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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s239672733
|
p02843
|
u972416428
| 2,000 | 1,048,576 |
Wrong Answer
| 117 | 15,120 | 150 |
AtCoder Mart sells 1000000 of each of the six items below: * Riceballs, priced at 100 yen (the currency of Japan) each * Sandwiches, priced at 101 yen each * Cookies, priced at 102 yen each * Cakes, priced at 103 yen each * Candies, priced at 104 yen each * Computers, priced at 105 yen each Takahashi wants to buy some of them that cost exactly X yen in total. Determine whether this is possible. (Ignore consumption tax.)
|
n = int(input().strip())
valid = { 0: True }
for i in range(1, n+1):
valid[i] = any(valid.get(i-j, False) for j in range(100, 106))
print (valid[n])
|
s267224660
|
Accepted
| 118 | 15,124 | 162 |
n = int(input().strip())
valid = { 0: True }
for i in range(1, n+1):
valid[i] = any(valid.get(i-j, False) for j in range(100, 106))
print (1 if valid[i] else 0)
|
s880381727
|
p03494
|
u831081653
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 236 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
a = list(map(int, input().split()))
count = 0
flag = 1
while flag == 1:
b = list(map(lambda x: x % 2, a))
if b.count(1) == 0:
a = list(map(lambda x: x / 2, a))
count += 1
else:
flag = 0
print(count)
|
s162710557
|
Accepted
| 19 | 3,060 | 271 |
n = int(input())
a = list(map(int, input().split()))
cnt = 0
flag = True
while flag:
ans = []
for i in a:
if i % 2 == 1:
flag = False
else:
ans.append(i/2)
if len(ans) == n:
cnt += 1
a = ans
print(cnt)
|
s516923132
|
p02678
|
u465900169
| 2,000 | 1,048,576 |
Wrong Answer
| 730 | 37,140 | 459 |
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
from collections import deque
N, M = map(int, input().split())
G = [[] for i in range(N)]
R = [0]*N
for m in range(M):
A,B = map(int, input().split())
G[A-1].append(B-1)
G[B-1].append(A-1)
d = deque()
for n in G[0]:
d.append(n)
R[n]=1
while len(d)>0:
i = d.popleft()
for j in range(len(G[i])):
if R[G[i][j]] == 0 and G[i][j]!=0:
d.append(G[i][j])
R[G[i][j]]=i+1
for n in range(1,N):
print(R[n])
|
s872932160
|
Accepted
| 822 | 37,208 | 558 |
import sys
from collections import deque
N, M = map(int, input().split())
G = [[] for i in range(N)]
R = [0]*N
for m in range(M):
A,B = map(int, input().split())
G[A-1].append(B-1)
G[B-1].append(A-1)
d = deque()
for n in G[0]:
d.append(n)
R[n]=1
while len(d)>0:
i = d.popleft()
for j in range(len(G[i])):
if R[G[i][j]] == 0 and G[i][j]!=0:
d.append(G[i][j])
R[G[i][j]]=i+1
for n in range(1,N):
if R[n]==0:
print("No")
sys.exit()
print("Yes")
for n in range(1,N):
print(R[n])
|
s875747782
|
p02398
|
u501414488
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,724 | 133 |
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
|
(a, b, c) = [int(i) for i in input().split()]
cnt = 0
for i in range(a, b + 1):
if (c % 1) == 0:
cnt = cnt + 1
print(cnt)
|
s014246180
|
Accepted
| 30 | 6,728 | 133 |
(a, b, c) = [int(i) for i in input().split()]
cnt = 0
for i in range(a, b + 1):
if (c % i) == 0:
cnt = cnt + 1
print(cnt)
|
s584852370
|
p03456
|
u629350026
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 204 |
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
a,b=map(int,input().split())
if b==100:
s=a*1000+b
elif b>=10:
s=a*100+b
else:
s=a*10+b
j=1
temp=0
while j**2<s:
if j**2==s:
print("Yes")
temp=1
break
j=j+1
if temp==0:
print("No")
|
s030241111
|
Accepted
| 18 | 3,064 | 205 |
a,b=map(int,input().split())
if b==100:
s=a*1000+b
elif b>=10:
s=a*100+b
else:
s=a*10+b
j=1
temp=0
while j**2<=s:
if j**2==s:
print("Yes")
temp=1
break
j=j+1
if temp==0:
print("No")
|
s654089398
|
p03548
|
u146346223
| 2,000 | 262,144 |
Wrong Answer
| 24 | 9,152 | 50 |
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())
print(x//((y+z)+z))
|
s478317909
|
Accepted
| 29 | 9,156 | 52 |
x,y,z=map(int,input().split())
print((x-z) // (y+z))
|
s217392048
|
p03162
|
u578850957
| 2,000 | 1,048,576 |
Wrong Answer
| 559 | 33,396 | 544 |
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
N = int(input())
siawaseList = [list(map(int,input().split())) for _ in range(N)]
dp = [[0]*3]*N
dp[0][0] = siawaseList[0][0]
dp[0][1] = siawaseList[0][1]
dp[0][2] = siawaseList[0][2]
for i in range(1,N):
dp[i][0] = max(dp[i-1][1]+siawaseList[i][0],dp[i-1][2]+siawaseList[i][0])
dp[i][1] = max(dp[i-1][0]+siawaseList[i][1],dp[i-1][2]+siawaseList[i][1])
dp[i][2] = max(dp[i-1][0]+siawaseList[i][2],dp[i-1][1]+siawaseList[i][2])
print(max(dp[N-1][0],dp[N-1][1],dp[N-1][2]))
|
s451474985
|
Accepted
| 634 | 47,344 | 576 |
N = int(input())
siawaseList = [list(map(int,input().split())) for _ in range(N)]
dp = [[0 for _ in range(3)] for __ in range(N)]
dp[0][0] = siawaseList[0][0]
dp[0][1] = siawaseList[0][1]
dp[0][2] = siawaseList[0][2]
for i in range(1,N):
dp[i][0] = max(dp[i-1][1]+siawaseList[i][0],dp[i-1][2]+siawaseList[i][0])
dp[i][1] = max(dp[i-1][0]+siawaseList[i][1],dp[i-1][2]+siawaseList[i][1])
dp[i][2] = max(dp[i-1][0]+siawaseList[i][2],dp[i-1][1]+siawaseList[i][2])
print(max(dp[N-1][0],dp[N-1][1],dp[N-1][2]))
|
s438032979
|
p03493
|
u655427167
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 227 |
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.
|
if __name__ == '__main__':
# line = str(input())
line = '000'
count = 0
if line[0] == '1':
count += 1
if line[1] == '1':
count += 1
if line[2] == '1':
count += 1
print(count)
|
s289183268
|
Accepted
| 19 | 2,940 | 209 |
if __name__ == '__main__':
line = str(input())
count = 0
if line[0] == '1':
count += 1
if line[1] == '1':
count += 1
if line[2] == '1':
count += 1
print(count)
|
s779459066
|
p03944
|
u074220993
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,132 | 432 |
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
W, H, N = map(int, input().split())
x_min, x_max, y_min, y_max = 0, H, 0, W
for i in range(N):
x, y, a = map(int, input().split())
if a == 1 and x > x_min:
x_min = x
elif a == 2 and x < x_max:
x_max = x
elif a == 3 and y > y_min:
y_min = y
elif a == 4 and y < y_max:
y_max = y
if x_max > x_min and y_max > y_min:
S = (x_max - x_min) * (y_max - y_min)
else:
S = 0
print(S)
|
s012017321
|
Accepted
| 28 | 9,156 | 313 |
W, H, N = map(int, input().split())
u, U, v, V = 0, W, 0, H
for _ in range(N):
x, y, a = map(int,input().split())
if a == 1:
u = max(u,x)
if a == 2:
U = min(U,x)
if a == 3:
v = max(v,y)
if a == 4:
V = min(V,y)
f = lambda x:x if x > 0 else 0
print(f(U-u)*f(V-v))
|
s310197064
|
p03407
|
u663710122
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 71 |
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
A, B, C = map(int, input().split())
print('Yes' if A+B == C else 'No')
|
s567497694
|
Accepted
| 17 | 2,940 | 71 |
A, B, C = map(int, input().split())
print('Yes' if A+B >= C else 'No')
|
s313620360
|
p03637
|
u781262926
| 2,000 | 262,144 |
Wrong Answer
| 159 | 14,052 | 281 |
We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective.
|
n, *A = map(int, open(0).read().split())
def f(q):
y = 0
while True:
q, r = divmod(q, 2)
if r == 1:
break
y += 1
return y
B = [0] * 3
for a in A:
b = f(a)
B[0] += b == 0
B[1] += b == 1
B[2] += b >= 2
print('Yes' if B[0] + B[1]%2 <= B[2] else 'No')
|
s787629313
|
Accepted
| 163 | 14,052 | 283 |
n, *A = map(int, open(0).read().split())
def f(x):
y = 0
q, r = divmod(x, 2)
while r == 0:
y += 1
q, r = divmod(q, 2)
return y
B = [0] * 3
for a in A:
b = f(a)
B[0] += b == 0
B[1] += b == 1
B[2] += b >= 2
print('Yes' if B[0] + (B[1]!=0) <= B[2]+1 else 'No')
|
s402186646
|
p02396
|
u801535964
| 1,000 | 131,072 |
Wrong Answer
| 130 | 5,612 | 98 |
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
i = 1
while True:
x = int(input())
if (x == 0):
break
print(f"Case {i}: {x}")
|
s750157838
|
Accepted
| 130 | 5,612 | 107 |
i = 1
while True:
x = int(input())
if (x == 0):
break
print(f"Case {i}: {x}")
i+=1
|
s424512301
|
p03371
|
u107091170
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,060 | 158 |
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
A,B,C,X,Y=map(int, input().split())
ans = 500000000
ans = min(ans, A*X+B*Y)
ans = min(ans, C*min(X,Y))
ans = min(ans, (A+B)*min(X,Y)+C*(abs(X-Y)))
print(ans)
|
s493062839
|
Accepted
| 110 | 3,064 | 155 |
A,B,C,X,Y=map(int, input().split())
ans = 50000000000
for i in range( max(X,Y)+1 ):
ans = min(ans, C*i*2 + A*(max(0,(X-i)))+B*(max(0,(Y-i))))
print(ans)
|
s405569920
|
p03477
|
u763280125
| 2,000 | 262,144 |
Wrong Answer
| 32 | 8,956 | 133 |
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
A, B, C, D = input().split()
if A + B > C + D:
print('Left')
elif A + B < C + D:
print('Right')
else:
print('Balanced')
|
s405798949
|
Accepted
| 22 | 9,032 | 157 |
I = input().split()
A, B, C, D = [int(i) for i in I]
if A + B > C + D:
print('Left')
elif A + B < C + D:
print('Right')
else:
print('Balanced')
|
s458246524
|
p03605
|
u696449926
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 197 |
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
a = list(map(int, input().split()))
if(len(a) == 2):
if(a[0] == 9):
print('Yes')
elif(a[1] == 9):
print('Yes')
else:
print('No')
else:
if(a[0] == 9):
print('Yes')
else:
print('No')
|
s450140829
|
Accepted
| 17 | 3,060 | 195 |
a = list(map(int, list(input())))
if(len(a) == 2):
if(a[0] == 9):
print('Yes')
elif(a[1] == 9):
print('Yes')
else:
print('No')
else:
if(a[0] == 9):
print('Yes')
else:
print('No')
|
s654520817
|
p03730
|
u234631479
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,444 | 119 |
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`.
|
a, b, c = map(int, input().split())
print(a,b,c)
for i in range(1,10000):
if a*i%b == c:
print("YES")
print("NO")
|
s727473774
|
Accepted
| 17 | 3,060 | 223 |
a, b, c = map(int, input().split())
L = []
for i in range(1, b+1):
L.append(a*i%b)
L2 = list(set(L))
for i in range(len(L2)):
for k in range(len(L2)):
if L2[i]+L2[k] == c:
print("YES")
quit()
print("NO")
|
s800370615
|
p03401
|
u118019047
| 2,000 | 262,144 |
Wrong Answer
| 2,108 | 12,252 | 398 |
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 = input().split()
b = 0
answer = []
for i in range(n):
b=0
if i == 0:
box = a[i]
a[i] = 0
else:
box = a[i]
a[i] = int(a[i-1])
b = int(a[0])
for j in range(n-1):
b += abs(int(a[j]) - int(a[j+1]))
b += abs(int(a[n-1]))
a[i] = box
answer.append(b)
print(a)
for k in range(len(answer)):
print(answer[k])
|
s174130196
|
Accepted
| 229 | 14,048 | 272 |
n=int(input())
a=list(map(int,input().split()))
a.insert(0,0)
a.insert(n+1,0)
total=0
for i in range(1,n+2):
total+=abs(a[i]-a[i-1])
for i in range(1,n+1):
ans=total-(abs(a[i]-a[i-1])+abs(a[i+1]-a[i]))+abs(a[i+1]-a[i-1])
print(ans)
|
s555435993
|
p03486
|
u642528832
| 2,000 | 262,144 |
Wrong Answer
| 26 | 8,960 | 142 |
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()
sr = sorted(s)
tr = sorted(t,reverse = True)
print(sr)
print(tr)
if sr < tr:
print('Yes')
else:
print('No')
|
s270461731
|
Accepted
| 28 | 9,004 | 144 |
s = input()
t = input()
sr = sorted(s)
tr = sorted(t,reverse = True)
#print(sr)
#print(tr)
if sr < tr:
print('Yes')
else:
print('No')
|
s187609331
|
p03543
|
u436110200
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 72 |
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**?
|
s=input()
print("YES" if s[0]==s[1]==s[2] or s[1]==s[2]==s[3] else "NO")
|
s229536255
|
Accepted
| 18 | 2,940 | 72 |
s=input()
print("Yes" if s[0]==s[1]==s[2] or s[1]==s[2]==s[3] else "No")
|
s495859459
|
p02422
|
u179070318
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 366 |
Write a program which performs a sequence of commands to a given string $str$. The command is one of: * print a b: print from the a-th character to the b-th character of $str$ * reverse a b: reverse from the a-th character to the b-th character of $str$ * replace a b p: replace from the a-th character to the b-th character of $str$ with p Note that the indices of $str$ start with 0.
|
string = input()
n = int(input())
for i in range(n):
inp = [x for x in input().split()]
a = int(inp[1])
b = int(inp[2])
if inp[0] == 'print':
print(string[a-1:b])
elif inp[0] == 'reverse':
string = string[:a-1] + string[a-1:b][::-1] +string[b:]
elif inp[0] == 'replace':
string = string.replace(string[a-1:b],inp[3])
|
s163878541
|
Accepted
| 20 | 5,620 | 359 |
string = input()
for _ in range(int(input())):
com = [x for x in input().split()]
c = com[0]
a,b = [int(com[i]) for i in range(1,3)]
if c == 'reverse':
string = string[:a] + string[a:b+1][::-1] + string[b+1:]
if c == 'print':
print(string[a:b+1])
if c == 'replace':
string = string[:a] + com[3] + string[b+1:]
|
s244254256
|
p04035
|
u591287669
| 2,000 | 262,144 |
Wrong Answer
| 224 | 14,152 | 361 |
We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i. At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly: * Choose a (connected) rope with a total length of at least L, then untie one of its knots. Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.
|
n,l=map(int,input().split())
arr=list(map(int,input().split()))
psbl=False
pos=0
for i in range(n-1):
if arr[i]+arr[i+1]>=l:
psbl=True
pos=i
break
print(pos)
if psbl:
print("Possible")
for i in range(0,pos):
print(i+1)
for i in range(n-2,pos,-1):
print(i+1)
print(pos+1)
else:
print("Impossible")
|
s089114431
|
Accepted
| 221 | 14,204 | 350 |
n,l=map(int,input().split())
arr=list(map(int,input().split()))
psbl=False
pos=0
for i in range(n-1):
if arr[i]+arr[i+1]>=l:
psbl=True
pos=i
break
if psbl:
print("Possible")
for i in range(0,pos):
print(i+1)
for i in range(n-2,pos,-1):
print(i+1)
print(pos+1)
else:
print("Impossible")
|
s079135750
|
p03598
|
u732870425
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 236 |
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 = int(input())
K = int(input())
X = list(map(int, input().split()))
sum_len = 0
for i in range(len(X)):
if abs(N-X[i]) < abs(K-X[i]):
sum_len += abs(N-X[i]) * 2
else:
sum_len += abs(K-X[i]) * 2
print(sum_len)
|
s920836998
|
Accepted
| 17 | 3,060 | 222 |
N = int(input())
K = int(input())
X = list(map(int, input().split()))
sum_len = 0
for i in range(len(X)):
if X[i] < abs(K-X[i]):
sum_len += X[i] * 2
else:
sum_len += abs(K-X[i]) * 2
print(sum_len)
|
s944062385
|
p02613
|
u409371339
| 2,000 | 1,048,576 |
Wrong Answer
| 149 | 16,324 | 387 |
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 _ in range(n)]
numAC = 0
numWA = 0
numTLE = 0
numRE = 0
for i in s:
if i == "AC":
numAC += 1
elif i == "WA":
numWA += 1
elif i == "TLE":
numTLE += 1
elif i == "RE":
numRE += 1
print("AC × " + str(numAC))
print("WA × " + str(numWA))
print("TLE × " + str(numTLE))
print("RE × " + str(numRE))
|
s479684347
|
Accepted
| 153 | 16,332 | 384 |
n = int(input())
s = [input() for _ in range(n)]
numAC = 0
numWA = 0
numTLE = 0
numRE = 0
for i in s:
if i == "AC":
numAC += 1
elif i == "WA":
numWA += 1
elif i == "TLE":
numTLE += 1
elif i == "RE":
numRE += 1
print("AC x " + str(numAC))
print("WA x " + str(numWA))
print("TLE x " + str(numTLE))
print("RE x " + str(numRE))
|
s411726502
|
p02578
|
u770018088
| 2,000 | 1,048,576 |
Wrong Answer
| 186 | 32,176 | 163 |
N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal.
|
N = int(input())
A = list(map(int, input().strip().split()))
ans = 0
for i in range(1, N):
ans += max(0, A[i] - A[i - 1])
A[i] = max(A[i], A[i - 1])
print(ans)
|
s088041686
|
Accepted
| 145 | 32,288 | 171 |
N = int(input())
A = list(map(int, input().strip().split()))
ans = 0
for i in range(1, N):
if A[i] < A[i - 1]:
ans += A[i - 1] - A[i]
A[i] = A[i - 1]
print(ans)
|
s123406077
|
p02392
|
u655879321
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,584 | 95 |
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
a, b, c = map(int, input().split())
if a > b and b > c:
print('Yes')
else:
print('No')
|
s723441003
|
Accepted
| 20 | 5,584 | 95 |
a, b, c = map(int, input().split())
if a < b and b < c:
print('Yes')
else:
print('No')
|
s508817499
|
p03192
|
u359474860
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 24 |
You are given an integer N that has exactly four digits in base ten. How many times does `2` occur in the base-ten representation of N?
|
N = input()
N.count("2")
|
s441181030
|
Accepted
| 17 | 2,940 | 31 |
N = input()
print(N.count("2"))
|
s350541139
|
p02612
|
u247680229
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,148 | 38 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N=int(input())
ans=N%1000
print(ans)
|
s279651929
|
Accepted
| 29 | 9,160 | 116 |
N=int(input())
if N%1000==0:
print(0)
exit()
elif N>=1000:
print(1000-N%1000)
elif N<1000:
print(1000-N)
|
s194481997
|
p02422
|
u480053997
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,548 | 316 |
Write a program which performs a sequence of commands to a given string $str$. The command is one of: * print a b: print from the a-th character to the b-th character of $str$ * reverse a b: reverse from the a-th character to the b-th character of $str$ * replace a b p: replace from the a-th character to the b-th character of $str$ with p Note that the indices of $str$ start with 0.
|
s = input()
for i in range(int(input())):
print(s)
q = input().split()
op, a, b = q[0], int(q[1]), int(q[2])
if op == 'print':
print(s[a : b+1])
elif op == 'reverse':
t = s[a : b+1]
s = s[:a] + t[::-1] + s[b+1:]
elif op == 'replace':
s = s[:a] + q[3] + s[b+1:]
|
s616408203
|
Accepted
| 60 | 7,728 | 303 |
s = input()
for i in range(int(input())):
q = input().split()
op, a, b = q[0], int(q[1]), int(q[2])
if op == 'print':
print(s[a : b+1])
elif op == 'reverse':
t = s[a : b+1]
s = s[:a] + t[::-1] + s[b+1:]
elif op == 'replace':
s = s[:a] + q[3] + s[b+1:]
|
s134481568
|
p03943
|
u143903328
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 176 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a = list(map(int, input().split()))
sum = a[0] + a[1] + a[2]
if sum - a[0] == sum / 2 or sum - a[0] == sum / 2 or sum - a[2] == sum / 2:
print("Yes")
else:
print("No")
|
s060720231
|
Accepted
| 17 | 2,940 | 176 |
a = list(map(int, input().split()))
sum = a[0] + a[1] + a[2]
if sum - a[0] == sum / 2 or sum - a[1] == sum / 2 or sum - a[2] == sum / 2:
print("Yes")
else:
print("No")
|
s656827717
|
p03569
|
u210827208
| 2,000 | 262,144 |
Wrong Answer
| 63 | 3,572 | 420 |
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.
|
from collections import Counter
S=input()
if S==S[::-1]:
print(0)
else:
X=Counter(S)
if 'x' in X.keys() and len(X.keys())==2:
cnt=0
ans=0
for s in S:
if s!='x':
cnt+=1
else:
ans+=1
if cnt==-(-(len(S)-X['x'])//2):
break
print(max(ans,X['x']-ans)-min(ans,X['x']-ans))
else:
print(0)
|
s841835623
|
Accepted
| 70 | 3,316 | 270 |
S=input()
n=len(S)
l=0
r=n-1
ans=0
while l+1<=r:
if S[l]=='x' and S[l]!=S[r]:
ans+=1
l+=1
elif S[r]=='x' and S[l]!=S[r]:
ans+=1
r-=1
elif S[l]==S[r]:
r-=1
l+=1
else:
ans=-1
break
print(ans)
|
s592557318
|
p03417
|
u626468554
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 52 |
There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region. The front and back sides of these cards can be distinguished, and initially every card faces up. We will perform the following operation once for each square contains a card: * For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square. It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed. Find the number of cards that face down after all the operations.
|
n,m = map(int,input().split())
n-=1
m-=1
print(n*m)
|
s885362255
|
Accepted
| 17 | 2,940 | 143 |
n,m = map(int,input().split())
if n>=2 and m>=2:
n-=2
m-=2
print(n*m)
elif n==1 and m==1:
print(1)
else:
print(max(n,m)-2)
|
s585170098
|
p02394
|
u748921161
| 1,000 | 131,072 |
Wrong Answer
| 50 | 7,568 | 211 |
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
input_str = input().split(' ')
W = int(input_str[0])
H = int(input_str[0])
x = int(input_str[0])
y = int(input_str[0])
r = int(input_str[0])
print('Yes' if x-r > 0 and x+r < W and y-r > 0 and y+r < H else 'No')
|
s974790994
|
Accepted
| 60 | 7,720 | 215 |
input_str = input().split(' ')
W = int(input_str[0])
H = int(input_str[1])
x = int(input_str[2])
y = int(input_str[3])
r = int(input_str[4])
print('Yes' if x-r >= 0 and x+r <= W and y-r >= 0 and y+r <= H else 'No')
|
s822560554
|
p03598
|
u153823221
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 269 |
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 = int(input())
k = int(input())
x = list(map(int, input().split()))
distance = []
for i in range(n):
if abs(x[i] - 0) < abs(x[i] - k):
distance.append(abs((x[i] - 0) * 2))
else:
distance.append(abs((x[i] - k) * 2))
print(distance)
print(sum(distance))
|
s631841480
|
Accepted
| 17 | 3,060 | 253 |
n = int(input())
k = int(input())
x = list(map(int, input().split()))
distance = []
for i in range(n):
if abs(x[i] - 0) < abs(x[i] - k):
distance.append(abs((x[i] - 0) * 2))
else:
distance.append(abs((x[i] - k) * 2))
print(sum(distance))
|
s322219170
|
p02613
|
u798260206
| 2,000 | 1,048,576 |
Wrong Answer
| 143 | 9,216 | 299 |
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())
AC = 0
WA = 0
TLE = 0
RE = 0
for i in range(N):
s = input()
if s == "AC":
AC += 1
elif s == "WA":
WA += 1
elif s == "TLE":
TLE +=1
else:
RE += 1
print("AC × " + str(AC))
print("WA × " + str(WA))
print("TLE × " + str(TLE))
print("RE × " + str(RE))
|
s131302717
|
Accepted
| 141 | 9,208 | 295 |
N = int(input())
AC = 0
WA = 0
TLE = 0
RE = 0
for i in range(N):
s = input()
if s == "AC":
AC += 1
elif s == "WA":
WA += 1
elif s == "TLE":
TLE +=1
else:
RE += 1
print("AC x " + str(AC))
print("WA x " + str(WA))
print("TLE x " + str(TLE))
print("RE x " + str(RE))
|
s657269100
|
p03795
|
u952708174
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 41 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
N = int(input())
print(800*N - 200*N//15)
|
s281926702
|
Accepted
| 18 | 2,940 | 43 |
N = int(input())
print(800*N - 200*(N//15))
|
s196277813
|
p03855
|
u786020649
| 2,000 | 262,144 |
Wrong Answer
| 740 | 138,032 | 938 |
There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities. We will say city A and B are _connected by roads_ if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define _connectivity by railways_ similarly. For each city, find the number of the cities connected to that city by both roads and railways.
|
import sys
from collections import deque
from collections import defaultdict
from collections import Counter
def conn(n,m,e):
d=dict(zip(range(1,n+1),range(1,n+1)))
c=0
for edge in e:
a=edge[0]
b=edge[1]
da=d[a]
db=d[b]
if da==m and db==m:
d[a]=c
d[b]=c
c+=1
else:
d[a]=min(da,db)
d[b]=min(da,db)
return d.values()
def main(n,k,l,e1,e2):
d1=conn(n,k,e1)
d2=conn(n,l,e2)
p=tuple(zip(iter(d1),iter(d2)))
print(p)
d=Counter(p)
print(d)
d[(k,l)]=1
print(' '.join([str(d[x]) for x in p]))
if __name__=='__main__':
ssr=sys.stdin.readline
n,k,l=map(int,ssr().strip().split())
e1=[]
e2=[]
for _ in range(k):
e1.append(tuple(map(int,ssr().strip().split())))
for _ in range(l):
e2.append(tuple(map(int,ssr().strip().split())))
main(n,k,l,e1,e2)
|
s102059299
|
Accepted
| 970 | 191,464 | 3,413 |
import sys
from collections import deque
from collections import defaultdict
from collections import Counter
def conn(n,m,e):
d=dict(zip(range(1,n+1),range(-1,(-1)*n-1,-1)))
td=defaultdict(lambda:deque([]))
c=1
for edge in e:
a=edge[0]
b=edge[1]
da=d[a]
db=d[b]
if da<0 and db<0:
d[a]=c
d[b]=c
td[c].append(a)
td[c].append(b)
c+=1
elif da>0 and db<0:
d[b]=da
td[d[a]].append(b)
elif da<0 and db>0:
d[a]=db
td[d[b]].append(a)
elif da>0 and db>0 and da!=db:
for x in td[db]:
d[x]=da
td[da].append(x)
return list(d.values())
# ed=defaultdict(lambda:deque())
# for edge in e:
# ed[edge[0]].append(edge[1])
# c=0
# s=[0]*n
# label=[0]*n
# if s[i-1]==0:
# c+=1
# label[c-1]=c
# stack=deque([i])
# while stack:
# w=stack.pop()
# s[w-1]=c
# while ed[w]:
# wn=ed[w].pop()
# if s[wn-1]==0:
# s[wn-1]=c
# if ed[wn]:
# stack.append(w)
# w=wn
# elif s[wn-1]<c:
# label[s[wn-1]-1]=c
# print(s)
# print(label)
def components(n,k,e):
ed=defaultdict(lambda:deque())
for edge in e:
ed[edge[0]].append(edge[1])
ed[edge[1]].append(edge[0])
c=0
s=[0]*n
stack=deque()
for i in range(1,n+1):
if s[i-1]==0:
c+=1
stack.clear()
stack.append(i)
while stack:
w=stack.pop()
s[w-1]=c
while ed[w]:
wn=ed[w].pop()
if s[wn-1]==0:
s[wn-1]=c
if ed[wn]:
stack.append(w)
w=wn
return [s[i] for i in range(n)]
def components2(n,k,e):
es=deque(set((edge[0],edge[1])) for edge in e)
c=0
s=deque()
stack=deque()
while es:
f=es.pop()
stack.clear()
while es:
l=len(es)
g=es.pop()
if g&f:
f|=g
else:
stack.append(g)
if l==len(stack):
s.append(f)
break
es=stack
t=[0]*n
c=0
while s:
for x in s.pop():
t[x-1]=c
c+=1
return t
def main(n,k,l,e1,e2):
d1=components(n,k,e1)
d2=components(n,l,e2)
p=tuple(zip(iter(d1),iter(d2)))
d=Counter(p)
# print(d1,d2,d,p)
print(' '.join([str(d[x]) for x in p]))
if __name__=='__main__':
ssr=sys.stdin.readline
n,k,l=map(int,ssr().strip().split())
e1=[]
e2=[]
for _ in range(k):
e1.append(tuple(map(int,ssr().strip().split())))
for _ in range(l):
e2.append(tuple(map(int,ssr().strip().split())))
main(n,k,l,e1,e2)
|
s099246027
|
p02608
|
u075317232
| 2,000 | 1,048,576 |
Time Limit Exceeded
| 2,205 | 8,860 | 937 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
def XYZTriplets():
num = int(input())
for i in range(num):
count = 0
x = 1
y = 1
z = 1
flag = True
while flag == True:
function = 0.5*( (x+y)*(x+y) + (y+z)*(y+z) + (z+x)*(z+x) ) - i
if function == 0:
if x == y and y == z and z == x:
count = count + 1
else:
count = count + 3
list_xyz = [x, y, z]
if list_xyz.index(max(list_xyz)) == 0:
x = x + 1
elif list_xyz.index(max(list_xyz)) == 1:
y = y + 1
else:
z = z + 1
if function > 0:
print(count)
break
if __name__ == '__main__':
XYZTriplets()
|
s772994456
|
Accepted
| 395 | 27,216 | 650 |
import math
import numpy as np
def XYZTriplets():
num = int(input())
list_num = []
count = [0]*num
for i in range(1, num+1):
list_num.append(i)
for x in range(1, int(math.sqrt(num))):
for y in range(1, int(math.sqrt(num))):
for z in range(1, int(math.sqrt(num))):
function = x*x + y*y + z*z + x*y + y*z + z*x
if function <= num and function > 0:
count[function-1] = count[function-1] + 1
for i in range(0, num):
print(count[i])
if __name__ == '__main__':
XYZTriplets()
|
s448432067
|
p04029
|
u888380104
| 2,000 | 262,144 |
Wrong Answer
| 24 | 9,124 | 49 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N = int(input())
x = N * (N * 1) // 2
print(x)
|
s467933123
|
Accepted
| 25 | 9,152 | 49 |
N = int(input())
x = N * (N + 1) // 2
print(x)
|
s363242157
|
p02664
|
u932370518
| 2,000 | 1,048,576 |
Wrong Answer
| 69 | 9,528 | 310 |
For a string S consisting of the uppercase English letters `P` and `D`, let the _doctoral and postdoctoral quotient_ of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3. We have a string T consisting of `P`, `D`, and `?`. Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
|
if __name__ == "__main__":
T = input()
ans = ""
for i in range(len(T)-1):
if T[i] == '?':
if T[i+1] == 'D':
ans += 'P'
else:
ans += 'D'
else:
ans += T[i]
if T[-1] == '?':
ans += 'D'
print(ans)
|
s155777421
|
Accepted
| 25 | 9,516 | 203 |
if __name__ == "__main__":
T = input()
ans = T
ans = ans.replace("P?", 'PD')
ans = ans.replace("??", 'PD')
ans = ans.replace("?D", 'PD')
ans = ans.replace("?", 'D')
print(ans)
|
s828952405
|
p03150
|
u494058663
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 3,060 | 255 |
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
import sys
S = list(input())
Ans_list=['k','e','y','e','n','c','e']
Ans = 0
for i in range(len(S)):
for j in range(i,len(S),+1):
if (S[:i:]+S[j::]) == Ans_list:
Ans = 1
print(Ans)
sys.exit()
print(Ans)
|
s254480392
|
Accepted
| 21 | 3,060 | 263 |
import sys
S = list(input())
Ans_list=['k','e','y','e','n','c','e']
Ans = 'NO'
for i in range(len(S)):
for j in range(i,len(S),+1):
if (S[:i:]+S[j::]) == Ans_list:
Ans = 'YES'
print(Ans)
sys.exit()
print(Ans)
|
s614863539
|
p02608
|
u821432765
| 2,000 | 1,048,576 |
Wrong Answer
| 854 | 9,372 | 240 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
N = int(input())
F = [0]*(N+1)
for x in range(1, 101):
for y in range(1, 101):
for z in range(1, 101):
c = x**2 + y**2 + z**2 + x*y + y*z + z*x
if (c <= N):
F[c] += 1
print(*F, sep='\n')
|
s487939477
|
Accepted
| 873 | 9,204 | 244 |
N = int(input())
F = [0]*(N+1)
for x in range(1, 101):
for y in range(1, 101):
for z in range(1, 101):
c = x**2 + y**2 + z**2 + x*y + y*z + z*x
if (c <= N):
F[c] += 1
print(*F[1:], sep='\n')
|
s257364945
|
p03814
|
u994521204
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,500 | 41 |
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s=input()
print(s.find('A')-s.rfind('Z'))
|
s202835233
|
Accepted
| 17 | 3,500 | 43 |
s=input()
print(s.rfind('Z')-s.find('A')+1)
|
s568760535
|
p03469
|
u600789379
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,004 | 31 |
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
print('2018/01' + input()[-2:])
|
s832959228
|
Accepted
| 29 | 9,044 | 32 |
print('2018/01/' + input()[-2:])
|
s089917790
|
p02865
|
u420399048
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 95 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
def f(N):
if N%2==0:
return N//2-1
else:
return N//2
N=int(input())
print(f(N))
|
s844946917
|
Accepted
| 17 | 2,940 | 103 |
def f(N):
if N%2==0:
return N//2-1
else:
return N//2
N=int(input())
print(f(N))
|
s078975031
|
p02850
|
u875291233
| 2,000 | 1,048,576 |
Wrong Answer
| 244 | 34,004 | 825 |
Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.
|
def top_sort_tree(g,root):
n = len(g)
parent = [-1]*n
top_order = [-1]*n
pos = 0
from collections import deque
q = deque([root])
while q:
v = q.pop()
top_order[pos] = v
pos += 1
for c in g[v]:
if c != parent[v]:
parent[c] = v
q.append(c)
return top_order
#####################################
# coding: utf-8
# Your code here!
import sys
read = sys.stdin.read
readline = sys.stdin.readline
readlines = sys.stdin.readlines
#n = int(input())
#ab = [[int(i) for i in readline().split()] for _ in range(n)]
n = int(input())
ab = map(int,read().split())
g = [[] for _ in range(n)]
for a,b in zip(ab,ab):
g[a-1].append(b-1)
g[b-1].append(a-1)
#print(g)
print(top_sort_tree(g,0)[0])
|
s542407510
|
Accepted
| 531 | 77,784 | 678 |
# coding: utf-8
# Your code here!
import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline
n = int(input())
g = [[] for _ in range(n)]
d = {}
for i in range(n-1):
a,b = [int(i) for i in readline().split()]
d[(a-1,b-1)] = i
g[a-1].append(b-1)
g[b-1].append(a-1)
def dfs(v,p,j): #p: parent of v
num = -1
for c in g[v]:
if c == p: continue
num += 1
col = (j+num)%m+1
if c > v: ans[d[(v,c)]] = col
else: ans[d[(c,v)]] = col
dfs(c,v,col)
m = max(len(i) for i in g)
ans = [0]*(n-1)
dfs(0,-1,0)
#print(g,m,ans)
print(m)
print(*ans,sep = "\n")
|
s439076150
|
p03470
|
u686036872
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 69 |
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())
list=[int(input()) for i in range(N)]
print(len(list))
|
s517493614
|
Accepted
| 20 | 3,060 | 95 |
N = int(input())
list=[]
for i in range(N):
list.append(int(input()))
print(len(set(list)))
|
s541758868
|
p03339
|
u072717685
| 2,000 | 1,048,576 |
Wrong Answer
| 319 | 40,176 | 547 |
There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions.
|
import sys
read = sys.stdin.read
from itertools import accumulate
def main():
n = int(input())
s = list(input())
e = [0] * n
w = [0] * n
for i1 in range(n):
e[i1] = s[i1] == 'E'
w[i1] = s[i1] == 'W'
e.insert(0, 0)
e.append(0)
w.insert(0, 0)
w.append(0)
ea = list(accumulate(e))
wa = list(accumulate(w))
rt = 0
for j1 in range(1, n):
rt = max(rt, ea[j1 - 1] + (wa[-1] - wa[j1]))
print(j1, rt)
r = n - rt - 1
print(r)
if __name__ == '__main__':
main()
|
s016524921
|
Accepted
| 173 | 51,768 | 568 |
import sys
read = sys.stdin.read
readlines = sys.stdin.readlines
from itertools import accumulate
def main():
n = int(input())
s = list(input())
enum = [1 if c == 'E' else 0 for c in s]
wnum = [1 if c == 'W' else 0 for c in s]
enuma = tuple(accumulate(enum))
wnuma = tuple(accumulate(wnum))
wnumm = wnuma[-1]
wnuma2 = [wnumm - wn for wn in wnuma]
num = max(wnuma2[0], enuma[-2])
for i1 in range(1, n - 1):
num = max(num, enuma[i1 - 1] + wnuma2[i1])
r = n - num - 1
print(r)
if __name__ == '__main__':
main()
|
s092992397
|
p03971
|
u545644875
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,956 | 517 |
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
x = list(map(int, input().split()))
s = list(input())
y = len(s)
rank = 0
frank = 0
for i in (0,y - 1):
if s[i] == "c":
print("NO")
elif s[i] == "b":
if rank < x[1] + x[2] and frank < x[2]:
rank += 1
frank += 1
print("YES")
else:
print("NO")
else:
if rank < x[1] + x[2]:
rank += 1
print("YES")
else:
print("NO")
|
s904565999
|
Accepted
| 112 | 4,708 | 489 |
li = list(map(int,input().split()))
s=list(input())
cnt = 0
b = 0
for i in s:
if i == "a":
if cnt < li[1] + li[2]:
cnt += 1
print("Yes")
else:
print("No")
elif i == "b":
b += 1
if cnt < li[1] + li[2]:
if b <= li[2]:
cnt += 1
print("Yes")
else:
print("No")
else:
print("No")
else:
#cnt -= 1
print("No")
|
s754710764
|
p00523
|
u847467233
| 8,000 | 262,144 |
Wrong Answer
| 30 | 5,608 | 607 |
JOI 君は妹のJOI 子ちゃんとJOI 美ちゃんと一緒におやつを食べようとしている．今日のおやつは3 人の大好物のバームクーヘンだ． バームクーヘンは下図のような円筒形のお菓子である．3 人に分けるために，JOI 君は半径方向に刃を3回入れて，これを3 つのピースに切り分けなければならない．ただしこのバームクーヘンは本物の木材のように固いので，刃を入れるのは簡単ではない．そのためこのバームクーヘンにはあらかじめ $N$ 個の切れ込みが入っており，JOI 君は切れ込みのある位置でのみ切ることができる．切れ込みに1 から $N$ まで時計回りに番号をふったとき，$1 \leq i \leq N - 1$ に対し， $i$ 番目の切れ込みと$i + 1$ 番目の切れ込みの間の部分の大きさは $A_i$ である．また $N$ 番目の切れ込みと1 番目の切れ込みの間の部分の大きさは $A_N$ である． 図1: バームクーヘンの例 $N = 6, A_1 = 1, A_2 = 5, A_3 = 4, A_4 = 5, A_5 = 2, A_6 = 4$
|
# Python3 2018.7.3 bal4u
import sys
from sys import stdin
input = stdin.readline
n = int(input())
a = [int(input()) for i in range(n)]
su = sum(a)
p = [0]*(n+1)
s = [0]*(n+1)
print("a=",a)
l, r = 0, su+1
while l + 1 < r:
m = (l+r) >> 1
for i in range(0, n):
if i == 0: s[i] = p[i] = 0
else:
p[i] = p[i-1] - 1
s[i] = s[i-1] - a[i-1]
while s[i] < m:
ii = i + p[i];
if ii >= n: ii -= n
s[i] += a[ii]
p[i] += 1
f = 0
for i in range(n):
ii = i + p[i];
if ii >= n: ii -= n
if su-s[i]-s[ii] >= m:
f = 1
break
if f: l = m
else: r = m
print(l)
|
s934696176
|
Accepted
| 9,330 | 18,072 | 593 |
# Python3 2018.7.3 bal4u
import sys
from sys import stdin
input = stdin.readline
n = int(input())
a = [int(input()) for i in range(n)]
su = sum(a)
p = [0]*(n+1)
s = [0]*(n+1)
l, r = 0, su+1
while l + 1 < r:
m = (l+r) >> 1
for i in range(0, n):
if i == 0: s[i] = p[i] = 0
else:
p[i] = p[i-1] - 1
s[i] = s[i-1] - a[i-1]
while s[i] < m:
ii = i + p[i];
if ii >= n: ii -= n
s[i] += a[ii]
p[i] += 1
f = 0
for i in range(n):
ii = i + p[i];
if ii >= n: ii -= n
if su-s[i]-s[ii] >= m:
f = 1
break
if f: l = m
else: r = m
print(l)
|
s504709197
|
p03729
|
u155251346
| 2,000 | 262,144 |
Wrong Answer
| 31 | 8,972 | 121 |
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
A, B, C = input().split()
if list(A[-1])==list(B[0]) and list(B[-1])==list(C[0]):
print("Yes")
else:
print("No")
|
s291858671
|
Accepted
| 31 | 9,088 | 128 |
A, B, C = input().split()
if A[-1]==B[0] and B[-1]==C[0]:
print("YES")
else:
print("NO")
|
s579561120
|
p03827
|
u629540524
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,092 | 156 |
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
n = int(input())
s = input()
x = 0
y = 0
for i in range(n):
if s[i] == 'I':
y += 1
else:
y -= 1
if y > 0:
x = y
print(x)
|
s435914044
|
Accepted
| 29 | 8,996 | 156 |
n = int(input())
s = input()
x = 0
y = 0
for i in range(n):
if s[i] == 'I':
y += 1
else:
y -= 1
if y > x:
x = y
print(x)
|
s750344090
|
p03721
|
u813102292
| 2,000 | 262,144 |
Wrong Answer
| 398 | 11,372 | 283 |
There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3.
|
#ABC061
n,k = (int(i) for i in input().split())
a = []
b = []
for i in range(n):
tmp = list(int(i) for i in input().split())
a.append(tmp[0])
b.append(tmp[1])
tmp = 0
tmp_i = 0
for i in range(n):
tmp += b[i]
if tmp>=n:
tmp_i = i
break
print(a[i])
|
s462813796
|
Accepted
| 463 | 15,064 | 435 |
#ABC061
N,K = (int(i) for i in input().split())
a = {}
for i in range(N):
tmp = list(int(i) for i in input().split())
if not tmp[0] in a.keys():
a[tmp[0]] = tmp[1]
else:
a[tmp[0]] += tmp[1]
ans_array = []
for k,v in a.items():
ans_array.append((k,v))
ans_array = sorted(ans_array)
count = 0
for i in range(N):
count += ans_array[i][1]
if count>=K:
print(ans_array[i][0])
break
|
s941623718
|
p02842
|
u343128979
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 2,940 | 310 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
import math
def inputs():
return [int(s) for s in input().split()]
def main():
N = int(input())
v = math.ceil(N / 1.08)
print(v)
print(math.floor(v * 1.08))
if math.floor(v * 1.08) == N:
return v
else:
return ':('
if __name__ == '__main__':
print(main())
|
s101461089
|
Accepted
| 19 | 2,940 | 266 |
import math
def inputs():
return [int(s) for s in input().split()]
def main():
N = int(input())
v = math.ceil(N / 1.08)
if math.floor(v * 1.08) == N:
return v
else:
return ':('
if __name__ == '__main__':
print(main())
|
s493119901
|
p02393
|
u661529494
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,548 | 51 |
Write a program which reads three integers, and prints them in ascending order.
|
A=list(map(int, input().split()))
A.sort()
print(A)
|
s520257652
|
Accepted
| 50 | 7,616 | 64 |
A=list(map(int, input().split()))
A.sort()
print(A[0],A[1],A[2])
|
s319068688
|
p03478
|
u033642300
| 2,000 | 262,144 |
Wrong Answer
| 33 | 3,776 | 421 |
Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).
|
def main():
N, A, B = map(int, input().split())
list = []
for i in range(N + 1):
if 10 < i:
temp = str(i)
ai = int(temp[0])
bi = int(temp[1])
if A - 1 < ai + bi < B + 1:
print(i)
list.append(i)
else:
if A - 1 < i < B + 1:
print(i)
list.append(i)
print(sum(list))
main()
|
s735236873
|
Accepted
| 41 | 9,172 | 253 |
def main():
N, A, B = map(int, input().split())
ans = 0
for i in range(N+1):
tw = str(i)
tn = 0
for j in range(len(tw)):
tn += int(tw[j])
if A <= tn <= B:
ans += i
print(ans)
main()
|
s691304510
|
p03504
|
u296150111
| 2,000 | 262,144 |
Wrong Answer
| 363 | 3,828 | 186 |
Joisino is planning to record N TV programs with recorders. The TV can receive C channels numbered 1 through C. The i-th program that she wants to record will be broadcast from time s_i to time t_i (including time s_i but not t_i) on Channel c_i. Here, there will never be more than one program that are broadcast on the same channel at the same time. When the recorder is recording a channel from time S to time T (including time S but not T), it cannot record other channels from time S-0.5 to time T (including time S-0.5 but not T). Find the minimum number of recorders required to record the channels so that all the N programs are completely recorded.
|
n,c=map(int,input().split())
time=[0]*(10**5)
for i in range(n):
s,t,c=map(int,input().split())
time[s-1]+=1
time[t-1]-=1
for i in range(10**5-1):
time[i+1]+=time[i]
print(max(time))
|
s202537498
|
Accepted
| 588 | 28,472 | 474 |
from itertools import accumulate
n,c=map(int,input().split())
time=[0]*(10**5+2)
stc=[]
for i in range(n):
s,t,c=map(int,input().split())
stc.append([s,t,c])
stc.sort(key=lambda x:(x[2],x[0]))
for i in range(n-1):
if stc[i][2]==stc[i+1][2]:
if stc[i][1]==stc[i+1][0]:
stc[i+1][0]=stc[i][0]
stc[i]=[-1]
lis=[]
for i in range(n):
if stc[i]!=[-1]:
lis.append(stc[i])
for i in range(len(lis)):
time[lis[i][0]-1]+=1
time[lis[i][1]]-=1
print(max(accumulate(time)))
|
s751306450
|
p03378
|
u368796742
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 140 |
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.
|
import bisect
n,m,x = map(int,input().split())
l = list(map(int,input().split()))
index = bisect.bisect_left(l,x)
print(min(index,n-index))
|
s380420199
|
Accepted
| 17 | 2,940 | 141 |
import bisect
n,m,x = map(int,input().split())
l = list(map(int,input().split()))
index = bisect.bisect_left(l,x)
print(min(index,m-index))
|
s527651887
|
p02612
|
u206570055
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,060 | 26 |
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.
|
print(int(input()) % 1000)
|
s440372611
|
Accepted
| 32 | 9,060 | 82 |
ans = int(input()) % 1000
if ans == 0:
print(0)
exit(0)
print(1000 - ans)
|
s036701518
|
p03861
|
u535659144
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 3,064 | 112 |
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?
|
x = list(map(int,input().split()))
a = 0
for i in range(x[0],x[1]):
if i % x[2]==0:
a = a+1
print(a)
|
s608016183
|
Accepted
| 17 | 2,940 | 142 |
x = list(map(int,input().split()))
a = 0
if x[0]%x[2]==0:
b = x[0]
else:
b = ((x[0] // x[2])+1)*x[2]
c = (x[1]-b)//x[2] + 1
print(c)
|
s592365724
|
p03997
|
u076498277
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
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, b, h = map(int,open(0).read().split())
print((a+b)* h / 2)
|
s883905321
|
Accepted
| 17 | 2,940 | 70 |
a, b, h = map(int,open(0).read().split())
print(round(((a+b)* h / 2)))
|
s196134744
|
p03573
|
u428012835
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 90 |
You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers.
|
num = input().split()
print(num)
for i in num:
if num.count(i) == 1:
print(i)
|
s539044223
|
Accepted
| 18 | 3,064 | 107 |
numbers = list(map(int, input().split()))
for i in numbers:
if numbers.count(i) == 1:
print(i)
|
s484591221
|
p03607
|
u294385082
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,112 | 146,072 | 151 |
You are playing the following game with Joisino. * Initially, you have a blank sheet of paper. * Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times. * Then, you are asked a question: How many numbers are written on the sheet now? The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?
|
n = int(input())
a = [int(input()) for i in range(n)]
lis = []
for i in a:
if i in lis:
a.remove(i)
else:
a.append(i)
print(len(lis))
|
s890565064
|
Accepted
| 214 | 19,852 | 235 |
n = int(input())
a = [int(input()) for i in range(n)]
dict = {}
count = 0
for i in a:
if i not in dict:
dict[i] = 1
elif i in dict:
dict[i]+= 1
for i in set(a):
if dict[i]%2!=0:
count += 1
print(count)
|
s731975591
|
p03693
|
u117078923
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 99 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
l=[int(x) for x in input().split()]
if (l[0]+l[1]+l[2])%4 == 0:
print("Yes")
else:
print("No")
|
s918793716
|
Accepted
| 17 | 2,940 | 83 |
r,g,b=map(int,input().split())
if (10*g+b)%4==0:
print('YES')
else:
print('NO')
|
s418420322
|
p03024
|
u767438459
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,036 | 70 |
Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s=input()
if s.count("x") >= 8:
print("No")
else:
print("Yes")
|
s153204231
|
Accepted
| 28 | 9,092 | 382 |
s=input()
if s.count("x") >= 8:
print("NO")
else:
print("YES")
|
s106997471
|
p03380
|
u761989513
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 14,052 | 229 |
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.
|
n = int(input())
a = list(map(int, input().split()))
half = max(a) / 2
sa = float("inf")
ans = 0
for i in a:
if i != max(a):
if abs(half - i) < sa:
ans = i
sa = abs(half - i)
print(a[0], ans)
|
s753554581
|
Accepted
| 70 | 14,428 | 227 |
n = int(input())
a = list(map(int, input().split()))
m = max(a)
half = m / 2
sa = float("inf")
ans = 0
for i in a:
if i != m:
if abs(half - i) < sa:
ans = i
sa = abs(half - i)
print(m, ans)
|
s178218259
|
p02397
|
u756468156
| 1,000 | 131,072 |
Wrong Answer
| 60 | 7,624 | 136 |
Write a program which reads two integers x and y, and prints them in ascending order.
|
while True:
xy_list = list(map(int, input().split()))
if xy_list[0] == xy_list[1] == 0:
break
print(sorted(xy_list))
|
s301953843
|
Accepted
| 60 | 7,528 | 137 |
while True:
xy_list = list(map(int, input().split()))
if xy_list[0] == xy_list[1] == 0:
break
print(*sorted(xy_list))
|
s859740593
|
p02692
|
u316649390
| 2,000 | 1,048,576 |
Wrong Answer
| 207 | 10,020 | 376 |
There is a game that involves three variables, denoted A, B, and C. As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or C then subtract 1 from the other; if s_i is `BC`, you must add 1 to B or C then subtract 1 from the other. After each choice, none of A, B, and C should be negative. Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.
|
N,A,B,C = map(int,input().split())
data = {
"A":A,
"B":B,
"C":C
}
select = []
for _ in range(N):
s = input()
if data[s[0]] == data[s[1]] ==0:
print("No")
exit(0)
elif data[s[0]] >= data[s[1]]:
data[s[0]] -= 1
data[s[1]] += 1
select.append(s[1])
else:
data[s[0]] += 1
data[s[1]] -= 1
select.append(s[0])
for i in select:
print(i)
|
s902732176
|
Accepted
| 225 | 17,056 | 645 |
N,A,B,C = map(int,input().split())
data = {
"A":A,
"B":B,
"C":C
}
select = []
S = list(input() for _ in range(N))
for i in range(N):
s = S[i]
if data[s[0]] == data[s[1]] ==0:
print("No")
exit(0)
elif data[s[0]] > data[s[1]]:
data[s[0]] -= 1
data[s[1]] += 1
select.append(s[1])
elif data[s[0]] < data[s[1]]:
data[s[0]] += 1
data[s[1]] -= 1
select.append(s[0])
elif i == N-1:
select.append(s[0])
elif s[0] in S[i+1]:
data[s[0]] += 1
data[s[1]] -= 1
select.append(s[0])
else:
data[s[0]] -= 1
data[s[1]] += 1
select.append(s[1])
print("Yes")
for i in select:
print(i)
|
s816740587
|
p03738
|
u498401785
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 119 |
You are given two positive integers A and B. Compare the magnitudes of these numbers.
|
a = input()
b = input()
if(a > b):
print("GRATER")
elif(a < b):
print("LESS")
elif(a == b):
print("EQUAL")
|
s524704592
|
Accepted
| 18 | 2,940 | 130 |
a = int(input())
b = int(input())
if(a > b):
print("GREATER")
elif(a < b):
print("LESS")
elif(a == b):
print("EQUAL")
|
s476355279
|
p03227
|
u846150137
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 48 |
You are given a string S of length 2 or 3 consisting of lowercase English letters. If the length of the string is 2, print it as is; if the length is 3, print the string after reversing it.
|
a=input()
print(a if len(a)==2 else reversed(a))
|
s143574195
|
Accepted
| 17 | 3,064 | 44 |
a=input()
print(a if len(a)==2 else a[::-1])
|
s331454505
|
p02742
|
u220394041
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 93 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
import math
h, w = map(int, input().split())
ans = math.ceil(h * w / 2)
print(AssertionError)
|
s278384429
|
Accepted
| 19 | 3,188 | 115 |
import math
h, w = map(int, input().split())
ans = math.ceil(h * w / 2)
if h == 1 or w == 1:
ans = 1
print(ans)
|
s204427826
|
p04012
|
u982517812
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 352 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
s = str(input())
t = list(s)
t.sort()
a = []
b = 1
for i in range(len(t)):
if i >= 1:
if t[i] == t[i-1]:
b = b + 1
else:
a.append(b)
b = 1
a.append(b)
for i in range(len(a)):
if a[i] % 2 != 0:
print("NO")
break
else:
if i == len(a) - 1:
print("YES")
|
s700912232
|
Accepted
| 17 | 3,064 | 352 |
s = str(input())
t = list(s)
t.sort()
a = []
b = 1
for i in range(len(t)):
if i >= 1:
if t[i] == t[i-1]:
b = b + 1
else:
a.append(b)
b = 1
a.append(b)
for i in range(len(a)):
if a[i] % 2 != 0:
print("No")
break
else:
if i == len(a) - 1:
print("Yes")
|
s061908612
|
p03557
|
u664907598
| 2,000 | 262,144 |
Wrong Answer
| 511 | 31,484 | 389 |
The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower. He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i. To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar. How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.
|
import bisect
import numpy as np
n = int(input())
a = sorted(list(map(int,input().split())))
b = sorted(list(map(int,input().split())))
c = sorted(list(map(int,input().split())))
memo = [0] * n
for i in range(n):
memo[i] = n - bisect.bisect_right(c,b[i])
memo.reverse()
memo1 = np.cumsum(memo)
ans = 0
for i in range(n):
ans += memo1[n-1-bisect.bisect_right(b,a[i])]
print(ans)
|
s409785468
|
Accepted
| 481 | 31,476 | 310 |
import bisect
import numpy as np
n = int(input())
a = sorted(list(map(int,input().split())))
b = sorted(list(map(int,input().split())))
c = sorted(list(map(int,input().split())))
ans = 0
for i in range(n):
s = bisect.bisect_left(a,b[i])
t = n - bisect.bisect_right(c,b[i])
ans += s*t
print(ans)
|
s353915173
|
p03693
|
u787562674
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 79 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r, g, b = map(int, input().split())
print("YES" if (g * 10 + b) % 4 else "NO")
|
s494290321
|
Accepted
| 17 | 2,940 | 84 |
r, g, b = map(int, input().split())
print("YES" if (g * 10 + b) % 4 == 0 else "NO")
|
s745601479
|
p03455
|
u407016024
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 90 |
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('EVEN')
else:
print('ODD')
|
s263729230
|
Accepted
| 17 | 2,940 | 90 |
a, b = map(int, input().split())
if (a*b)%2 == 0:
print('Even')
else:
print('Odd')
|
s585225507
|
p03110
|
u902462889
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 258 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
N = int(input())
lst_1 = []
ans = 0
for n in range(N):
lst_1.append(input().split())
print(lst_1)
for n in range(N):
money = 0
money = float(lst_1[n][0])
if lst_1[n][1] == "BTC":
money = money * 380000
ans += money
print(ans)
|
s618561035
|
Accepted
| 17 | 3,060 | 244 |
N = int(input())
lst_1 = []
ans = 0
for n in range(N):
lst_1.append(input().split())
for n in range(N):
money = 0
money = float(lst_1[n][0])
if lst_1[n][1] == "BTC":
money = money * 380000
ans += money
print(ans)
|
s000327428
|
p02393
|
u327546577
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,584 | 100 |
Write a program which reads three integers, and prints them in ascending order.
|
a, b, c = map(int, input().split())
li = sorted([a, b, c])
print("%(li[0])s %(li[1])s %(li[2])s ")
|
s271115184
|
Accepted
| 20 | 5,600 | 94 |
a, b, c = map(int, input().split())
li = sorted([a, b, c])
print(f"{li[0]} {li[1]} {li[2]}")
|
s213723341
|
p03434
|
u566619532
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 346 |
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
# -*- coding: utf-8 -*-
a = int(input())
b = input()
b = map(int, b.split())
print(type(b))
b = sorted(b)
b = b[::-1]
c = b[::2]
c_number = len(c)
d = b[1::2]
d_number = len(d)
c_sum = 0
d_sum = 0
for i in range(c_number):
c_sum += c[i]
for g in range(d_number):
d_sum += d[g]
print(c_sum - d_sum)
|
s972393102
|
Accepted
| 17 | 3,064 | 596 |
# -*- coding: utf-8 -*-
a = int(input())
b = input()
b = map(int, b.split())
b = sorted(b)
b = b[::-1]
c = b[::2]
c_number = len(c)
d = b[1::2]
d_number = len(d)
c_sum = 0
d_sum = 0
for i in range(c_number):
c_sum += c[i]
for g in range(d_number):
d_sum += d[g]
print(c_sum - d_sum)
|
s955186021
|
p02694
|
u010379708
| 2,000 | 1,048,576 |
Wrong Answer
| 2,206 | 9,172 | 100 |
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
x=100
target = int(input())
count=0
while count<target:
x*=1.01
x = x//1
count+=1
print(count)
|
s027210526
|
Accepted
| 25 | 9,168 | 96 |
t=int(input())
ans=0
start=100
while start<t:
start += int(start/100)
ans+=1
print(ans)
|
s179738709
|
p03730
|
u469281291
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 261 |
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`.
|
a, b, c = map(int, input().split())
rem = a % b
tmp = 0
fa = a
ans = "No"
if(rem!=0):
while (tmp != rem):
a += fa
tmp = a % b
if (tmp == c):
ans = "Yes"
break
else:
if(c==0):
ans = "Yes"
print(ans)
|
s667243156
|
Accepted
| 17 | 3,064 | 261 |
a, b, c = map(int, input().split())
rem = a % b
tmp = 0
fa = a
ans = "NO"
if(rem!=0):
while (tmp != rem):
a += fa
tmp = a % b
if (tmp == c):
ans = "YES"
break
else:
if(c==0):
ans = "YES"
print(ans)
|
s297421400
|
p03565
|
u853900545
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 344 |
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 = input()
t = input()
S = len(s)
T = len(t)
ans= ''
for i in range(S-T):
for j in range(T):
if s[i+j] == t[j] or s[i+j] == '?':
c = 1
continue
else:
c = 0
if c :
ans = s[0:i] + t + s[i+T:]
if ans:
ans = ans.replace('?','a')
print(ans)
else:
print('UNRESTORABLE')
|
s501602415
|
Accepted
| 17 | 3,064 | 338 |
s = input()
t = input()
S = len(s)
T = len(t)
ans= ''
for i in range(S-T+1):
c = 1
for j in range(T):
if s[i+j] == t[j] or s[i+j] == '?':
continue
else:
c = 0
if c :
ans = s[0:i] + t + s[i+T:]
if ans:
ans = ans.replace('?','a')
print(ans)
else:
print('UNRESTORABLE')
|
s801130838
|
p03417
|
u175590965
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 154 |
There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region. The front and back sides of these cards can be distinguished, and initially every card faces up. We will perform the following operation once for each square contains a card: * For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square. It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed. Find the number of cards that face down after all the operations.
|
n,m = map(int,input().split())
if n > m:
n,m = m,n
if n == 1 and m == 1:
print(1)
elif n == 1 and m>1:
print(m-2)
else:
print((n-2*(m-2)))
|
s162347515
|
Accepted
| 17 | 3,060 | 156 |
n,m = map(int,input().split())
if n > m:
n,m = m,n
if n == 1 and m == 1:
print(1)
elif n == 1 and m > 1:
print(m-2)
else:
print((n-2)*(m-2))
|
s835080642
|
p02615
|
u514118270
| 2,000 | 1,048,576 |
Wrong Answer
| 125 | 31,460 | 131 |
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())
l=sorted(list(map(int,input().split())))
l.reverse()
s=l[0]-N%2*l[N//2]
for i in range(N//2):
s+=l[i+1]*2
print(s)
|
s958214113
|
Accepted
| 129 | 31,552 | 133 |
N=int(input())
l=sorted(list(map(int,input().split())))
l.reverse()
s=l[0]+N%2*l[N//2]
for i in range(N//2-1):
s+=l[i+1]*2
print(s)
|
s591666013
|
p02612
|
u811535874
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 8,916 | 32 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(n % 1000)
|
s316430284
|
Accepted
| 26 | 9,048 | 87 |
n = int(input())
if n % 1000 == 0:
print(0)
else:
print((n// 1000 + 1) * 1000 - n)
|
s779324242
|
p03645
|
u794173881
| 2,000 | 262,144 |
Wrong Answer
| 992 | 52,840 | 592 |
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
from collections import deque
n,m = map(int,input().split())
tree = [[] for i in range(n)]
for i in range(m):
tmp_a,tmp_b = map(int,input().split())
tree[tmp_a-1].append(tmp_b-1)
tree[tmp_b-1].append(tmp_a-1)
print(tree)
def bfs():
q = deque()
visited =[False]*n
visited[0]=True
for elem in tree[0]:
q.append([elem,1])
while q:
i,cnt = q.popleft()
if i == n-1 and cnt <=2 :
print("POSSIBLE")
exit()
if not visited[i]:
visited[i] = True
if cnt<=1:
for elem in tree[i]:
q.append([elem,cnt+1])
bfs()
print("IMPOSSIBLE")
|
s818458045
|
Accepted
| 903 | 49,868 | 587 |
from collections import deque
n,m = map(int,input().split())
tree = [[] for i in range(n)]
for i in range(m):
tmp_a,tmp_b = map(int,input().split())
tree[tmp_a-1].append(tmp_b-1)
tree[tmp_b-1].append(tmp_a-1)
def bfs():
q = deque()
visited =[False]*n
visited[0]=True
for elem in tree[0]:
q.append([elem,1])
while q:
i,cnt = q.popleft()
if i == n-1 and cnt <=2 :
print("POSSIBLE")
exit()
if not visited[i]:
visited[i] = True
if cnt<=1:
for elem in tree[i]:
q.append([elem,cnt+1])
bfs()
print("IMPOSSIBLE")
|
s812653938
|
p03759
|
u767438459
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,092 | 85 |
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c=map(int, input().split())
if b-a == c-b:
print("Yes")
else:
print("No")
|
s346799869
|
Accepted
| 26 | 9,080 | 85 |
a,b,c=map(int, input().split())
if b-a == c-b:
print("YES")
else:
print("NO")
|
s058482421
|
p02259
|
u064313887
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,520 | 468 |
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 show(nums):
for i in range(len(nums)):
if i != len(nums) - 1:
print(nums[i],end=' ')
else:
print(nums[i])
def bubbleSort(A,N):
flag = 1
count = 0
while flag:
flag = 0
for j in range(N-1,0,-1):
if A[j] < A[j-1]:
tmp = A[j]
A[j] = A[j-1]
A[j-1] = tmp
flag = 1
count += 1
show(A)
print(count)
|
s752145786
|
Accepted
| 20 | 5,612 | 503 |
def show(nums):
for i in range(len(nums)):
if i != len(nums)-1:
print(nums[i],end=' ')
else:
print(nums[i])
N = int(input())
A = list(map(int,input().split()))
count = 0
flag = 1
for i in range(N):
while flag:
flag = 0
for j in range(N-1,i,-1):
if A[j] < A[j-1]:
tmp = A[j]
A[j] = A[j-1]
A[j-1] = tmp
count += 1
flag = 1
show(A)
print(count)
|
s022480434
|
p03501
|
u263691873
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 64 |
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
N, A, B = map(int, input().split())
print(N*A+B if N*A<B else B)
|
s407434275
|
Accepted
| 17 | 2,940 | 62 |
N, A, B = map(int, input().split())
print(N*A if N*A<B else B)
|
s742807706
|
p02414
|
u150984829
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 144 |
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula: \\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\] where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
|
import sys
e=[list(map(int,e.split()))for e in sys.stdin]
n=e[0][0]+1
for c in e[:n]:print(*[sum(s*t for s,t in zip(c,l))for l in zip(*e[n:])])
|
s236430198
|
Accepted
| 110 | 6,588 | 147 |
import sys
e=[list(map(int,x.split()))for x in sys.stdin];n=e[0][0]+1
for c in e[1:n]:print(*[sum([s*t for s,t in zip(c,l)])for l in zip(*e[n:])])
|
s026281891
|
p00009
|
u766477342
| 1,000 | 131,072 |
Wrong Answer
| 910 | 31,376 | 379 |
Write a program which reads an integer n and prints the number of prime numbers which are less than or equal to n. A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7.
|
def mk_table(n):
res = [1] * (n + 1)
res[:2] = 0, 0
for i in range(2, n):
if i ** 2 > n:
break
if res[i] == 1:
for j in range(2, n // i + 1, i):
res[i * j] = 0
return res
tbl = mk_table(999999)
try:
while 1:
print(len([x for x in tbl[:int(input())+1] if x == 1]))
except Exception:
pass
|
s838174276
|
Accepted
| 820 | 24,296 | 372 |
def mk_table(n):
res = [1] * (n + 1)
res[:2] = 0, 0
for i in range(2, n):
if i ** 2 > n:
break
if res[i] == 1:
for j in range(i*2, n + 1, i):
res[j] = 0
return res
tbl = mk_table(999999)
try:
while 1:
print(len([x for x in tbl[:int(input())+1] if x == 1]))
except Exception:
pass
|
s020442088
|
p02613
|
u460615319
| 2,000 | 1,048,576 |
Wrong Answer
| 154 | 9,184 | 198 |
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 = []
judges = {'AC':0,'WA':0,'TLE':0,'RE':0}
for i in range(n):
s = input().strip()
judges[s] += 1
for judge,count in judges.items():
print(f"{judge} ✖️ {count}")
|
s700198296
|
Accepted
| 146 | 9,184 | 193 |
n = int(input())
s = []
judges = {'AC':0,'WA':0,'TLE':0,'RE':0}
for i in range(n):
s = input().strip()
judges[s] += 1
for judge,count in judges.items():
print(f"{judge} x {count}")
|
s192625440
|
p03943
|
u690781906
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 119 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c = map(int, input().split())
if a + b + c - max(a, b, c) == max(a, b, c):
print('YES')
else:
print('NO')
|
s153722310
|
Accepted
| 17 | 2,940 | 120 |
a, b, c = map(int, input().split())
if a + b + c - max(a, b, c) == max(a, b, c):
print('Yes')
else:
print('No')
|
s612318241
|
p02260
|
u433154529
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,616 | 285 |
Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠ mini.
|
n=int(input())
a=[int(i) for i in input().split(" ")]
x=0
for i in range(n):
mi=i
for j in range(i,n):
print(a[j],a[mi])
if a[j] < a[mi]:
mi = j
if i != mi:
x+=1
a[i],a[mi]=a[mi],[i]
print(" ".join([str(i) for i in a]))
print(x)
|
s868363417
|
Accepted
| 20 | 7,776 | 276 |
n=int(input())
a=[int(i) for i in input().split(" ")]
x=0
for i in range(n):
mi = i
for j in range(i, n):
if a[j] < a[mi]:
mi = j
if i != mi:
x += 1
a[i], a[mi] = a[mi] , a[i]
print(" ".join([str(i) for i in a]))
print(x)
|
s714142253
|
p03695
|
u811000506
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 299 |
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
N = int(input())
a = list(map(int,input().split()))
li = [10**9]*N
overrate = 0
for i in range(N):
if a[i]>=3200:
overrate += 1
continue
else:
li[i] = a[i]//400
kind = len(list(set(li)))
mn = min(kind-min(overrate,1),8)
mx = min(8,max(kind-1+overrate,1))
print(mn,mx)
|
s765021459
|
Accepted
| 18 | 3,060 | 335 |
N = int(input())
a = list(map(int,input().split()))
li = [0]*8
overrate = kind = 0
for i in range(N):
if a[i]>=3200:
overrate += 1
continue
else:
li[a[i]//400] += 1
for j in range(8):
if li[j]!=0:
kind += 1
if kind==0:
print(1,kind+overrate)
else:
print(min(8,kind),kind+overrate)
|
s713006723
|
p03044
|
u722535636
| 2,000 | 1,048,576 |
Wrong Answer
| 438 | 34,160 | 170 |
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
n=int(input())
t=[0]*n
q=[list(map(int,input().split())) for i in range(n-1)]
for u,v,w in q:
if w%2==0:
t[v-1],t[u-1]=1,1
for i in range(n):
print(t[i])
|
s473121313
|
Accepted
| 688 | 79,356 | 322 |
import sys
sys.setrecursionlimit(100100)
n=int(input())
route=[[] for i in range(n)]
for i in range(n-1):
u,v,w=map(int,input().split())
route[u-1].append((v-1,w))
route[v-1].append((u-1,w))
ans=[-1]*n
def dfs(p,d):
ans[p]=d
for i,j in route[p]:
if ans[i]<0:dfs(i,(d+j)%2)
dfs(0,0)
for i in range(n):
print(ans[i])
|
s251584184
|
p02409
|
u884012707
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,608 | 320 |
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
N=int(input())
l=[[[0 for i in range(10)]for j in range(3)]for k in range(4)]
for i in range(N):
b, f, r, v = map(lambda x: int(x)-1, input().split())
print(b,f,r,v)
l[b][f][r]+=v+1
for j in range(4):
for k in l[j]:
print(" ".join(map(str, k)))
if j < 3:
print("####################")
|
s298281598
|
Accepted
| 30 | 7,724 | 305 |
N=int(input())
l=[[[0 for i in range(10)]for j in range(3)]for k in range(4)]
for i in range(N):
b, f, r, v = map(lambda x: int(x)-1, input().split())
l[b][f][r]+=v+1
for j in range(4):
for k in l[j]:
print(" "+" ".join(map(str, k)))
if j < 3:
print("####################")
|
s722846831
|
p02936
|
u254871849
| 2,000 | 1,048,576 |
Wrong Answer
| 798 | 49,700 | 886 |
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
# 2019-11-17 10:52:21(JST)
import sys
# import collections
# import math
# from string import ascii_lowercase, ascii_uppercase, digits
# from bisect import bisect_left as bi_l, bisect_right as bi_r
import itertools
# from functools import reduce
# import operator as op
# import re
# import heapq
# import array
# from scipy.misc import comb # (default: exact=False)
# import numpy as np
def main():
n, q = [int(x) for x in sys.stdin.readline().split()]
AB = [[] for _ in range(n+1)]
for _ in range(n-1):
a, b = [int(x) for x in sys.stdin.readline().split()]
AB[a].append(b)
ans = [0 for _ in range(n+1)]
for _ in range(q):
p, x = [int(x) for x in sys.stdin.readline().split()]
ans[p] += x
for a in range(1, n+1):
for b in AB[a]:
ans[b] += ans[a]
print(ans)
if __name__ == "__main__":
main()
|
s057189406
|
Accepted
| 878 | 74,680 | 665 |
import sys
n, q = map(int, sys.stdin.readline().split())
g = [[] for _ in range(n+1)]
for _ in range(n-1):
a, b = map(int, sys.stdin.readline().split())
g[a].append(b)
g[b].append(a)
px = zip(*[map(int, sys.stdin.read().split())] * 2)
def main():
value = [0] * (n + 1)
for p, x in px:
value[p] += x
stack = [1]
par = [None] * (n + 1)
while stack:
u = stack.pop()
for v in g[u]:
if v != par[u]:
par[v] = u
value[v] += value[u]
stack.append(v)
return value[1:]
if __name__ == '__main__':
ans = main()
print(*ans, sep=' ')
|
s906460887
|
p02407
|
u093488647
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 136 |
Write a program which reads a sequence and prints it in the reverse order.
|
num = int(input())
data = input().split()
output = []
lengh = len(data)
for tmp in range(lengh):
output += data.pop(-1)
print(output)
|
s588681694
|
Accepted
| 20 | 5,600 | 182 |
num = int(input())
data = list(map(int, input().split()))
lengh = len(data)
for tmp in range(lengh):
if tmp == lengh-1:
print(data.pop(-1))
else:
print(data.pop(-1), end=" ")
|
s405670634
|
p03007
|
u922952729
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 14,428 | 329 |
There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.
|
import bisect
N=int(input())
A=[int(i) for i in input().split(" ")]
A=sorted(A)
traceback=[]
for i in range(N-1):
if i%2==0:
x=A[0]
y=A[-1]
else:
x=A[-1]
y=A[0]
A=A[1:-1:]
bisect.insort_left(A,x-y)
traceback.append((x,y))
print(A[0])
for t in traceback:
print(t[0],t[1])
|
s551874493
|
Accepted
| 216 | 13,964 | 433 |
import bisect
N=int(input())
A=[int(i) for i in input().split(" ")]
A=sorted(A)
traceback=[]
if A[0]>=0:
minus=[A[0]]
plus=A[1::]
elif A[-1]<=0:
minus=A[:-1:]
plus=[A[-1]]
else:
index=bisect.bisect_left(A,0)
minus=A[:index:]
plus=A[index::]
print(sum(plus)-sum(minus))
m=minus.pop(0)
for p in plus[:-1:]:
print(m,p)
m=m-p
print(plus[-1],m)
p=plus[-1]-m
for m in minus:
print(p,m)
p=p-m
|
s109630772
|
p03854
|
u099022530
| 2,000 | 262,144 |
Wrong Answer
| 75 | 3,956 | 310 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
s = input()
dp = [0] * (len(s) + 1)
dp[0] = 1
words = ["dream","dreamer","erase","eraser"]
done = "No"
for i in range(len(s)):
if dp[i] == 0:
continue
for w in words:
if s[i:i+len(w)] == w:
dp[i+len(w)] = 1
if dp[len(s)] == 1:
done = "Yes"
print(done)
|
s671459876
|
Accepted
| 73 | 3,956 | 310 |
S = input()
dp = [0] * (len(S) + 1)
dp[0] = 1
words = ["dream", "dreamer", "erase", "eraser"]
done = 'NO'
for i in range(len(S)):
if dp[i] == 0:
continue
for w in words:
if S[i:i+len(w)] == w:
dp[i+len(w)] = 1
if dp[len(S)] == 1:
done = 'YES'
print(done)
|
s640350875
|
p03760
|
u820351940
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 98 |
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
a = input()
b = input()
"".join(map("".join, zip(a, b))) + (a[-1] if len(a) is not len(b) else "")
|
s262190185
|
Accepted
| 18 | 2,940 | 115 |
a = input()
b = input()
print("".join(map("".join, zip(list(a), list(b)))) + (a[-1] if len(a) != len(b) else ""))
|
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