // 100
i10 = (i % 100) // 10
i1 = (i % 100) % 10
if a <= i100+i10+i1 <= b:
ans_list.append(i)
elif 1000 <= i <= 9999:
i1000 = i // 1000
i100 = (i % 1000) // 100
i10 = ((i % 1000) % 1000) // 10
i1 = ((i % 1000) % 1000) % 10
if a <= i1000 +i100 +i10 +i1 <= b:
ans_list.append(i)
else:
if a <= 1 <= b:
ans_list.append(i)
|
s470501000
|
Accepted
| 36
| 9,392
| 670
|
n,a,b = map(int, input().split())
ans_list = []
for i in range(1, n+1):
if i <= 9:
if a <= i <= b:
ans_list.append(i)
elif 10 <= i <= 99:
i10 = i // 10
i1 = i % 10
if a <= i10+i1 <= b:
ans_list.append(i)
elif 100 <= i <= 999:
i100 = i // 100
i10 = (i % 100) // 10
i1 = (i % 100) % 10
if a <= i100+i10+i1 <= b:
ans_list.append(i)
elif 1000 <= i <= 9999:
i1000 = i // 1000
i100 = (i % 1000) // 100
i10 = ((i % 1000) % 100) // 10
i1 = ((i % 1000) % 100) % 10
if a <= i1000 +i100 +i10 +i1 <= b:
ans_list.append(i)
else:
if a <= 1 <= b:
ans_list.append(i)
print(sum(ans_list))
|
s037160588
|
p03555
|
u446711904
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 66
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
a=list(input())
b=list(input())
print('NYOE S'[a==b.reverse()::2])
|
s304043698
|
Accepted
| 17
| 2,940
| 68
|
a=list(input())
b=list(input())
b.reverse()
print('NYOE S'[a==b::2])
|
s203995674
|
p03049
|
u507613674
| 2,000
| 1,048,576
|
Wrong Answer
| 45
| 4,336
| 849
|
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.
|
from collections import defaultdict
from string import ascii_uppercase
import sys, bisect, math
stdin = sys.stdin
read_int = lambda : list(map(int,stdin.readline().split()))
read_str = lambda : stdin.readline().rstrip()
N = read_int()[0]
S = [read_str() for _ in range(N)]
def solve():
_a = 0
b_ = 0
ba = 0
in_ab = 0
# search
for i in range(N):
if S[i][0] == 'B' and S[i][-1:] == 'A':
ba += 1
elif S[i][0] == 'B':
b_ += 1
elif S[i][-1:] == 'A':
_a += 1
for j in range(len(S[i]) - 1):
if S[i][j: j + 2] == 'AB':
in_ab += 1
ans = 0
ans += in_ab + ba - 1 + max(0, min(_a - 1, b_ - 1))
print(in_ab, ba, ans)
if _a > 0: ans += 1
if b_ > 0: ans += 1
return ans
if __name__ == "__main__":
print(solve())
|
s440723311
|
Accepted
| 45
| 4,336
| 928
|
from collections import defaultdict
from string import ascii_uppercase
import sys, bisect, math
stdin = sys.stdin
read_int = lambda : list(map(int,stdin.readline().split()))
read_str = lambda : stdin.readline().rstrip()
N = read_int()[0]
S = [read_str() for _ in range(N)]
def solve():
_a = 0
b_ = 0
ba = 0
in_ab = 0
# search
for i in range(N):
if S[i][0] == 'B' and S[i][-1:] == 'A':
ba += 1
elif S[i][0] == 'B':
b_ += 1
elif S[i][-1:] == 'A':
_a += 1
for j in range(len(S[i]) - 1):
if S[i][j: j + 2] == 'AB':
in_ab += 1
ans = 0
if ba == 0:
ans += in_ab + min(_a, b_)
return ans
ans += in_ab + max(ba - 1,0) + max(0, min(_a - 1, b_ - 1))
# print(in_ab, ba, ans)
if _a > 0: ans += 1
if b_ > 0: ans += 1
return ans
if __name__ == "__main__":
print(solve())
|
s117891509
|
p03545
|
u391475811
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,192
| 760
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
s=list(input().strip())
if int(s[0])+int(s[1])+int(s[2])+int(s[3])==7:
print(s[0]+"+"+s[1]+"+"+s[2]+"+"+s[3])
elif int(s[0])-int(s[1])-int(s[2])-int(s[3])==7:
print(s[0]+"-"+s[1]+"-"+s[2]+"-"+s[3])
elif int(s[0])+int(s[1])-int(s[2])-int(s[3])==7:
print(s[0]+"+"+s[1]+"-"+s[2]+"-"+s[3])
elif int(s[0])+int(s[1])+int(s[2])-int(s[3])==7:
print(s[0]+"+"+s[1]+"+"+s[2]+"-"+s[3])
elif int(s[0])-int(s[1])+int(s[2])-int(s[3])==7:
print(s[0]+"-"+s[1]+"+"+s[2]+"-"+s[3])
elif int(s[0])-int(s[1])+int(s[2])+int(s[3])==7:
print(s[0]+"-"+s[1]+"+"+s[2]+"+"+s[3])
elif int(s[0])+int(s[1])-int(s[2])+int(s[3])==7:
print(s[0]+"+"+s[1]+"-"+s[2]+"+"+s[3])
elif int(s[0])-int(s[1])-int(s[2])+int(s[3])==7:
print(s[0]+"-"+s[1]+"-"+s[2]+"+"+s[3])
else:
print(-1)
|
s649851197
|
Accepted
| 18
| 3,192
| 800
|
s=list(input().strip())
if int(s[0])+int(s[1])+int(s[2])+int(s[3])==7:
print(s[0]+"+"+s[1]+"+"+s[2]+"+"+s[3]+"=7")
elif int(s[0])-int(s[1])-int(s[2])-int(s[3])==7:
print(s[0]+"-"+s[1]+"-"+s[2]+"-"+s[3]+"=7")
elif int(s[0])+int(s[1])-int(s[2])-int(s[3])==7:
print(s[0]+"+"+s[1]+"-"+s[2]+"-"+s[3]+"=7")
elif int(s[0])+int(s[1])+int(s[2])-int(s[3])==7:
print(s[0]+"+"+s[1]+"+"+s[2]+"-"+s[3]+"=7")
elif int(s[0])-int(s[1])+int(s[2])-int(s[3])==7:
print(s[0]+"-"+s[1]+"+"+s[2]+"-"+s[3]+"=7")
elif int(s[0])-int(s[1])+int(s[2])+int(s[3])==7:
print(s[0]+"-"+s[1]+"+"+s[2]+"+"+s[3]+"=7")
elif int(s[0])+int(s[1])-int(s[2])+int(s[3])==7:
print(s[0]+"+"+s[1]+"-"+s[2]+"+"+s[3]+"=7")
elif int(s[0])-int(s[1])-int(s[2])+int(s[3])==7:
print(s[0]+"-"+s[1]+"-"+s[2]+"+"+s[3]+"=7")
else:
print(-1)
|
s478047325
|
p03351
|
u940102677
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 130
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a, b, c, d = map(int, input().split())
if abs(a-c) <= d or (abs(a-b) <= d and abs(a-c) <=d):
print("Yes")
else:
print("No")
|
s324323639
|
Accepted
| 17
| 3,064
| 127
|
a, b, c, d = map(int, input().split())
if abs(a-c)<= d or (abs(a-b)<= d and abs(b-c)<=d):
print("Yes")
else:
print("No")
|
s627204380
|
p02606
|
u654240084
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,020
| 47
|
How many multiples of d are there among the integers between L and R (inclusive)?
|
l,r,d = map(int,input().split())
print(r/d-l/d)
|
s021310478
|
Accepted
| 27
| 9,140
| 54
|
l,r,d = map(int,input().split())
print(r//d-(l-1)//d)
|
s775855604
|
p03470
|
u597017430
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 160
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
n = int(input())
d = [0]*n
for i in range(n):
d[i] = int(input())
d.sort
count = 1
for i in range(n-1):
if d[i]>d[i+1]:
count += 1
print(count)
print(d)
|
s725387305
|
Accepted
| 17
| 3,060
| 167
|
n = int(input())
d = [0]*n
for i in range(n):
d[i] = int(input())
d.sort(reverse = True)
count = 1
for i in range(n-1):
if d[i]>d[i+1]:
count += 1
print(count)
|
s503984707
|
p02262
|
u637322311
| 6,000
| 131,072
|
Wrong Answer
| 20
| 5,612
| 709
|
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 read_n_lows_input(n):
Alist=[int(input()) for i in range(n)]
return Alist
def print_list(A):
print(*A, sep=" ")
def print_list_multi_low(A):
for i in A:
print(i)
def insertion_sort(A, n, g, cnt):
for i in range(g-1, n):
v = A[i]
j = i - g
while j >= 0 and A[j] > v:
A[j+g] = A[j]
j -= g
cnt += 1
A[j+g] = v
return A, cnt
def shell_sort(A, n):
cnt = 0
G = [8, 4, 2, 1]
m = len(G)
print(m)
print_list(G)
for i in range(m):
A, cnt = insertion_sort(A, n, G[i], cnt)
print(cnt)
print_list_multi_low(A)
n=int(input())
A = read_n_lows_input(n)
shell_sort(A, n)
|
s402292914
|
Accepted
| 17,750
| 45,520
| 804
|
def read_n_lows_input(n):
Alist=[int(input()) for i in range(n)]
return Alist
def print_list(A):
print(*A, sep=" ")
def print_list_multi_low(A):
for i in A:
print(i)
def insertion_sort(A, n, g, cnt):
for i in range(g-1, n):
v = A[i]
j = i - g
while j >= 0 and A[j] > v:
A[j+g] = A[j]
j -= g
cnt += 1
A[j+g] = v
return A, cnt
def shell_sort(A, n):
cnt = 0
G0 = [797161, 265720, 88573, 29524, 9841, 3280, 1093, 364, 121, 40, 13, 4, 1]
G = [i for i in G0 if i <= n]
m = len(G)
print(m)
print_list(G)
for i in range(m):
A, cnt = insertion_sort(A, n, G[i], cnt)
print(cnt)
print_list_multi_low(A)
n=int(input())
A = read_n_lows_input(n)
shell_sort(A, n)
|
s904262777
|
p03592
|
u418199806
| 2,000
| 262,144
|
Wrong Answer
| 1,050
| 10,036
| 435
|
We have a grid with N rows and M columns of squares. Initially, all the squares are white. There is a button attached to each row and each column. When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa. When a button attached to a column is pressed, the colors of all the squares in that column are inverted. Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.
|
N, M, K = map(int, input().split())
if K == 0 :
print('Yes')
exit()
for j in range(1, M):
print("----", j)
cnt = 0
for i in range(N * j, N*M, M-j):
if cnt == 0:
print(i)
if K == i:
print('Yes')
exit()
else :
print(i - j)
if K == i - j:
print('Yes')
exit()
cnt += 1
print('No')
|
s236155388
|
Accepted
| 298
| 2,940
| 189
|
X, Y, K = map(int, input().split())
for Xi in range(0, X+1):
for Yi in range(0, Y+1):
if Yi * (X-Xi) + Xi * (Y-Yi) == K:
print("Yes")
exit()
print("No")
|
s696988121
|
p00067
|
u756426560
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,708
| 511
|
地勢を示す縦 12, 横 12 のマスからなる平面図があります。おのおののマスは白か黒に塗られています。白は海を、黒は陸地を表します。二つの黒いマスが上下、あるいは左右に接しているとき、これらは地続きであるといいます。この平面図では、黒いマス一つのみ、あるいは地続きの黒いマスが作る領域を「島」といいます。例えば下図には、5 つの島があります。 ■■■■□□□□■■■■ ■■■□□□□□■■■■ ■■□□□□□□■■■■ ■□□□□□□□■■■■ □□□■□□□■□□□□ □□□□□□■■■□□□ □□□□□■■■■■□□ ■□□□■■■■■■■□ ■■□□□■■■■■□□ ■■■□□□■■■□□□ ■■■■□□□■□□□□ □□□□□□□□□□□□ マスのデータを読み込んで、島の数を出力するプログラムを作成してください。
|
def remove_linked(x, y, DATA):
DATA[x][y]=0# clear
move = [[1, 0], [0, 1], [-1, 0], [0, -1]]#left, right, down, up
for i, j in move:
nx,ny=x+i,y+j#next x and y
if -1<nx<12 and -1<ny<12 and DATA[nx][ny]:
DATA=remove_linked(nx, ny, DATA)
return DATA
DATA = [[int(x) for x in list(input())] for _ in range(12)]
count = 0
for x in range(12):
for y in range(12):
if DATA[x][y]:
count += 1
DATA = remove_linked(x, y, DATA)
print (count)
|
s860695270
|
Accepted
| 30
| 7,732
| 590
|
def remove_linked(x, y, DATA):
DATA[x][y]=0# clear
move = [[1, 0], [0, 1], [-1, 0], [0, -1]]#left, right, down, up
for i, j in move:
nx,ny=x+i,y+j#next x and y
if -1<nx<12 and -1<ny<12 and DATA[nx][ny]:
DATA=remove_linked(nx, ny, DATA)
return DATA
while 1:
DATA = [[int(x) for x in list(input())] for _ in range(12)]
count = 0
for x in range(12):
for y in range(12):
if DATA[x][y]:
count += 1
DATA = remove_linked(x, y, DATA)
print (count)
try:input()
except:break
|
s943967693
|
p02409
|
u340503368
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,624
| 430
|
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.
|
rooms = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]
n = int(input())
for l in range(n):
a,b,c,d = map(int, input().split())
rooms[a-1][b-1][c-1] += d
for i in range(4):
for j in range(3):
for k in range(10):
print(rooms[i][j][k], end = "")
if k != 9:
print(" ", end="")
print()
print("####################")
|
s810258102
|
Accepted
| 20
| 5,620
| 371
|
rooms = [[[0 for i in range(10)] for j in range(3)] for k in range(4)]
n = int(input())
for l in range(n):
a,b,c,d = map(int, input().split())
rooms[a-1][b-1][c-1] += d
for i in range(4):
for j in range(3):
for k in range(10):
print(" " + str(rooms[i][j][k]), end = "")
print()
if i != 3:
print("####################")
|
s139636470
|
p03485
|
u361826811
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 244
|
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.
|
import sys
readline = sys.stdin.readline
readlines = sys.stdin.readlines
sys.setrecursionlimit(10 ** 7)
n, a = map(int, readline().rstrip().split())
print((n+a+1)/2 if (n+a)%2==1 else (n+a)/2)
|
s095104364
|
Accepted
| 17
| 3,060
| 261
|
import sys
import itertools
# import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
A, B = map(int, readline().split())
print((A+B+1)//2)
|
s784883225
|
p03550
|
u242679311
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 156
|
We have a deck consisting of N cards. Each card has an integer written on it. The integer on the i-th card from the top is a_i. Two people X and Y will play a game using this deck. Initially, X has a card with Z written on it in his hand, and Y has a card with W written on it in his hand. Then, starting from X, they will alternately perform the following action: * Draw some number of cards from the top of the deck. Then, discard the card in his hand and keep the last drawn card instead. Here, at least one card must be drawn. The game ends when there is no more card in the deck. The score of the game is the absolute difference of the integers written on the cards in the two players' hand. X will play the game so that the score will be maximized, and Y will play the game so that the score will be minimized. What will be the score of the game?
|
n,z,w = map(int,input().split())
a = list(map(int,input().split()))
if n == 1:
print(abs(a[0]-w))
else:
print(min(abs(a[n-1]-w),abs(a[n-1]-a[n-2])))
|
s761610788
|
Accepted
| 19
| 3,188
| 156
|
n,z,w = map(int,input().split())
a = list(map(int,input().split()))
if n == 1:
print(abs(a[0]-w))
else:
print(max(abs(a[n-1]-w),abs(a[n-1]-a[n-2])))
|
s988849377
|
p03370
|
u404556828
| 2,000
| 262,144
|
Wrong Answer
| 278
| 3,060
| 213
|
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 = [int(x) for x in input().split()]
list01=[]
for i in range(N):
j = int(input())
list01.append(j)
i = 0
mass = sum(list01)
while mass < X:
i += 1
mass = sum(list01) + min(list01) * i
print(i -1)
|
s662476278
|
Accepted
| 17
| 3,060
| 320
|
N, X = [int(x) for x in input().split()]
list01 = []
for i in range(N):
i = int(input())
list01.append(i)
min_01 = 0
for i in range(len(list01)):
if min_01 == 0:
min_01 = list01[i]
elif list01[i] < min_01:
min_01 = list01[i]
for i in range(len(list01)):
X -= list01[i]
print(X // min_01 + N)
|
s835117058
|
p02397
|
u105694406
| 1,000
| 131,072
|
Wrong Answer
| 50
| 7,604
| 119
|
Write a program which reads two integers x and y, and prints them in ascending order.
|
while 1:
a, b = map(int, input().split())
if b < a:
print(b, a)
else:
print(a, b)
if a == 0 and b == 0:
break
|
s505671897
|
Accepted
| 50
| 7,580
| 124
|
while 1:
a, b = map(int, input().split())
if a == 0 and b == 0:
break
if b < a:
print(b, a)
else:
print(a, b)
|
s663243680
|
p03024
|
u223904637
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 138
|
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s=list(input())
o=0
z=len(s)
for i in range(len(s)):
if s[i]=='o':
o+=1
if o+(15-z)>=8:
print('Yes')
else:
print('No')
|
s331132676
|
Accepted
| 17
| 2,940
| 139
|
s=list(input())
o=0
z=len(s)
for i in range(len(s)):
if s[i]=='o':
o+=1
if o+(15-z)>=8:
print('YES')
else:
print('NO')
|
s975617562
|
p03080
|
u728047357
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 254
|
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.
|
def main():
n = int(input()) // 2
numR = 0
for c in input():
if c != 'B':
continue
numR += 1
if numR > n:
print("Yes")
return
print("No")
if __name__ == '__main__':
main()
|
s076995579
|
Accepted
| 17
| 2,940
| 254
|
def main():
n = int(input()) // 2
numR = 0
for c in input():
if c == 'B':
continue
numR += 1
if numR > n:
print("Yes")
return
print("No")
if __name__ == '__main__':
main()
|
s539203801
|
p03963
|
u089230684
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 97
|
There are N balls placed in a row. AtCoDeer the deer is painting each of these in one of the K colors of his paint cans. For aesthetic reasons, any two adjacent balls must be painted in different colors. Find the number of the possible ways to paint the balls.
|
try:
n = int(input())
k = int(input())
print(k * (k - 1) ** (n - 1))
except:
pass
|
s724576572
|
Accepted
| 17
| 2,940
| 76
|
import math
k,m=list(map(int,input().split()))
res=m*pow(m-1,k-1)
print(res)
|
s589437628
|
p03457
|
u838559130
| 2,000
| 262,144
|
Wrong Answer
| 397
| 3,064
| 315
|
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())
flg = True
t, x, y = 0, 0, 0
for i in range(n):
nt, nx, ny = map(int, input().split())
dist = abs(x - nx) + abs(y - ny)
dt = nt - t
if dist > dt or (dist - dt) % 2 == 1:
print("NO")
flg = False
break
t, x, y = nt, nx, ny
if flg == True:
print("YES")
|
s568182588
|
Accepted
| 374
| 3,064
| 315
|
n = int(input())
flg = True
t, x, y = 0, 0, 0
for i in range(n):
nt, nx, ny = map(int, input().split())
dist = abs(x - nx) + abs(y - ny)
dt = nt - t
if dist > dt or (dist - dt) % 2 == 1:
print("No")
flg = False
break
t, x, y = nt, nx, ny
if flg == True:
print("Yes")
|
s214548558
|
p02612
|
u953020348
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,140
| 30
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
x=input()
print(int(x[-3:]))
|
s354882095
|
Accepted
| 27
| 9,144
| 62
|
x=int(input())
x=x%1000
x=1000-x
if x==1000:
x=0
print(x)
|
s765593335
|
p02259
|
u581154076
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 411
|
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode.
|
def main():
n = int(input())
a = []
for _ in range(n):
a.append(int(input()))
count = 0
flag = True
while flag:
flag = False
for j in range(n-1, 0, -1):
if a[j] < a[j-1]:
a[j], a[j-1] = a[j-1], a[j]
flag = True
count += 1
print(a)
print(count)
if __name__ == '__main__':
main()
|
s516853714
|
Accepted
| 20
| 5,608
| 388
|
def main():
n = int(input())
a = [int(i) for i in input().split()]
count = 0
flag = True
while flag:
flag = False
for j in range(n-1, 0, -1):
if a[j] < a[j-1]:
a[j], a[j-1] = a[j-1], a[j]
flag = True
count += 1
print(*a)
print(count)
if __name__ == '__main__':
main()
|
s660905395
|
p04029
|
u560301743
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 39
|
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N = int(input())
print(N * (N + 1) / 2)
|
s244993965
|
Accepted
| 17
| 2,940
| 44
|
N = int(input())
print(int(N * (N + 1) / 2))
|
s709762444
|
p03478
|
u077127204
| 2,000
| 262,144
|
Wrong Answer
| 27
| 3,296
| 104
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
N, A, B = map(int, input().split())
sum([n for n in range(1, N + 1) if A <= sum(map(int, str(n))) <= B])
|
s923109558
|
Accepted
| 29
| 3,296
| 111
|
N, A, B = map(int, input().split())
print(sum([n for n in range(1, N + 1) if A <= sum(map(int, str(n))) <= B]))
|
s039071249
|
p03476
|
u608088992
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 4,504
| 516
|
We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i ≤ x ≤ r_i.
|
import math
Primes = {2}
Sim = [0 for i in range(100002)]
for i in range(3, 100001):
if i % 2 == 0:
Sim[i] = Sim[i-1]
else:
for p in Primes:
if i % p == 0:
Sim[i] = Sim[i-1]
break
else:
Primes |= {i}
if (i+1)//2 in Primes:
Sim[i] = Sim[i-1] + 1
else:
Sim[i] = Sim[i-1]
Q = int(input())
for i in range(Q):
l, r = map(int, input().split())
print(Sim[r] - Sim[l-1])
|
s442117689
|
Accepted
| 254
| 5,432
| 669
|
import sys
def solve():
input = sys.stdin.readline
isPrime = [True] * (1 + 10**5)
isPrime[0] = isPrime[1] = False
Like = [0] * (1 + 10 ** 5)
for i in range(2, 1 + 10 ** 5):
if isPrime[i]:
k = 2 * i
while k <= 10 ** 5:
isPrime[k] = False
k += i
if i % 2 == 1 and isPrime[(i + 1) // 2]: Like[i] = Like[i-1] + 1
else: Like[i] = Like[i-1]
else: Like[i] = Like[i-1]
Q = int(input())
for _ in range(Q):
l, r = map(int, input().split())
print(Like[r] - Like[l-1])
return 0
if __name__ == "__main__":
solve()
|
s236461176
|
p02647
|
u894521144
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 53,604
| 673
|
We have N bulbs arranged on a number line, numbered 1 to N from left to right. Bulb i is at coordinate i. Each bulb has a non-negative integer parameter called intensity. When there is a bulb of intensity d at coordinate x, the bulb illuminates the segment from coordinate x-d-0.5 to x+d+0.5. Initially, the intensity of Bulb i is A_i. We will now do the following operation K times in a row: * For each integer i between 1 and N (inclusive), let B_i be the number of bulbs illuminating coordinate i. Then, change the intensity of each bulb i to B_i. Find the intensity of each bulb after the K operations.
|
import numpy as np
from math import log
def main(N, K, A):
A = np.array(A, dtype = int)
if K > log(N):
A = [N for _ in range(N)]
else:
for k in range(K):
B = np.zeros(N, dtype = int)
for x in range(N):
if A[x] >= N - 1:
B += 1
else:
minimum = max(0, x - A[x])
maximum = min(N, x + A[x] + 1)
B[minimum:maximum] += 1
A = B
print(' '.join([str(i) for i in A]))
if __name__ == '__main__':
N, K = list(map(int, input().split()))
A = list(map(int, input().split()))
main(N, K, A)
|
s717349019
|
Accepted
| 910
| 137,332
| 1,008
|
import numpy as np
from numba import njit
def main(N, K, A):
@njit(cache = True)
def rep(N, A):
B = np.zeros(N + 1, dtype=np.int64)
for x in range(N):
minimum = max(0, x - A[x])
maximum = min(N, x + A[x] + 1)
B[minimum] += 1
B[maximum] -= 1
return np.cumsum(B)[:-1]
A = np.array(A, dtype=np.int64)
for k in range(K):
B = rep(N, A)
if np.array_equal(A, B):
break
else:
A = B
return A
if __name__ == '__main__':
N, K = list(map(int, input().split()))
A = list(map(int, input().split()))
print(' '.join([str(i) for i in main(N, K, A)]))
|
s347183714
|
p04029
|
u251017754
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 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*(N+1)/2)
|
s346251551
|
Accepted
| 17
| 2,940
| 34
|
N = int(input())
print(N*(N+1)//2)
|
s191454989
|
p03401
|
u858523893
| 2,000
| 262,144
|
Wrong Answer
| 209
| 14,172
| 549
|
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()))
# calc dif
a_zero = [0,] + a + [0,]
a_d = []
for i in range(1, N + 2) :
a_d.append(abs(a_zero[i] - a_zero[i - 1]))
a_d_sum = sum(a_d)
for i in range(N) :
if i == N - 1 :
if a[i] >= a[i - 1] :
print(a_d_sum)
else :
print(a_d_sum - a_d[i] * 2)
else :
if abs(a[i]) <= abs(a[i + 1]) and ((a[i] >= 0 and a[i + 1] >= 0) or (a[i] < 0 and a[i + 1] < 0)):
print(a_d_sum)
else :
print(a_d_sum - a_d[i] * 2)
|
s465960374
|
Accepted
| 270
| 14,164
| 620
|
N = int(input())
a = [int(x) for x in input().split()]
d = []
prev = 0
for i in range(N + 1) :
if i == 0 : d.append(abs(a[i]))
elif i == N : d.append(abs(a[i - 1]))
else : d.append(abs(prev - a[i]))
if i != N : prev = a[i]
totsum = sum(d)
for i in range(N) :
# for first
if i == 0 :
x, y, z = 0, a[i], a[i + 1]
elif i == N - 1 :
x, y, z = a[i - 1], a[i], 0
else :
x, y, z = a[i - 1], a[i], a[i + 1]
if x <= y <= z or x >= y >= z :
print(totsum)
else :
print(totsum - abs(d[i] + d[i + 1]) + abs(d[i] - d[i + 1]))
|
s245040026
|
p03555
|
u338225045
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 184
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
S1 = input()
S2 = input()
isMatch = True
for i in range( len(S1) ):
if S1[i] != S2[-i]:
isMatch = False
break
if isMatch: print( 'YES' )
else: print( 'NO' )
|
s944140727
|
Accepted
| 20
| 2,940
| 186
|
S1 = input()
S2 = input()
isMatch = True
for i in range( len(S1) ):
if S1[i] != S2[-i-1]:
isMatch = False
break
if isMatch: print( 'YES' )
else: print( 'NO' )
|
s489647861
|
p03377
|
u059684735
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 146
|
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:
print('No')
elif a == x:
print('Yes')
elif a+b >= x:
print('Yes')
else:
print('No')
|
s058480011
|
Accepted
| 17
| 2,940
| 95
|
a, b, x = map(int, input().split())
if a <= x <= a + b:
print('YES')
else:
print('NO')
|
s771152288
|
p03861
|
u946386741
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,316
| 108
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a, b, x = map(int, input().split())
result = 0
if a is 0:
result = 1
result += (b-a)//x
print(result)
|
s206497645
|
Accepted
| 18
| 2,940
| 131
|
a, b, x = map(int, input().split())
def f(n):
if n >= 0:
return n//x+1
else:
return 0
print(f(b)-f(a-1))
|
s765718369
|
p03448
|
u663014688
| 2,000
| 262,144
|
Wrong Answer
| 47
| 3,060
| 231
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
ans = 0
for i in range(A):
for j in range(B):
for k in range(C):
if 500 * i + 100 * j + 50 * k == X:
ans += 1
print(ans)
|
s536932283
|
Accepted
| 50
| 3,060
| 239
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
ans = 0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
if (500 * i + 100 * j + 50 * k) == X:
ans += 1
print(ans)
|
s795410675
|
p03729
|
u124498235
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 33
|
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a = str(input())
print("ABC" + a)
|
s422619631
|
Accepted
| 17
| 2,940
| 105
|
a, b, c = map(str,input().split())
if a[-1] == b[0] and b[-1] == c[0]:
print ("YES")
else:
print ("NO")
|
s424344227
|
p03139
|
u994988729
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 72
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
n,a,b=map(int,input().split())
and_=abs(a-b)
or_=a+b-n
print(and_, or_)
|
s575065876
|
Accepted
| 17
| 2,940
| 79
|
n,a,b=map(int,input().split())
and_=min(a,b)
or_=max(0, a+b-n)
print(and_, or_)
|
s662605662
|
p03545
|
u405660020
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 207
|
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
s=input()
op=['+', '-']
for op1 in op:
for op2 in op:
for op3 in op:
tmp=s[0]+op1+s[1]+op2+s[2]+op3+s[3]
if eval(tmp)==7:
print(tmp)
break
|
s037066864
|
Accepted
| 18
| 2,940
| 197
|
s=input()
op=['+', '-']
for op1 in op:
for op2 in op:
for op3 in op:
tmp=s[0]+op1+s[1]+op2+s[2]+op3+s[3]
if eval(tmp)==7:
ans=tmp
print(ans+'=7')
|
s381765983
|
p02605
|
u729133443
| 3,000
| 1,048,576
|
Wrong Answer
| 903
| 71,508
| 330
|
M-kun is a brilliant air traffic controller. On the display of his radar, there are N airplanes numbered 1, 2, ..., N, all flying at the same altitude. Each of the airplanes flies at a constant speed of 0.1 per second in a constant direction. The current coordinates of the airplane numbered i are (X_i, Y_i), and the direction of the airplane is as follows: * if U_i is `U`, it flies in the positive y direction; * if U_i is `R`, it flies in the positive x direction; * if U_i is `D`, it flies in the negative y direction; * if U_i is `L`, it flies in the negative x direction. To help M-kun in his work, determine whether there is a pair of airplanes that will collide with each other if they keep flying as they are now. If there is such a pair, find the number of seconds after which the first collision will happen. We assume that the airplanes are negligibly small so that two airplanes only collide when they reach the same coordinates simultaneously.
|
d=[[],[],[],[]]
for t in[*open(0)][1:]:*z,u=t.split();d[ord(u)%5]+=[*map(int,z)],
I=N=9e9
P,M='x+y,','x-y,'
for*A,S in(3,0,'x,y,'),(1,2,'y,x,'),(0,2,P+M),(1,3,P+M),(1,0,M+P),(3,2,M+P):
n=p=-I
for x,y,r in eval('sorted((%sQ)for Q,P in enumerate(A)for x,y in d[P])'%S):
if r:n,p=x,y
elif x==n:N=min(N,y-p)
print(N%I*5or'SAFE')
|
s845274450
|
Accepted
| 2,812
| 72,464
| 318
|
I,*U=N,*D=9**9,
P,*L='x+y,',
M,*R='x-y,',
for t in[*open(0)][1:]:*z,u=t.split();exec(u+'+=[*map(int,z)],')
for*A,S in(D,U,'x,y,'),(L,R,'y,x,'),(U,R,P+M),(L,D,P+M),(L,U,M+P),(D,R,M+P):
for x,y,r in eval('sorted((%sQ)for Q,P in enumerate(A)for x,y in P)'%S):
if r:S,p=x,y
elif x==S:N=min(N,y-p)
print(N%I*5or'SAFE')
|
s796978614
|
p03474
|
u981864683
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,008
| 197
|
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
A,B = map(int,input().split())
S = input()
count = "No"
if S[:A+1].isnumeric():
if S[A+1] == "-":
if len(S[A+2:]) == B and S[A+2:].isnumeric():
count = "Yes"
print(count)
|
s145046340
|
Accepted
| 27
| 9,104
| 193
|
A,B = map(int,input().split())
S = input()
count = "No"
if S[:A].isnumeric():
if S[A] == "-":
if len(S[A+1:]) == B and S[A+1:].isnumeric():
count = "Yes"
print(count)
|
s580583546
|
p03024
|
u688126754
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 139
|
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
S = input()
lose = 0
for i in range(len(S)):
if S[i] == "x":
lose += 1
if lose >= 8:
print("No")
else:
print("Yes")
|
s252601722
|
Accepted
| 17
| 2,940
| 139
|
S = input()
lose = 0
for i in range(len(S)):
if S[i] == "x":
lose += 1
if lose >= 8:
print("NO")
else:
print("YES")
|
s589362962
|
p03827
|
u821251381
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 105
|
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
S=input()
max_v =0
v =0
for i in S:
if i=="I":
v+=0
if v>max_v:
max_v=v
else:
v-=0
|
s399897773
|
Accepted
| 17
| 2,940
| 127
|
N=input()
S=input()
max_v =0
v =0
for i in S:
if i=="I":
v+=1
if v>max_v:
max_v=v
else:
v-=1
print(max_v)
|
s298005393
|
p03369
|
u729133443
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 41
|
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+list(input()).count('○')*100)
|
s418003240
|
Accepted
| 27
| 8,932
| 33
|
print(700+input().count('o')*100)
|
s826563505
|
p02613
|
u807241701
| 2,000
| 1,048,576
|
Wrong Answer
| 177
| 16,188
| 368
|
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())
X = []
for i in range(N):
X.append(input())
AC = 0
WA = 0
TLE = 0
RE = 0
for s in X:
print(s)
if s == 'AC':
AC += 1
elif s == 'WA':
WA += 1
elif s == 'TLE':
TLE += 1
elif s == 'RE':
RE += 1
print('AC x ' + str(AC))
print('WA x ' + str(WA))
print('TLE x ' + str(TLE))
print('RE x ' + str(RE))
|
s682519161
|
Accepted
| 153
| 16,236
| 355
|
N = int(input())
X = []
for i in range(N):
X.append(input())
AC = 0
WA = 0
TLE = 0
RE = 0
for s in X:
if s == 'AC':
AC += 1
elif s == 'WA':
WA += 1
elif s == 'TLE':
TLE += 1
elif s == 'RE':
RE += 1
print('AC x ' + str(AC))
print('WA x ' + str(WA))
print('TLE x ' + str(TLE))
print('RE x ' + str(RE))
|
s299817668
|
p03469
|
u266171694
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 34
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
s = input()
print('2017' + s[4::])
|
s365918576
|
Accepted
| 17
| 2,940
| 33
|
s = input()
print('2018' + s[4:])
|
s615358116
|
p03478
|
u870297120
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 98
|
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
N,A,B = map(int, input().split())
x = sum(i for i in range(1, N) if A <= i//10+i%10 <= B)
print(x)
|
s903033329
|
Accepted
| 32
| 2,940
| 117
|
N,A,B = map(int, input().split())
x = sum(i for i in range(1, N+1) if A <= sum(map(int, list(str(i)))) <= B)
print(x)
|
s181532470
|
p03795
|
u853900545
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 38
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n=int(input())
print(800*n-50*(n//15))
|
s293611861
|
Accepted
| 17
| 2,940
| 40
|
n=int(input())
print(800*n-200*(n//15))
|
s072942873
|
p03740
|
u931938233
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,064
| 72
|
Alice and Brown loves games. Today, they will play the following game. In this game, there are two piles initially consisting of X and Y stones, respectively. Alice and Bob alternately perform the following operation, starting from Alice: * Take 2i stones from one of the piles. Then, throw away i of them, and put the remaining i in the other pile. Here, the integer i (1≤i) can be freely chosen as long as there is a sufficient number of stones in the pile. The player who becomes unable to perform the operation, loses the game. Given X and Y, determine the winner of the game, assuming that both players play optimally.
|
X,Y=map(int,input().split())
print('Alice' if abs(X-Y) < 2 else 'Brown')
|
s424355613
|
Accepted
| 28
| 9,008
| 72
|
X,Y=map(int,input().split())
print('Alice' if abs(X-Y) > 1 else 'Brown')
|
s786700688
|
p03730
|
u637451088
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 168
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
import sys
a, b, c = map(int, input().split())
flag = False
for i in range(a, b*a+1, a):
if i % b == c:
flag = True
print("Yes") if flag else print("No")
|
s978597977
|
Accepted
| 18
| 2,940
| 168
|
import sys
a, b, c = map(int, input().split())
flag = False
for i in range(a, b*a+1, a):
if i % b == c:
flag = True
print("YES") if flag else print("NO")
|
s231862353
|
p03836
|
u060455496
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 521
|
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.
|
def main():
sx,sy,tx,ty = map(int,input().split())
path =''
ly,lx = abs(ty-sy),abs(tx-sx)
path += 'U' * ly
path += 'R' * lx
path += 'D' * ly
path += 'L' * lx
path += 'L' + 'U' * (ly+1)
path += 'R' * (lx+1) + 'D'
path += 'R' +'U'*(ly+1)
path += 'L'*(lx+1) +'U'
print(path)
if __name__ == '__main__':
main()
|
s290354220
|
Accepted
| 18
| 3,064
| 521
|
def main():
sx,sy,tx,ty = map(int,input().split())
path =''
ly,lx = abs(ty-sy),abs(tx-sx)
path += 'U' * ly
path += 'R' * lx
path += 'D' * ly
path += 'L' * lx
path += 'L' + 'U' * (ly+1)
path += 'R' * (lx+1) + 'D'
path += 'R' +'D'*(ly+1)
path += 'L'*(lx+1) +'U'
print(path)
if __name__ == '__main__':
main()
|
s815683714
|
p03494
|
u081714930
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,205
| 9,040
| 172
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n=int(input())
n_length=list(map(int,input().split()))
counter=0
while all(i%2==0 for i in n_length):
n_list=[i/2 for i in n_length]
counter=+1
print(counter)
|
s196426985
|
Accepted
| 29
| 9,176
| 170
|
n = int(input())
a = list(map(int, input().split()))
cnt = 0
while True:
if [i for i in a if i % 2 == 1]:
break
a = [i/2 for i in a]
cnt += 1
print(cnt)
|
s142818488
|
p03555
|
u751863169
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 324
|
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
#!/usr/bin/env python
def main():
matrix = []
for _ in range(2):
matrix.append(list(input()))
if matrix[0][0] == matrix[1][2] \
and matrix[0][1] == matrix[1][1] \
and matrix[0][2] == matrix[1][0]:
print("Yes")
else:
print("No")
if __name__ == '__main__':
main()
|
s496231228
|
Accepted
| 18
| 3,060
| 302
|
def main():
matrix = []
for _ in range(2):
matrix.append(list(input()))
if matrix[0][0] == matrix[1][2] \
and matrix[0][1] == matrix[1][1] \
and matrix[0][2] == matrix[1][0]:
print("YES")
else:
print("NO")
if __name__ == '__main__':
main()
|
s383531925
|
p03598
|
u385244248
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 76
|
There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots.
|
N,K,*x =map(int,open(0).read().split())
print(sum([min(i,K-i) for i in x]))
|
s828470417
|
Accepted
| 17
| 2,940
| 78
|
N,K,*x =map(int,open(0).read().split())
print(sum(2*[min(i,K-i) for i in x]))
|
s521323737
|
p03456
|
u452198080
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 103
|
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=input().split()
d=int(a[0]+a[1])
x=int(math.sqrt(d))
x=x*x
print("YES" if x==d else "No")
|
s343771493
|
Accepted
| 17
| 2,940
| 103
|
import math
a=input().split()
d=int(a[0]+a[1])
x=int(math.sqrt(d))
x=x*x
print("Yes" if x==d else "No")
|
s416245615
|
p03997
|
u616188005
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 67
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2)
|
s986576134
|
Accepted
| 17
| 2,940
| 82
|
a = int(input())
b = int(input())
h = int(input())
ans = (a+b)*h/2
print(int(ans))
|
s901684161
|
p02833
|
u732870425
| 2,000
| 1,048,576
|
Time Limit Exceeded
| 2,104
| 2,940
| 134
|
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
N = int(input())
ans = 0
i = 0
while True:
x = (2 * 5 * (5 ** i))
if x <= N:
ans += N // x
i += 1
print(ans)
|
s386567186
|
Accepted
| 17
| 2,940
| 182
|
N = int(input())
ans = 0
if N % 2 == 0:
i = 0
while True:
x = (2 * 5 * (5 ** i))
if x > N:
break
ans += N // x
i += 1
print(ans)
|
s970579740
|
p03380
|
u177040005
| 2,000
| 262,144
|
Wrong Answer
| 82
| 14,052
| 700
|
Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order. From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized. If there are multiple pairs that maximize the value, any of them is accepted.
|
import bisect
N = int(input())
A = sorted(list(map(int,input().split())))
if N == 2:
print(A[-1],A[0])
exit()
r_val = A[-1]//2
r_ind = bisect.bisect_left(A,r_val,0,N-1)
# if abs(A[-1]/2 - float(A[r_ind])) < abs(A[-1]/2 - float(A[min(r_ind+1,N-2)])):
# r_ind = r_ind + 1
#print(A[r_ind])
ans = [A[-1],A[r_ind]]
# print(A)
# for i,n in enumerate(A[:-1]):
# print(i,n,n//2)
# r_val = n // 2
# if abs(n/2 - A[r_ind]) > abs(n/2 - A[r_ind+1]):
# r_ind = r_ind + 1
# print(r_val,'r_ind = ',r_ind,A[r_ind])
# r = A[r_ind]
# print(n,r,comb(n,r))
# if max_comb < comb(n,r):
# ans = [n,r]
print(ans)
|
s670754201
|
Accepted
| 127
| 14,428
| 234
|
N = int(input())
A = sorted(list(map(int,input().split())))
n = A[-1]
A = A[:-1]
r_val = n//2
diff_min = float(n)
for a in A:
diff = abs(n/2 - float(a))
if diff_min > diff:
diff_min = diff
r = a
print(n,r)
|
s681377390
|
p03478
|
u382303205
| 2,000
| 262,144
|
Wrong Answer
| 72
| 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 = map(int,input().split())
count = 0
for i in range(1, n + 1):
tmp = i
x = 0
for j in range(5)[::-1]:
x += tmp // (10**j)
tmp = tmp - (tmp // (10**j))*(10**j)
if a <= x <= b:
count +=1
print(count)
|
s906756215
|
Accepted
| 69
| 3,060
| 257
|
n,a,b = map(int,input().split())
sum = 0
for i in range(1, n + 1):
tmp = i
x = 0
for j in range(5)[::-1]:
x += tmp // (10**j)
tmp = tmp - (tmp // (10**j))*(10**j)
if a <= x <= b:
sum += i
print(sum)
|
s877451487
|
p02795
|
u760961723
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 64
|
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,W,N = [int(input()) for i in range(3)]
print(min(N//H,N//W))
|
s142846344
|
Accepted
| 18
| 2,940
| 98
|
import math
H,W,N = [int(input()) for i in range(3)]
print(min(math.ceil(N/H),math.ceil(N/W)))
|
s452022980
|
p03394
|
u052499405
| 2,000
| 262,144
|
Wrong Answer
| 27
| 5,488
| 822
|
Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers. She thinks that a set S = \\{a_{1}, a_{2}, ..., a_{N}\\} of **distinct** positive integers is called **special** if for all 1 \leq i \leq N, the gcd (greatest common divisor) of a_{i} and the sum of the remaining elements of S is **not** 1. Nagase wants to find a **special** set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a **special** set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000.
|
import math
k = int(input())
if k == 3:
print("3 6 9")
elif k == 4:
print("3 9 2 4")
elif k == 6:
print("3 9 2 4 8 10")
elif k <= 15002:
prints = set([3, 9])
k -= 2
for i in range(int(math.ceil(k / 3))):
prints.add(i*6+2)
prints.add(i*6+4)
prints.add(i*6+6)
if 3 - k % 3 == 1:
prints.remove(6)
if 3 - k % 3 == 2:
prints.remove(6)
prints.remove(12)
print(" ".join([str(item) for item in prints]))
else:
prints = set()
for i in range(5000):
prints.add(i*6+2)
prints.add(i*6+4)
prints.add(i*6+6)
k -= 15000
for i in range(int(math.ceil(k / 2))):
prints.add(i*12+3)
prints.add(i*12+9)
if 3 - k % 3 == 1:
prints.remove(6)
print(" ".join([str(item) for item in prints]))
|
s879798496
|
Accepted
| 26
| 5,488
| 823
|
import math
k = int(input())
if k == 3:
print("2 5 63")
elif k == 4:
print("3 9 2 4")
elif k == 6:
print("3 9 2 4 8 10")
elif k <= 15002:
prints = set([3, 9])
k -= 2
for i in range(int(math.ceil(k / 3))):
prints.add(i*6+2)
prints.add(i*6+4)
prints.add(i*6+6)
if 3 - k % 3 == 1:
prints.remove(6)
if 3 - k % 3 == 2:
prints.remove(6)
prints.remove(12)
print(" ".join([str(item) for item in prints]))
else:
prints = set()
for i in range(5000):
prints.add(i*6+2)
prints.add(i*6+4)
prints.add(i*6+6)
k -= 15000
for i in range(int(math.ceil(k / 2))):
prints.add(i*12+3)
prints.add(i*12+9)
if 2 - k % 2 == 1:
prints.remove(6)
print(" ".join([str(item) for item in prints]))
|
s159123908
|
p03130
|
u191635495
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 247
|
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
a1, b1 = map(int, input().split())
a2, b2 = map(int, input().split())
a3, b3 = map(int, input().split())
l = [a1, a2, a3, b1, b2, b3]
res = [sum([1 for _ in l if _ == __]) for __ in [1,2,3,4]]
if 3 in res:
print("No")
else:
print("Yes")
|
s205602946
|
Accepted
| 17
| 3,060
| 248
|
a1, b1 = map(int, input().split())
a2, b2 = map(int, input().split())
a3, b3 = map(int, input().split())
l = [a1, a2, a3, b1, b2, b3]
res = [sum([1 for _ in l if _ == __]) for __ in [1,2,3,4]]
if 3 in res:
print("NO")
else:
print("YES")
|
s662403884
|
p03611
|
u188138642
| 2,000
| 262,144
|
Wrong Answer
| 69
| 20,680
| 186
|
You are given an integer sequence of length N, a_1,a_2,...,a_N. For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing. After these operations, you select an integer X and count the number of i such that a_i=X. Maximize this count by making optimal choices.
|
n = int(input())
p = list(map(int, input().split()))
i=0
ans =0
while i<n:
if p[i]==i+1:
ans += 1
i+=1
if not (i==n-2 and p[i+1]==i+2):
i += 1
print(ans)
|
s637659426
|
Accepted
| 114
| 31,696
| 195
|
from collections import Counter
n = int(input())
a = list(map(int, input().split()))
c = []
for i in a:
c.append(i)
c.append(i-1)
c.append(i+1)
c = Counter(c)
print(max(c.values()))
|
s869699348
|
p03037
|
u094932051
| 2,000
| 1,048,576
|
Wrong Answer
| 373
| 27,332
| 301
|
We have N ID cards, and there are M gates. We can pass the i-th gate if we have one of the following ID cards: the L_i- th, (L_i+1)-th, ..., and R_i-th ID cards. How many of the ID cards allow us to pass all the gates alone?
|
while True:
try:
N, M = map(int, input().split())
gate = [list(map(int, input().split())) for i in range(M)]
L, R = 0, 10**5+5
for l, r in gate:
L = max(L, l)
R = min(R, r)
print(R-L+1)
except:
break
|
s541041364
|
Accepted
| 346
| 3,956
| 418
|
while True:
try:
N, M = map(int, input().split())
dp = [0]*(10**5+5)
for i in range(M):
L, R = map(int, input().split())
dp[L-1] += 1
dp[R] -= 1
cnt = 0
ans = 0
for i in range(N):
cnt += dp[i]
if cnt == M:
ans += 1
print(ans)
except:
break
|
s795897639
|
p03854
|
u591287669
| 2,000
| 262,144
|
Wrong Answer
| 72
| 3,188
| 236
|
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()
while len(s)>0:
if s[-5:]=='erase' or s[-5:]=='dream':
s=s[:-5]
elif s[-6:]=='eraser':
s=s[:-6]
elif s[-7:]=='dreamer':
s=s[:-7]
else:
print('No')
exit(0)
print('Yes')
|
s047869423
|
Accepted
| 72
| 3,316
| 236
|
s = input()
while len(s)>0:
if s[-5:]=='erase' or s[-5:]=='dream':
s=s[:-5]
elif s[-6:]=='eraser':
s=s[:-6]
elif s[-7:]=='dreamer':
s=s[:-7]
else:
print('NO')
exit(0)
print('YES')
|
s036461251
|
p03565
|
u465101448
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 766
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
S = list(input())[:-1]
T = list(input())[:-1]
ans_list=[]
for i in range(len(S)-len(T)+1):
S_temp=S.copy()
serch_S=S_temp[i:i+len(T)]
# print(serch_S)
count=0
for t in range(len(T)):
if serch_S[t] in [T[t],'?']:
# print(serch_S[t])
count+=1
else:
break
if count==len(T):
S_temp[i:i+len(T)]=T
for s in range(len(S_temp)):
if S_temp[s]=='?':
S_temp[s]='a'
S_temp_str=''
for s_t in S_temp:
S_temp_str+=s_t
ans_list.append(S_temp_str)
if len(ans_list)>0:
ans_list.sort()
ans=ans_list[0]
else:
ans='UNRESTORABLE'
print(ans)
|
s923185424
|
Accepted
| 18
| 3,064
| 703
|
S = list(input())
T = list(input())
ans_list=[]
for i in range(len(S)-len(T)+1):
S_temp=S.copy()
serch_S=S_temp[i:i+len(T)]
count=0
for t in range(len(T)):
if serch_S[t] in [T[t],'?']:
count+=1
else:
break
if count==len(T):
S_temp[i:i+len(T)]=T
for s in range(len(S_temp)):
if S_temp[s]=='?':
S_temp[s]='a'
S_temp_str=''
for s_t in S_temp:
S_temp_str+=s_t
ans_list.append(S_temp_str)
if len(ans_list)>0:
ans_list.sort()
ans=ans_list[0]
else:
ans='UNRESTORABLE'
print(ans)
|
s463175084
|
p03352
|
u870518235
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,384
| 40
|
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
X = int(input())
print(int(X**0.5-1)**2)
|
s616481319
|
Accepted
| 33
| 9,176
| 235
|
X = int(input())
lst = []
for i in range(1,101):
for j in range(2,11):
if i**j > 1000:
break
else:
lst.append(i**j)
for k in range(X):
if X - k in lst:
print(X-k)
break
|
s798467450
|
p04043
|
u995062424
| 2,000
| 262,144
|
Wrong Answer
| 16
| 2,940
| 133
|
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 = input()
if a.count('5') == 2:
if a.count('7') == 1:
print('yes')
else:
print('no')
else:
print('no')
|
s052420118
|
Accepted
| 16
| 2,940
| 141
|
a = input().split()
if a.count('5') == 2:
if a.count('7') == 1:
print('YES')
else:
print('NO')
else:
print('NO')
|
s075118071
|
p02618
|
u312078744
| 2,000
| 1,048,576
|
Wrong Answer
| 127
| 27,360
| 658
|
AtCoder currently hosts three types of contests: ABC, ARC, and AGC. As the number of users has grown, in order to meet the needs of more users, AtCoder has decided to increase the number of contests to 26 types, from AAC to AZC. For convenience, we number these 26 types as type 1 through type 26. AtCoder wants to schedule contests for D days so that user satisfaction is as high as possible. For every day, AtCoder will hold exactly one contest, and each contest will end on that day. The satisfaction is calculated as follows. * The satisfaction at the beginning of day 1 is 0. Satisfaction can be negative. * Holding contests increases satisfaction. The amount of increase will vary depending on a variety of factors. Specifically, we know in advance that holding a contest of type i on day d will increase the satisfaction by s_{d,i}. * If a particular type of contest is not held for a while, the satisfaction decreases. Each contest type i has an integer c_i, and at the end of each day d=1,2,...,D, the satisfaction decreases as follows. Let \mathrm{last}(d,i) be the last day before day d (including d) on which a contest of type i was held. If contests of type i have never been held yet, we define \mathrm{last}(d,i)=0. At the end of day d, the satisfaction decreases by \sum _{i=1}^{26}c_i \times (d-\mathrm{last}(d,i)). Please schedule contests on behalf of AtCoder. If the satisfaction at the end of day D is S, you will get a score of \max(10^6 + S, 0). There are 50 test cases, and the score of a submission is the total scores for each test case. You can make submissions multiple times, and the highest score among your submissions will be your score.
|
import sys
import numpy as np
from operator import mul
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
d = int(readline())
c = list(map(int, readline().split()))
s = []
for _ in range(d):
add = list(map(int, readline().split()))
s.append(add)
#t = [0] * d
day = 0
satisfy = 0
sumC = sum(c)
count = np.zeros(26)
for i in range(d):
#t = int(readline()) - 1
tarLis = s[day]
tar = max(tarLis)
t = tarLis.index(tar)
up = s[day][t]
count += 1
count[t] = 0
sub = map(mul, count, c)
satisfy += (up - sum(sub))
# satisfy
print(int(satisfy))
day += 1
|
s348286315
|
Accepted
| 132
| 27,240
| 686
|
import sys
import numpy as np
from operator import mul
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
d = int(readline())
c = list(map(int, readline().split()))
s = []
for _ in range(d):
add = list(map(int, readline().split()))
s.append(add)
#t = [0] * d
day = 0
satisfy = 0
sumC = sum(c)
count = np.zeros(26)
for i in range(d):
#t = int(readline()) - 1
tarLis = s[day]
tar = max(tarLis)
t = tarLis.index(tar) + 1
print(t)
day += 1
'''
up = s[day][t]
count += 1
count[t] = 0
sub = map(mul, count, c)
satisfy += (up - sum(sub))
# satisfy
print(satisfy)
'''
|
s616208360
|
p02397
|
u092736322
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 226
|
Write a program which reads two integers x and y, and prints them in ascending order.
|
for i in range(3000):
k=input().split()
a=int(k[0])
b=int(k[0])
if a==0 and b==0:
break
else:
if a<b:
print(str(a)+" "+str(b))
else:
print(str(b)+" "+str(a))
|
s110052992
|
Accepted
| 50
| 5,612
| 226
|
for i in range(3000):
k=input().split()
a=int(k[0])
b=int(k[1])
if a==0 and b==0:
break
else:
if a<b:
print(str(a)+" "+str(b))
else:
print(str(b)+" "+str(a))
|
s682298333
|
p03569
|
u354915818
| 2,000
| 262,144
|
Wrong Answer
| 2,108
| 21,516
| 709
|
We have a string s consisting of lowercase English letters. Snuke can perform the following operation repeatedly: * Insert a letter `x` to any position in s of his choice, including the beginning and end of s. Snuke's objective is to turn s into a palindrome. Determine whether the objective is achievable. If it is achievable, find the minimum number of operations required.
|
# coding: utf-8
# Here your code !
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon Oct 9 22:12:40 2017
@author: kitagenbakurou
"""
import numpy as np
import copy
def insert(pos, s, x):
return x.join([s[:pos], s[pos:] ])
s = input()
N = len(s)
i = 0
cnt = 0
flag = True
while i < N - i - 1 :
if s[i] != s[N - i - 1] :
if s[i] == "x" :
cnt += 1
s = insert(N - i , s , "x")
i = -1
elif s[N - i - 1] == "x" :
cnt += 1
s = insert(i , s , "x")
i = -1
else :
flag = False
break
N = len(s)
i += 1
print(s)
if flag :
print(cnt)
else :
print(-1)
|
s387956002
|
Accepted
| 204
| 12,724
| 741
|
# coding: utf-8
# Here your code !
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon Oct 9 22:12:40 2017
@author: kitagenbakurou
"""
import numpy as np
import copy
def insert(pos, s, x):
return x.join([s[:pos], s[pos:] ])
s = input()
N = len(s)
i = 0
j = N - 1
cnt = 0
flag = True
while i < j :
if s[i] != s[j] :
if s[i] == "x" :
cnt += 1
i += 1
j += 0
elif s[j] == "x" :
cnt += 1
i +=0
j -= 1
else :
flag = False
break
else :
i += 1
j -= 1
if flag :
print(cnt)
else :
print(-1)
|
s985290545
|
p03494
|
u818542539
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 221
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
As = [int(c) for c in input().split()]
counter = 0
w = 0
while w == 0:
for i in range(N):
if As[i] % 2 != 0:
w = 1
As[i] = int(As[i] / 2)
counter += 1
print(counter)
|
s959623897
|
Accepted
| 18
| 3,060
| 254
|
N = int(input())
As = [int(c) for c in input().split()]
counter = 0
while True:
for i in range(N):
if As[i] % 2 != 0:
break
As[i] = int(As[i] / 2)
else:
counter += 1
continue
break
print(counter)
|
s571430169
|
p03409
|
u064408584
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 434
|
On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i). A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point. At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.
|
n=int(input())
a=[]
b=[]
for i in range(n):
a.append(list(map(int,input().split())))
for i in range(n):
b.append(list(map(int,input().split())))
an=sorted(a,reverse=True)
bn=sorted(b,reverse=True)
print(an,bn)
count=0
for i in range(len(bn)):
for j in range(len(an)):
# print(an[j],bn[i])
if bn[i][0]>an[j][0] and bn[i][1]>an[j][1]:
count+=1
an.pop(j)
break
print(count)
|
s578003937
|
Accepted
| 19
| 3,064
| 444
|
n=int(input())
a=[]
b=[]
for i in range(n):
a.append(list(map(int,input().split())))
for i in range(n):
b.append(list(map(int,input().split())))
an=sorted(a, key=lambda x:-x[1])
bn=sorted(b)#, key=lambda x:(x[0],x[1]))
count=0
for i in range(len(bn)):
for j in range(len(an)):
# print(an[j],bn[i])
if bn[i][0]>an[j][0] and bn[i][1]>an[j][1]:
count+=1
an.pop(j)
break
print(count)
|
s689664470
|
p03351
|
u682730715
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 91
|
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a, b, c, d = map(int, input().split())
if abs(a-c) <= d:
print('Yes')
else:
print('No')
|
s743729473
|
Accepted
| 17
| 2,940
| 132
|
a, b, c, d = map(int, input().split())
if abs(a-c) <= d or (abs(a-b) <= d and abs(b-c) <= d):
print('Yes')
else:
print('No')
|
s871184672
|
p02612
|
u353548710
| 2,000
| 1,048,576
|
Wrong Answer
| 31
| 9,140
| 36
|
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.
|
pay = int(input())
print( pay%1000 )
|
s912028521
|
Accepted
| 27
| 9,144
| 72
|
pay = int(input())%1000
if pay == 0:
print(0)
else:
print(1000- pay)
|
s681292541
|
p03067
|
u118867138
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 89
|
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
A,B,C = map(int, input().split())
if C<A and B>A:
print("Yes")
else:
print("No")
|
s302343587
|
Accepted
| 17
| 3,060
| 124
|
A,B,C = map(int, input().split())
if C<A and B<C:
print("Yes")
elif C>A and B>C:
print("Yes")
else:
print("No")
|
s381906873
|
p02612
|
u631579948
| 2,000
| 1,048,576
|
Wrong Answer
| 33
| 9,076
| 56
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
a=int(input())
while a>=1000:
a-=1000
print(a)
|
s956696448
|
Accepted
| 34
| 9,144
| 53
|
a=int(input())
while a>1000:
a-=1000
print(1000-a)
|
s744096848
|
p03455
|
u752552310
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,316
| 84
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
if a*b%2 == 0:
print("Odd")
else:
print("Even")
|
s860633973
|
Accepted
| 17
| 2,940
| 86
|
a, b = map(int, input().split())
if (a*b)%2 != 0:
print("Odd")
else:
print("Even")
|
s205871339
|
p00003
|
u316584871
| 1,000
| 131,072
|
Wrong Answer
| 40
| 5,596
| 191
|
Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.
|
for i in range(int(input())):
llist = list(map(int,input().split()))
llist.sort()
if(llist[2]**2 == llist[0]**2 + llist[1]**2):
print('Yes')
else:
print('No')
|
s176331758
|
Accepted
| 40
| 5,600
| 191
|
for i in range(int(input())):
llist = list(map(int,input().split()))
llist.sort()
if(llist[2]**2 == llist[0]**2 + llist[1]**2):
print('YES')
else:
print('NO')
|
s010685490
|
p03997
|
u862757671
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 73
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a + b) * h / 2)
|
s637964990
|
Accepted
| 17
| 2,940
| 78
|
a = int(input())
b = int(input())
h = int(input())
print(int((a + b) * h / 2))
|
s313198438
|
p03548
|
u960171798
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 68
|
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
x,y,z = map(int, input().split())
ans = 1+(x-y-2*z)/(y+z)
print(ans)
|
s003339515
|
Accepted
| 17
| 2,940
| 69
|
x,y,z = map(int, input().split())
ans = 1+(x-y-2*z)//(y+z)
print(ans)
|
s507050134
|
p03493
|
u331464808
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 99
|
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.
|
A = list(map(int, input().split()))
res = 0
for i in range(len(A)):
res =res + A[i]
print(res)
|
s136664200
|
Accepted
| 17
| 2,940
| 58
|
s = input()
res = 0
for v in s:
res += int(v)
print(res)
|
s121805812
|
p03360
|
u226873514
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 94
|
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?
|
a = list(map(int, input().split()))
k = int(input())
a.sort()
print(a[2] ** k + a[1] + a[0])
|
s267360033
|
Accepted
| 17
| 2,940
| 100
|
a = list(map(int, input().split()))
k = int(input())
a.sort()
print(a[2] * (2 ** k) + a[1] + a[0])
|
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