message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Try guessing the statement from this picture <http://tiny.cc/ogyoiz>.
You are given two integers A and B, calculate the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true; conc(a, b) is the concatenation of a and b (for example, conc(12, 23) = 1223, conc(100, 11) = 10011). a and b should not contain leading zeroes.
Input
The first line contains t (1 ≤ t ≤ 100) — the number of test cases.
Each test case contains two integers A and B (1 ≤ A, B ≤ 10^9).
Output
Print one integer — the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true.
Example
Input
3
1 11
4 2
191 31415926
Output
1
0
1337
Note
There is only one suitable pair in the first test case: a = 1, b = 9 (1 + 9 + 1 ⋅ 9 = 19).
Submitted Solution:
```
#-----------------
# cook your dish here
#############-----------------
try:#######
w=int(input())#####
########
for _ in range(w):###
#######
x,y=map( int,input().split() )#######
#######
e=len( str( y ) )#######
######
z=0########
#######
if ((10**e) - 2) < y:########
z += e#####
else:############
z += e-1#########
######
##########
ans = x*z ######
#######
print(ans)#######
except:
pass
#############
###########
#############
``` | instruction | 0 | 80,064 | 5 | 160,128 |
Yes | output | 1 | 80,064 | 5 | 160,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Try guessing the statement from this picture <http://tiny.cc/ogyoiz>.
You are given two integers A and B, calculate the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true; conc(a, b) is the concatenation of a and b (for example, conc(12, 23) = 1223, conc(100, 11) = 10011). a and b should not contain leading zeroes.
Input
The first line contains t (1 ≤ t ≤ 100) — the number of test cases.
Each test case contains two integers A and B (1 ≤ A, B ≤ 10^9).
Output
Print one integer — the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true.
Example
Input
3
1 11
4 2
191 31415926
Output
1
0
1337
Note
There is only one suitable pair in the first test case: a = 1, b = 9 (1 + 9 + 1 ⋅ 9 = 19).
Submitted Solution:
```
exec(int(input()) * "a, b = map(int, input().split()); print(a * len(str(b + 1)) - a); ")
``` | instruction | 0 | 80,065 | 5 | 160,130 |
Yes | output | 1 | 80,065 | 5 | 160,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Try guessing the statement from this picture <http://tiny.cc/ogyoiz>.
You are given two integers A and B, calculate the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true; conc(a, b) is the concatenation of a and b (for example, conc(12, 23) = 1223, conc(100, 11) = 10011). a and b should not contain leading zeroes.
Input
The first line contains t (1 ≤ t ≤ 100) — the number of test cases.
Each test case contains two integers A and B (1 ≤ A, B ≤ 10^9).
Output
Print one integer — the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true.
Example
Input
3
1 11
4 2
191 31415926
Output
1
0
1337
Note
There is only one suitable pair in the first test case: a = 1, b = 9 (1 + 9 + 1 ⋅ 9 = 19).
Submitted Solution:
```
t = int(input())
for _ in range(t):
a, b = map(int, input().split())
print(a*(len(str(b+1))-1))
``` | instruction | 0 | 80,066 | 5 | 160,132 |
Yes | output | 1 | 80,066 | 5 | 160,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Try guessing the statement from this picture <http://tiny.cc/ogyoiz>.
You are given two integers A and B, calculate the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true; conc(a, b) is the concatenation of a and b (for example, conc(12, 23) = 1223, conc(100, 11) = 10011). a and b should not contain leading zeroes.
Input
The first line contains t (1 ≤ t ≤ 100) — the number of test cases.
Each test case contains two integers A and B (1 ≤ A, B ≤ 10^9).
Output
Print one integer — the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true.
Example
Input
3
1 11
4 2
191 31415926
Output
1
0
1337
Note
There is only one suitable pair in the first test case: a = 1, b = 9 (1 + 9 + 1 ⋅ 9 = 19).
Submitted Solution:
```
t = int(input())
for i in range(t):
a, b = input().split()
a = int(a)
n = len(b)
if int(b[-1]) < 9:
n-=1
print(a*n)
``` | instruction | 0 | 80,067 | 5 | 160,134 |
No | output | 1 | 80,067 | 5 | 160,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Try guessing the statement from this picture <http://tiny.cc/ogyoiz>.
You are given two integers A and B, calculate the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true; conc(a, b) is the concatenation of a and b (for example, conc(12, 23) = 1223, conc(100, 11) = 10011). a and b should not contain leading zeroes.
Input
The first line contains t (1 ≤ t ≤ 100) — the number of test cases.
Each test case contains two integers A and B (1 ≤ A, B ≤ 10^9).
Output
Print one integer — the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true.
Example
Input
3
1 11
4 2
191 31415926
Output
1
0
1337
Note
There is only one suitable pair in the first test case: a = 1, b = 9 (1 + 9 + 1 ⋅ 9 = 19).
Submitted Solution:
```
import math
n = int(input())
for _ in range(n):
a,b = input().split()
if int(a)>int(b):
print("0")
else:
l = (len(b)-1)
if l==0:
print("1")
else:
temp = int(a)*l
print(temp)
``` | instruction | 0 | 80,068 | 5 | 160,136 |
No | output | 1 | 80,068 | 5 | 160,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Try guessing the statement from this picture <http://tiny.cc/ogyoiz>.
You are given two integers A and B, calculate the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true; conc(a, b) is the concatenation of a and b (for example, conc(12, 23) = 1223, conc(100, 11) = 10011). a and b should not contain leading zeroes.
Input
The first line contains t (1 ≤ t ≤ 100) — the number of test cases.
Each test case contains two integers A and B (1 ≤ A, B ≤ 10^9).
Output
Print one integer — the number of pairs (a, b) such that 1 ≤ a ≤ A, 1 ≤ b ≤ B, and the equation a ⋅ b + a + b = conc(a, b) is true.
Example
Input
3
1 11
4 2
191 31415926
Output
1
0
1337
Note
There is only one suitable pair in the first test case: a = 1, b = 9 (1 + 9 + 1 ⋅ 9 = 19).
Submitted Solution:
```
def Deadline():
for _ in range(int(input())):
a,b = list(map(int, input().split(' ')))
if a >= b:
print(0)
continue
else:
t = 0
c = 0
while t < b:
if (t*10 + 9) <= b:
t = (t*10 + 9)
c += 1
else:
break
print(a*c)
Deadline()
``` | instruction | 0 | 80,069 | 5 | 160,138 |
No | output | 1 | 80,069 | 5 | 160,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains two integers row, col (0 ≤ row, col ≤ 63), separated by a single space.
Output
Output "IN" or "OUT".
Examples
Input
0 0
Output
OUT
Input
27 0
Output
IN
Input
0 27
Output
OUT
Input
27 27
Output
IN | instruction | 0 | 80,071 | 5 | 160,142 |
Tags: *special, dfs and similar, geometry, implementation
Correct Solution:
```
L = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0], [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0], [0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0], [0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0], [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0], [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
a, b = input().split()
a = int(a)
b = int(b)
if L[a][b] == 1:
print('IN')
else:
print('OUT')
``` | output | 1 | 80,071 | 5 | 160,143 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains two integers row, col (0 ≤ row, col ≤ 63), separated by a single space.
Output
Output "IN" or "OUT".
Examples
Input
0 0
Output
OUT
Input
27 0
Output
IN
Input
0 27
Output
OUT
Input
27 27
Output
IN | instruction | 0 | 80,072 | 5 | 160,144 |
Tags: *special, dfs and similar, geometry, implementation
Correct Solution:
```
t=""" + + ++++++ +
+ +++ + + + + +++++
+++ + + ++ + + +
++++ + + +++++ ++++ + +++ +
++ + + ++++ + + + ++
+ ++ +++++ +++ +++ + + ++ + + +
+++ + + + + ++++ ++ ++ ++ ++
++++++ ++ ++ +++ +++ +++ ++ + + +++
++++++++++++ + + + ++ ++ ++ + +++
++++++++++++++++++++ +++++++ ++++ ++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++++++++++++++++++++++++++
++++++++++++ + ++++++++++++++ + + +++++++++++++
++ +++++++++ + +++++++++++++++ + ++++++++
++++ +++++ ++ + +++++++++++++++ ++ ++ + ++ ++
+++++++ + +++ +++ +++++++++++ + + ++ ++++++
+++++++++ + + ++ ++ +++++++++ ++ +++++ + +++++++++
++++++++++++ ++ +++ +++ +++++++ + + +++ +++++++++++
+++++++++ + +++ +++ ++++++++ + +++++ + + + ++++++++
++++++++++++ +++ + + + + ++++++ + + ++ +++ +++++++++
+++++++++++ + + ++ +++ +++ ++++ + +++ ++++ +++ + +++++++++
+++++++++++ ++++++ ++++ + + + + + + ++++ +++ +++++++++
+++++++++++ + + +++ +++ + + ++ ++ + + + + +++++++++
++++++++++++ ++ ++++ ++ +++ + +++ ++++ + ++++++ ++++++++++
+++++++++++ +++ + + +++ + +++ ++++ + +++ + +++++++++
+++++++++++++ ++ +++ ++ +++ ++ + ++ ++ + + +++++++++
++++++++++++++ ++ + + + + + + + ++ + ++ ++ ++++++++++++
++++++++++++ ++++ + + + + ++++ ++ + + +++ +++ +++++++++++
+++++++++++++ + + +++ ++++ + ++ + ++ +++++++++++++
++++++++++++++++ +++ + ++ ++ ++ ++ + +++ + + ++++++++++++
++++++++++++++++++ + ++++++ +++++++ + ++++++++++++++++
++++++++++++++++ + +++++ + +++++++++++ + ++++++++++++++
++++++++++++++++++++++++++++++ ++++++++++++++++++++++++++++
+++++++++++++++++++++++++++ + + +++++++++++++++++++++++++++++
+++++++++++++++++++++++++ + ++++++++++++++++++++++++++++++++
++++++++++++++++++++++++ + +++++++++++++++++++++++++++++
+++++++ +++++++++++++++++++ +++++++++++++++++++++++++++++++
+++++++ + ++++++++++++++++ +++++++++++++++++++++ +++++++
+++++++ + +++++++++++ + +++++++++++ +++++ + ++++++
+++++++ + + + + + ++ + ++ + + + ++ ++ + +++++++
++++++++ ++ + ++ ++ + + +++ ++++++++
+++++++ + ++++ ++++ +++ +++ +++ +++++ + +++++++++
++++++++++ + + + + ++++ +++ ++ + + +++++++++
++++++++ ++ ++ + + + + +++ + ++++++++++
+++++++++ ++ +++++ ++ + + + ++ +++++++++++++
+++++++++ +++ + + +++ + +++ + +++++++++++
+++++++++ ++++ +++ ++ ++++ ++++ ++++++++++++
+++++++++ ++ + + + + ++++++++++++++
++++++++++++ + ++++ ++ ++ + ++ ++++++++++++
+++++++++++++ ++++ + ++++++++++++
++++++++++++++++++++ +++++++++++++++++
++++++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++++
++++++++++++++++++++++++++++++
++++++++++++++++++++++++
++++++++++++++++++++
++++++++++++ """.split("\n")
i,j=[int(k) for k in input().split(" ")]
print("IN" if t[i][j]=="+" else "OUT")
``` | output | 1 | 80,072 | 5 | 160,145 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains two integers row, col (0 ≤ row, col ≤ 63), separated by a single space.
Output
Output "IN" or "OUT".
Examples
Input
0 0
Output
OUT
Input
27 0
Output
IN
Input
0 27
Output
OUT
Input
27 27
Output
IN | instruction | 0 | 80,073 | 5 | 160,146 |
Tags: *special, dfs and similar, geometry, implementation
Correct Solution:
```
x,y = [int(i) for i in input().split()]
im = [[ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 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0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]]
if im[y][x] == 1:
print("OUT")
else:
print("IN")
``` | output | 1 | 80,073 | 5 | 160,147 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains two integers row, col (0 ≤ row, col ≤ 63), separated by a single space.
Output
Output "IN" or "OUT".
Examples
Input
0 0
Output
OUT
Input
27 0
Output
IN
Input
0 27
Output
OUT
Input
27 27
Output
IN | instruction | 0 | 80,074 | 5 | 160,148 |
Tags: *special, dfs and similar, geometry, implementation
Correct Solution:
```
#from PIL import Image
res = [[64 for i in range(64)] for j in range(64)]
res[0]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[1]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[2]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[3]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[4]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[5]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[6]=[0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[7]=[0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0]
res[8]=[0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0]
res[9]=[0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0]
res[10]=[0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0]
res[11]=[0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,0,1,0,0,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0]
res[12]=[0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,0,0,0,1,0,0,0,1,1,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0]
res[13]=[0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,0,1,0,1,0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0]
res[14]=[0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,0,1,1,1,0,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,1,0,1,1,1,1,1,1,1,1,0,0,0,0,0]
res[15]=[0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,0,1,0,0,1,0,1,0,0,1,1,1,1,1,1,1,1,1,0,1,1,0,0,0,1,1,1,1,1,1,1,1,0,0,0,0,0]
res[16]=[0,0,0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,1,1,1,1,0,1,1,0,0,1,1,1,1,1,1,1,1,0,0,1,1,1,0,0,0,1,1,1,1,1,1,1,0,0,0,0]
res[17]=[0,0,0,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,0,0,0,1,1,0,1,0,1,1,1,1,1,1,1,0,0,1,0,0,1,0,0,1,1,1,1,1,1,1,0,0,0]
res[18]=[0,0,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,0,0,0,0,0,1,0,1,1,1,0,0,0,0,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,0,1,1,1,1,1,1,0,0,0]
res[19]=[0,0,0,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,0,1,1,0,1,1,1,1,1,1,0,1,0,1,0,1,1,1,1,1,1,1,1,0,0,1,1,1,0,1,0,1,1,1,1,1,1,1,0,0,0]
res[20]=[0,0,1,1,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,0,0,1,1,1,1,0,0,0,0,0,0,0,1,0,0,1,1,1,1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0]
res[21]=[0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,1,0,1,0,0,1,0,1,0,1,0,1,1,1,1,0,1,1,1,1,1,1,1,0,0,1,0,1,0,0,1,0,1,1,1,1,1,1,1,0,0]
res[22]=[0,1,1,1,0,1,0,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,0,0,1,1,1,1,1,1,0,0,1,0,0,0,1,1,1,1,1,1,1,0,1,1,1,1,1,0,1,0,0,1,1,1,1,1,1,1,0]
res[23]=[0,0,0,0,0,1,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,0,0,1,1,1,0,1,0,0,1,1,1,1,0,1,0,1,1,1,1,1,1,1,0,0,1,0,0,1,1,1,0,1,1,1,1,1,1,1,1,0]
res[24]=[0,1,1,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,0,1,0,1,0,1,1,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,0,0,0,0,0,1,1,1,1,1,1,0]
res[25]=[0,1,0,0,0,0,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,0,0,0,0,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,1,1,0,1,1,0,1,1,1,1,1,1,0]
res[26]=[1,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,0,0,1,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,1,1,1,0,0,1,1,0,1,0,0,0,1,0,1,1,1,1,1,1,1]
res[27]=[0,0,0,1,0,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,1,0,1,0,0,1,1,1,1,1,1,1,1,0,1,1,1,1,0,1,0,0,0,0,1,1,1,0,1,1,1,1,1,1,1]
res[28]=[1,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,1,0,1,0,1,1,0,0,1,1,1,1,1,0,0,0,1,1,0,0,0,1,1,0,1,0,1,0,1,1,1,1,1,1,1,1,1]
res[29]=[0,0,0,1,0,0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,0,0,0,0,0,1,0,1,1,0,0,0,1,1,0,1,1,0,0,1,0,0,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1]
res[30]=[1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,1,0,0,0,0,0,1,1,1,0,1,0,0,1,0,0,1,1,1,1,1,1,1,1,1]
res[31]=[1,0,0,0,0,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,1,0,0,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1]
res[32]=[1,1,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,0,0,1,0,1,1,0,1,1,1,0,0,1,0,0,0,0,1,0,0,0,1,1,1,1,1,1,1]
res[33]=[1,0,1,0,1,0,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,0,0,0,1,0,1,0,0,1,1,1,1,1,1,0,1,1,1,1,0,0,0,1,0,0,1,1,1,1,1,1]
res[34]=[1,0,0,0,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,0,0,1,1,1,1,1,0,1,1,1,1,1,1,0,0,0,1,0,0,0,1,1,1,0,1,1,1,1,1,1,1]
res[35]=[1,1,1,1,1,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,0,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,1,1,0,0,0,1,1,1,1,1,1,1]
res[36]=[0,0,0,1,0,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,1,0,0,0,0,0,1,0,1,1,1,1,1,1,1,1,1]
res[37]=[1,1,0,1,0,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,0,1,1,1,0,0,0,1,1,1,1,1,1,1]
res[38]=[0,1,0,1,0,0,0,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,0,0,0,1,0,0,0,1,0,1,1,1,1,1,1,1,1,1,1,0,1,1,0,0,1,0,0,1,0,1,1,1,1,1,1,0]
res[39]=[0,1,0,0,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,0,0,0,1,1,1,0,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,0,0,1,0,0,1,1,1,1,1,0,1,1,1,1,1,1,0]
res[40]=[0,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,0,1,1,0,1,1,1,0,0,0,1,1,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,0,1,0,0,0,1,1,1,1,1,1,0]
res[41]=[0,1,0,0,0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,1,0,0,1,1,0,1,1,1,1,1,1,1,0]
res[42]=[0,0,0,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,0,0,1,0,1,0,1,1,1,1,1,1,1,1,1,1,0,0]
res[43]=[0,0,1,1,0,0,1,1,1,0,1,1,1,1,1,1,1,1,1,0,0,1,0,1,1,1,0,1,0,0,1,1,1,0,1,1,1,0,0,1,1,1,1,1,1,0,0,1,1,0,0,0,0,0,1,1,1,1,1,1,1,1,0,0]
res[44]=[0,0,0,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,0,1,0,1,1,0,0,0,0,0,1,0,1,1,0,1,1,1,1,1,1,1,0,0,1,0,1,1,0,1,1,1,1,1,1,1,1,0,0,0]
res[45]=[0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,1,0,1,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,0,0,1,0,1,1,1,1,1,1,1,1,0,0,0]
res[46]=[0,0,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,1,1,0,1,0,0,1,1,1,0,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,0,0,0]
res[47]=[0,0,0,0,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,1,0,0,0,1,1,1,0,1,1,0,1,1,1,0,0,0,1,1,1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0]
res[48]=[0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,0,0,1,0,0,1,0,0,0,1,0,0,1,0,1,1,1,1,1,1,1,0,1,1,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0]
res[49]=[0,0,0,0,0,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,1,1,0,1,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,0,0,1,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0]
res[50]=[0,0,0,0,0,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,0,0,0,1,0,0,1,0,0,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0]
res[51]=[0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,1,0,1,0,1,1,0,0,0,1,0,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0]
res[52]=[0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,0,0,0,1,1,1,0,0,1,0,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0]
res[53]=[0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,0,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0]
res[54]=[0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,0,0,0,0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0]
res[55]=[0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0]
res[56]=[0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0]
res[57]=[0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[58]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[59]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[60]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[61]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[62]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
res[63]=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
y, x = map(int, input().split())
print("OUT" if res[x][y] == 0 else "IN")
'''
def newImg():
img = Image.new('RGB', (64, 64))
for i in range(64):
for j in range(64):
img.putpixel((i, j), (res[i][j], res[i][j], res[i][j]))
img.save('sqr.png')
return img
wallpaper = newImg()
wallpaper.show()
'''
``` | output | 1 | 80,074 | 5 | 160,149 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains two integers row, col (0 ≤ row, col ≤ 63), separated by a single space.
Output
Output "IN" or "OUT".
Examples
Input
0 0
Output
OUT
Input
27 0
Output
IN
Input
0 27
Output
OUT
Input
27 27
Output
IN | instruction | 0 | 80,075 | 5 | 160,150 |
Tags: *special, dfs and similar, geometry, implementation
Correct Solution:
```
data = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 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1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 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0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
size = 64
y, x = map(int, input().split(' '))
print('IN' if data[size * y + x] else 'OUT')
``` | output | 1 | 80,075 | 5 | 160,151 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains two integers row, col (0 ≤ row, col ≤ 63), separated by a single space.
Output
Output "IN" or "OUT".
Examples
Input
0 0
Output
OUT
Input
27 0
Output
IN
Input
0 27
Output
OUT
Input
27 27
Output
IN | instruction | 0 | 80,077 | 5 | 160,154 |
Tags: *special, dfs and similar, geometry, implementation
Correct Solution:
```
s = "AAAAAAAAAAAAAAAAAAAAAAAAAABABABBBBBBABAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABABBBABABABAABABBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBABAAABAAABBABAAAABAABAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBABABABBBBBAAAABBBBABABBBABAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBAAAAAABABAAAAABBBBAAAABABABABBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABAABBABBBBBABBBABBBABAABABBAAABAABABAAAAAAAAAAAAAAAAAAAAAAAAAAABBBAABAABAAAAAABABAAABBBBAABBABBABBABBAAAAAAAAAAAAAAAAAAAAAAAABBBBBBABBABBABBBABBBABBBAAAABBAABAABAAABBBAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBBBAAAABAAAAABAAABABBAABBABBABAABBBAAAAAAAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBABBBBBBBAABBBBAABBBBBBBBBBAAAAAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAABBBBBBBBBBBBAAABAABBBBBBBBBBBBBBAAABABABBBBBBBBBBBBBAAAAAAABBAAAAABBBBBBBBBAABAAABBBBBBBBBBBBBBBABAAAAAAABBBBBBBBAAAAAAAAAABBBBAAAAABBBBBAAABBABAAABBBBBBBBBBBBBBBABBABBAABABBAAAAAAABBAAABBBBBBBAAAAABAAABBBAABBBAABBBBBBBBBBBABAABAAABBAAAAAAAAABBBBBBAABBBBBBBBBAAAAABAABAABBABBAAABBBBBBBBBABBABBBBBAABAAAABBBBBBBBBAABBBBBBBBBBBBABBABBBAAAAABBBAABBBBBBBAAABAABAAAABBBABBBBBBBBBBBAABBBBBBBBBAAAAABAAABBBABBBAAABBBBBBBBABABBBBBABAAABAABABBBBBBBBABBBBBBBBBBBBABBBABAABAABAAABAABBBBBBAABAABAAAABBABBBAAABBBBBBBBBBBBBBBBBBBBAAABAABABBABBBABBBABBBBABABBBABBBBAABBBAAABABBBBBBBBBBBBBBBBBBBBABBBBBBABBBBAAAABAABABAAAABABAABABBBBAAABBBABBBBBBBBBBBBBBBBBBBBAAAAABAABAABBBABBBAAABABABBABBABAABAAABAABAABBBBBBBBBBBBBBBBBBBBBABBABBBBABBABBBAAABABBBAAAAABBBBABABBBBBBABBBBBBBBBBBBBBBBBBBBBAAABBBAABAABAAABBBABAABBBABBBBAABABBBAAABAAABBBBBBBBBBBBBBBBBBBBBBAAAAABBABBBABBABBBABBAAAAABAABBABBAABAAABABBBBBBBBBBBBBBBBBBBBBBBABBABAAAABABAAAABABAABABABBABAAABBABBABBBBBBBBBBBBBBBBBBBBBBBBAAAABBBBABABABABABBBBABBABAABABBBAABBBAAABBBBBBBBBBBBBBBBBBBBBBBBABAABAAABBBAAABBBBAAABAABBAAAABAABBAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAABBBAAABABBABBABBABBAABABBBABAABAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAABABBBBBBAAAAABBBBBBBAAABAAABBBBBBBBBBBBBBBBABBBBBBBBBBBBBBBBAABAAAABBBBBABABBBBBBBBBBBAAABAABBBBBBBBBBBBBBAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBAABBBBBBBBBBBBBBBBBBBBBBBBBBBABAABABBBBBBBBBBBBBBBBBBBBBBBBBBBBBAABBBBBBBBBBBBBBBBBBBBBBBBBABAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAABBBBBBBBBBBBBBBBBBBBBBBBAAABAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAABBBBBBBAABBBBBBBBBBBBBBBBBBBABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAABBBBBBBABABBBBBBBBBBBBBBBBAAABBBBBBBBBBBBBBBBBBBBBABBBBBBBAAAAAABBBBBBBAAABAAABBBBBBBBBBBAABABBBBBBBBBBBABBBBBABAAAABBBBBBAAAAAABBBBBBBABABABAABABAAABBABABBAABABAABABBAAABBAAAAABABBBBBBBAAAAAAABBBBBBBBAAABBAAAAABAAAAAAABBAAABBAAAAAABABAABBBABBBBBBBBAAAAAAAAABBBBBBBABAABBBBABBBBABBBAABBBAAAABBBABBBBBABABBBBBBBBBAAAAAAAAAABBBBBBBBBBABAABAABAAABABBBBABBBABBABAAAAABAAABBBBBBBBBAAAAAAAAAAABBBBBBBBAAAABBABBABAAAABAAABABAAAABBBABAAABBBBBBBBBBAAAAAAAAAAAAABBBBBBBBBABBAAABBBBBABBABABAAABABBAAABBBBBBBBBBBBBAAAAAAAAAAAAAABBBBBBBBBAABBBAABABAAABBBAAABABBBAABAAABBBBBBBBBBBAAAAAAAAAAAAAAABBBBBBBBBAAABBBBAAABBBABBABBBBABBBBABBBBBBBBBBBBAAAAAAAAAAAAAAAAABBBBBBBBBABBAAAABABAAAAAABAAAABABBBBBBBBBBBBBBAAAAAAAAAAAAAAAAAAABBBBBBBBBBBBABABBBBABBABBABABBAABBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBBBBAAAABBBBAAAABAAAABBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBABBBBBBBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAABBBBBBBBBBBBAAAAAAAAAAAAAAAAAAAAAAAAAA"
(a,b) = map(int, input().split())
if(s[64*a + b] == 'A'):
print('OUT')
else:
print('IN')
``` | output | 1 | 80,077 | 5 | 160,155 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains two integers row, col (0 ≤ row, col ≤ 63), separated by a single space.
Output
Output "IN" or "OUT".
Examples
Input
0 0
Output
OUT
Input
27 0
Output
IN
Input
0 27
Output
OUT
Input
27 27
Output
IN | instruction | 0 | 80,078 | 5 | 160,156 |
Tags: *special, dfs and similar, geometry, implementation
Correct Solution:
```
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
i, j = tuple(map(int, input().strip().split()))
print("IN" if img[i][j] else "OUT")
``` | output | 1 | 80,078 | 5 | 160,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains two integers row, col (0 ≤ row, col ≤ 63), separated by a single space.
Output
Output "IN" or "OUT".
Examples
Input
0 0
Output
OUT
Input
27 0
Output
IN
Input
0 27
Output
OUT
Input
27 27
Output
IN
Submitted Solution:
```
lis = [['O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'I', 'O', 'I', 'O', 'I', 'I', 'I', 'I', 'I', 'I', 'O', 'I', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O'], ['O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'I', 'O', 'I', 'I', 'I', 'O', 'I', 'O', 'I', 'O', 'I', 'O', 'O', 'I', 'O', 'I', 'I', 'I', 'I', 'I', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O'], ['O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'I', 'I', 'I', 'O', 'I', 'O', 'O', 'O', 'I', 'O', 'O', 'O', 'I', 'I', 'O', 'I', 'O', 'O', 'O', 'O', 'I', 'O', 'O', 'I', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O', 'O'], ['O', 'O', 'O', 'O', 'O', 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a,b = map(int,input().split())
if lis[a][b] == "I":
print ("IN")
else:
print ("OUT")
``` | instruction | 0 | 80,082 | 5 | 160,164 |
Yes | output | 1 | 80,082 | 5 | 160,165 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains two integers row, col (0 ≤ row, col ≤ 63), separated by a single space.
Output
Output "IN" or "OUT".
Examples
Input
0 0
Output
OUT
Input
27 0
Output
IN
Input
0 27
Output
OUT
Input
27 27
Output
IN
Submitted Solution:
```
import math
def rMap():
return map(int,input().split())
def rList():
return list(map(int,input().split()))
l = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0], [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0], [0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0], [0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0], [0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0], [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0], [0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0], [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]
a,b = rMap()
print("IN" if l[a][b] == 1 else "OUT")
``` | instruction | 0 | 80,083 | 5 | 160,166 |
No | output | 1 | 80,083 | 5 | 160,167 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains two integers row, col (0 ≤ row, col ≤ 63), separated by a single space.
Output
Output "IN" or "OUT".
Examples
Input
0 0
Output
OUT
Input
27 0
Output
IN
Input
0 27
Output
OUT
Input
27 27
Output
IN
Submitted Solution:
```
row, col = map(int, input().split())
if row == 0:
print('OUT')
else:
print('IN')
``` | instruction | 0 | 80,084 | 5 | 160,168 |
No | output | 1 | 80,084 | 5 | 160,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains two integers row, col (0 ≤ row, col ≤ 63), separated by a single space.
Output
Output "IN" or "OUT".
Examples
Input
0 0
Output
OUT
Input
27 0
Output
IN
Input
0 27
Output
OUT
Input
27 27
Output
IN
Submitted Solution:
```
x=list(map(int,input().split( )))
if(x[0]==0)and(x[1]==0):
print("OUT")
if(x[0]!=0)and(x[1]==0):
print("IN")
if(x[0]==0)and(x[1]!=0):
print("OUT")
if(x[0]!=0)and(x[1]!=0):
print("OUT")
``` | instruction | 0 | 80,086 | 5 | 160,172 |
No | output | 1 | 80,086 | 5 | 160,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a multiset S. Over all pairs of subsets A and B, such that:
* B ⊂ A;
* |B| = |A| - 1;
* greatest common divisor of all elements in A is equal to one;
find the sum of ∑_{x ∈ A}{x} ⋅ ∑_{x ∈ B}{x}, modulo 998 244 353.
Input
The first line contains one integer m (1 ≤ m ≤ 10^5): the number of different values in the multiset S.
Each of the next m lines contains two integers a_i, freq_i (1 ≤ a_i ≤ 10^5, 1 ≤ freq_i ≤ 10^9). Element a_i appears in the multiset S freq_i times. All a_i are different.
Output
Print the required sum, modulo 998 244 353.
Examples
Input
2
1 1
2 1
Output
9
Input
4
1 1
2 1
3 1
6 1
Output
1207
Input
1
1 5
Output
560
Note
A multiset is a set where elements are allowed to coincide. |X| is the cardinality of a set X, the number of elements in it.
A ⊂ B: Set A is a subset of a set B.
In the first example B=\{1\}, A=\{1,2\} and B=\{2\}, A=\{1, 2\} have a product equal to 1⋅3 + 2⋅3=9. Other pairs of A and B don't satisfy the given constraints.
Submitted Solution:
```
t=int(input())
for i in range(t):
n,m=map(int,input().split())
print((23-n)*60+(60-m))
``` | instruction | 0 | 80,145 | 5 | 160,290 |
No | output | 1 | 80,145 | 5 | 160,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
a is an array of n positive integers, all of which are not greater than n.
You have to process q queries to this array. Each query is represented by two numbers p and k. Several operations are performed in each query; each operation changes p to p + ap + k. There operations are applied until p becomes greater than n. The answer to the query is the number of performed operations.
Input
The first line contains one integer n (1 ≤ n ≤ 100000).
The second line contains n integers — elements of a (1 ≤ ai ≤ n for each i from 1 to n).
The third line containts one integer q (1 ≤ q ≤ 100000).
Then q lines follow. Each line contains the values of p and k for corresponding query (1 ≤ p, k ≤ n).
Output
Print q integers, ith integer must be equal to the answer to ith query.
Example
Input
3
1 1 1
3
1 1
2 1
3 1
Output
2
1
1
Note
Consider first example:
In first query after first operation p = 3, after second operation p = 5.
In next two queries p is greater than n after the first operation.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
from collections import Counter
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
import heapq
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
# n = int(input())
# ls = list(map(int, input().split()))
# n, k = map(int, input().split())
# n =int(input())
#arr=[(i,x) for i,x in enum]
#arr.sort(key=lambda x:x[0])
#print(arr)
# e=list(map(int, input().split()))
from collections import Counter
#print("\n".join(ls))
#print(os.path.commonprefix(ls[0:2]))
#n=int(input())
from bisect import bisect_right
#for _ in range(int(input())):
#n=int(input())
#arr = list(map(int, input().split()))
#for _ in range(int(input())):
#n, k = map(int, input().split())
import bisect
import math
#n=int(input())
#arr = list(map(int, input().split()))
n=int(input())
arr = list(map(int, input().split()))
dp=[[-1]*(300) for i in range(n+1)]
for i in range(1,300):
for j in range(n-1,-1,-1):
if j+arr[j]+i>n:
dp[j][i]=1
elif j+arr[j]+i==n:
dp[j][i]=2
else:
dp[j][i] = 1 + dp[j + arr[j] + i][i]
for i in range(int(input())):
p, k = map(int, input().split())
if p==n:
print(1)
elif k<299:
print(dp[p][k])
else:
cnt = 0
while p <=n:
p = p + k + arr[p - 1]
cnt += 1
print(cnt)
``` | instruction | 0 | 80,506 | 5 | 161,012 |
No | output | 1 | 80,506 | 5 | 161,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
a is an array of n positive integers, all of which are not greater than n.
You have to process q queries to this array. Each query is represented by two numbers p and k. Several operations are performed in each query; each operation changes p to p + ap + k. There operations are applied until p becomes greater than n. The answer to the query is the number of performed operations.
Input
The first line contains one integer n (1 ≤ n ≤ 100000).
The second line contains n integers — elements of a (1 ≤ ai ≤ n for each i from 1 to n).
The third line containts one integer q (1 ≤ q ≤ 100000).
Then q lines follow. Each line contains the values of p and k for corresponding query (1 ≤ p, k ≤ n).
Output
Print q integers, ith integer must be equal to the answer to ith query.
Example
Input
3
1 1 1
3
1 1
2 1
3 1
Output
2
1
1
Note
Consider first example:
In first query after first operation p = 3, after second operation p = 5.
In next two queries p is greater than n after the first operation.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
from collections import Counter
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
import heapq
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
# n = int(input())
# ls = list(map(int, input().split()))
# n, k = map(int, input().split())
# n =int(input())
#arr=[(i,x) for i,x in enum]
#arr.sort(key=lambda x:x[0])
#print(arr)
# e=list(map(int, input().split()))
from collections import Counter
#print("\n".join(ls))
#print(os.path.commonprefix(ls[0:2]))
#n=int(input())
from bisect import bisect_right
#for _ in range(int(input())):
#n=int(input())
#arr = list(map(int, input().split()))
#for _ in range(int(input())):
#n, k = map(int, input().split())
import bisect
import math
#n=int(input())
#arr = list(map(int, input().split()))
n=int(input())
arr = list(map(int, input().split()))
dp=[[-1]*(300) for i in range(n+1)]
for i in range(1,290):
for j in range(n-1,-1,-1):
if j+arr[j]+i>n:
dp[j][i]=1
elif j+arr[j]+i==n:
dp[j][i]=2
else:
dp[j][i] = 1 + dp[j + arr[j] + i][i]
for i in range(int(input())):
p, k = map(int, input().split())
if k<290:
print(dp[k][p])
else:
cnt = 0
while p <=n:
p = p + k + arr[p - 1]
cnt += 1
print(cnt)
``` | instruction | 0 | 80,507 | 5 | 161,014 |
No | output | 1 | 80,507 | 5 | 161,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
a is an array of n positive integers, all of which are not greater than n.
You have to process q queries to this array. Each query is represented by two numbers p and k. Several operations are performed in each query; each operation changes p to p + ap + k. There operations are applied until p becomes greater than n. The answer to the query is the number of performed operations.
Input
The first line contains one integer n (1 ≤ n ≤ 100000).
The second line contains n integers — elements of a (1 ≤ ai ≤ n for each i from 1 to n).
The third line containts one integer q (1 ≤ q ≤ 100000).
Then q lines follow. Each line contains the values of p and k for corresponding query (1 ≤ p, k ≤ n).
Output
Print q integers, ith integer must be equal to the answer to ith query.
Example
Input
3
1 1 1
3
1 1
2 1
3 1
Output
2
1
1
Note
Consider first example:
In first query after first operation p = 3, after second operation p = 5.
In next two queries p is greater than n after the first operation.
Submitted Solution:
```
def process(array, p, k):
n = len(array)
n1 = array[p - 1] + k
n2 = p
if n <= n2:
return 1
r = (n - n2) // n1
return r + 1
if __name__ == "__main__":
input()
arr = list(map(int, input().split()))
k = int(input())
for i in range(k):
p, k = map(int, input().split())
print(process(arr, p, k))
``` | instruction | 0 | 80,508 | 5 | 161,016 |
No | output | 1 | 80,508 | 5 | 161,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
a is an array of n positive integers, all of which are not greater than n.
You have to process q queries to this array. Each query is represented by two numbers p and k. Several operations are performed in each query; each operation changes p to p + ap + k. There operations are applied until p becomes greater than n. The answer to the query is the number of performed operations.
Input
The first line contains one integer n (1 ≤ n ≤ 100000).
The second line contains n integers — elements of a (1 ≤ ai ≤ n for each i from 1 to n).
The third line containts one integer q (1 ≤ q ≤ 100000).
Then q lines follow. Each line contains the values of p and k for corresponding query (1 ≤ p, k ≤ n).
Output
Print q integers, ith integer must be equal to the answer to ith query.
Example
Input
3
1 1 1
3
1 1
2 1
3 1
Output
2
1
1
Note
Consider first example:
In first query after first operation p = 3, after second operation p = 5.
In next two queries p is greater than n after the first operation.
Submitted Solution:
```
def process(array, p, k):
n = len(array)
n1 = array[p - 1]
n2 = p + k
r = (n - n2) // n1
return max(r + 1, 1)
if __name__ == "__main__":
input()
arr = list(map(int, input().split()))
k = int(input())
for i in range(k):
p, k = map(int, input().split())
print(process(arr, p, k))
``` | instruction | 0 | 80,509 | 5 | 161,018 |
No | output | 1 | 80,509 | 5 | 161,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
You are given a sorted in increasing order singly linked list. You should find the minimum integer in the list which is greater than or equal to x.
More formally, there is a singly liked list built on an array of n elements. Element with index i contains two integers: valuei is the integer value in this element, and nexti that is the index of the next element of the singly linked list (or -1, if the current element is the last). The list is sorted, i.e. if nexti ≠ - 1, then valuenexti > valuei.
You are given the number of elements in the list n, the index of the first element start, and the integer x.
You can make up to 2000 queries of the following two types:
* ? i (1 ≤ i ≤ n) — ask the values valuei and nexti,
* ! ans — give the answer for the problem: the minimum integer, greater than or equal to x, or ! -1, if there are no such integers. Your program should terminate after this query.
Write a program that solves this problem.
Input
The first line contains three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109) — the number of elements in the list, the index of the first element and the integer x.
Output
To print the answer for the problem, print ! ans, where ans is the minimum integer in the list greater than or equal to x, or -1, if there is no such integer.
Interaction
To make a query of the first type, print ? i (1 ≤ i ≤ n), where i is the index of element you want to know information about.
After each query of type ? read two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
It is guaranteed that if nexti ≠ - 1, then valuenexti > valuei, and that the array values give a valid singly linked list with start being the first element.
Note that you can't ask more than 1999 queries of the type ?.
If nexti = - 1 and valuei = - 1, then it means that you asked more queries than allowed, or asked an invalid query. Your program should immediately terminate (for example, by calling exit(0)). You will receive "Wrong Answer", it means that you asked more queries than allowed, or asked an invalid query. If you ignore this, you can get other verdicts since your program will continue to read from a closed stream.
Your solution will get "Idleness Limit Exceeded", if you don't print anything or forget to flush the output, including the final answer.
To flush you can use (just after printing a query and line end):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* For other languages see documentation.
Hacks format
For hacks, use the following format:
In the first line print three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109).
In the next n lines print the description of the elements of the list: in the i-th line print two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
The printed structure should be a valid singly linked list. In particular, it should be possible to reach all elements from start by following links nexti, and the last element end should have -1 in the nextend.
Example
Input
5 3 80
97 -1
58 5
16 2
81 1
79 4
Output
? 1
? 2
? 3
? 4
? 5
! 81
Note
You can read more about singly linked list by the following link: <https://en.wikipedia.org/wiki/Linked_list#Singly_linked_list>
The illustration for the first sample case. Start and finish elements are marked dark. <image>
Submitted Solution:
```
from random import sample
SIZE = 950
def R():
return map(int, input().split())
def ask(i):
print('?', i, flush=True)
v, nxt = R()
if v < 0:
exit()
return v, nxt
n, s, x = R()
v, nxt = ask(s)
if v < x:
q = range(1, n + 1)
if n > SIZE:
q = sample(q, SIZE)
try:
v, nxt = max(filter(lambda t: t[0] < x, map(ask, q)))
except ValueError:
pass
while v < x and nxt >= 1:
v, nxt = ask(nxt)
print('!', v if v >= x else -1)
``` | instruction | 0 | 80,534 | 5 | 161,068 |
Yes | output | 1 | 80,534 | 5 | 161,069 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
You are given a sorted in increasing order singly linked list. You should find the minimum integer in the list which is greater than or equal to x.
More formally, there is a singly liked list built on an array of n elements. Element with index i contains two integers: valuei is the integer value in this element, and nexti that is the index of the next element of the singly linked list (or -1, if the current element is the last). The list is sorted, i.e. if nexti ≠ - 1, then valuenexti > valuei.
You are given the number of elements in the list n, the index of the first element start, and the integer x.
You can make up to 2000 queries of the following two types:
* ? i (1 ≤ i ≤ n) — ask the values valuei and nexti,
* ! ans — give the answer for the problem: the minimum integer, greater than or equal to x, or ! -1, if there are no such integers. Your program should terminate after this query.
Write a program that solves this problem.
Input
The first line contains three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109) — the number of elements in the list, the index of the first element and the integer x.
Output
To print the answer for the problem, print ! ans, where ans is the minimum integer in the list greater than or equal to x, or -1, if there is no such integer.
Interaction
To make a query of the first type, print ? i (1 ≤ i ≤ n), where i is the index of element you want to know information about.
After each query of type ? read two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
It is guaranteed that if nexti ≠ - 1, then valuenexti > valuei, and that the array values give a valid singly linked list with start being the first element.
Note that you can't ask more than 1999 queries of the type ?.
If nexti = - 1 and valuei = - 1, then it means that you asked more queries than allowed, or asked an invalid query. Your program should immediately terminate (for example, by calling exit(0)). You will receive "Wrong Answer", it means that you asked more queries than allowed, or asked an invalid query. If you ignore this, you can get other verdicts since your program will continue to read from a closed stream.
Your solution will get "Idleness Limit Exceeded", if you don't print anything or forget to flush the output, including the final answer.
To flush you can use (just after printing a query and line end):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* For other languages see documentation.
Hacks format
For hacks, use the following format:
In the first line print three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109).
In the next n lines print the description of the elements of the list: in the i-th line print two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
The printed structure should be a valid singly linked list. In particular, it should be possible to reach all elements from start by following links nexti, and the last element end should have -1 in the nextend.
Example
Input
5 3 80
97 -1
58 5
16 2
81 1
79 4
Output
? 1
? 2
? 3
? 4
? 5
! 81
Note
You can read more about singly linked list by the following link: <https://en.wikipedia.org/wiki/Linked_list#Singly_linked_list>
The illustration for the first sample case. Start and finish elements are marked dark. <image>
Submitted Solution:
```
from random import sample
def R():
return map(int, input().split())
def ask(i):
print('?', i, flush=True)
v, nxt = R()
if v < 0:
exit()
return v, nxt
def ans(v):
print('!', v)
exit()
n, s, x = R()
d = [None] * (n + 1)
f = [False] * (n + 1)
mv = -1
i = s
count = 950
q = range(1, n + 1)
if n > count:
q = sample(q, count)
if s not in q:
q[0] = s
for i in q:
if not f[i]:
f[i] = True
v, nxt = ask(i)
if v == x or i == s and v > x:
ans(v)
if nxt != -1:
f[nxt] = True
elif v < x:
ans(-1)
if v < x:
nv = d[nxt]
if nv is None:
if v > mv:
mv, mnxt = v, nxt
elif nv > x:
ans(nv)
d[i] = v
while mv < x and mnxt >= 1:
mv, mnxt = (d[mnxt], None) if d[mnxt] else ask(mnxt)
ans(mv if mv >= x else -1)
``` | instruction | 0 | 80,535 | 5 | 161,070 |
Yes | output | 1 | 80,535 | 5 | 161,071 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
You are given a sorted in increasing order singly linked list. You should find the minimum integer in the list which is greater than or equal to x.
More formally, there is a singly liked list built on an array of n elements. Element with index i contains two integers: valuei is the integer value in this element, and nexti that is the index of the next element of the singly linked list (or -1, if the current element is the last). The list is sorted, i.e. if nexti ≠ - 1, then valuenexti > valuei.
You are given the number of elements in the list n, the index of the first element start, and the integer x.
You can make up to 2000 queries of the following two types:
* ? i (1 ≤ i ≤ n) — ask the values valuei and nexti,
* ! ans — give the answer for the problem: the minimum integer, greater than or equal to x, or ! -1, if there are no such integers. Your program should terminate after this query.
Write a program that solves this problem.
Input
The first line contains three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109) — the number of elements in the list, the index of the first element and the integer x.
Output
To print the answer for the problem, print ! ans, where ans is the minimum integer in the list greater than or equal to x, or -1, if there is no such integer.
Interaction
To make a query of the first type, print ? i (1 ≤ i ≤ n), where i is the index of element you want to know information about.
After each query of type ? read two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
It is guaranteed that if nexti ≠ - 1, then valuenexti > valuei, and that the array values give a valid singly linked list with start being the first element.
Note that you can't ask more than 1999 queries of the type ?.
If nexti = - 1 and valuei = - 1, then it means that you asked more queries than allowed, or asked an invalid query. Your program should immediately terminate (for example, by calling exit(0)). You will receive "Wrong Answer", it means that you asked more queries than allowed, or asked an invalid query. If you ignore this, you can get other verdicts since your program will continue to read from a closed stream.
Your solution will get "Idleness Limit Exceeded", if you don't print anything or forget to flush the output, including the final answer.
To flush you can use (just after printing a query and line end):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* For other languages see documentation.
Hacks format
For hacks, use the following format:
In the first line print three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109).
In the next n lines print the description of the elements of the list: in the i-th line print two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
The printed structure should be a valid singly linked list. In particular, it should be possible to reach all elements from start by following links nexti, and the last element end should have -1 in the nextend.
Example
Input
5 3 80
97 -1
58 5
16 2
81 1
79 4
Output
? 1
? 2
? 3
? 4
? 5
! 81
Note
You can read more about singly linked list by the following link: <https://en.wikipedia.org/wiki/Linked_list#Singly_linked_list>
The illustration for the first sample case. Start and finish elements are marked dark. <image>
Submitted Solution:
```
from random import sample
def R():
return map(int, input().split())
def ask(i):
print('?', i, flush=True)
v, nxt = R()
if v < 0:
exit()
return v, nxt
def ans(v):
print('!', v)
exit()
n, s, x = R()
mv = -1
i = s
count = 950
q = range(1, n + 1)
if n > count:
q = sample(q, count)
if s not in q:
q[0] = s
for i in q:
v, nxt = ask(i)
if v == x or i == s and v > x:
ans(v)
if v < x:
if nxt < 0:
ans(-1)
if v > mv:
mv, mnxt = v, nxt
while mv < x and mnxt >= 1:
mv, mnxt = ask(mnxt)
ans(mv if mv >= x else -1)
``` | instruction | 0 | 80,536 | 5 | 161,072 |
Yes | output | 1 | 80,536 | 5 | 161,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
You are given a sorted in increasing order singly linked list. You should find the minimum integer in the list which is greater than or equal to x.
More formally, there is a singly liked list built on an array of n elements. Element with index i contains two integers: valuei is the integer value in this element, and nexti that is the index of the next element of the singly linked list (or -1, if the current element is the last). The list is sorted, i.e. if nexti ≠ - 1, then valuenexti > valuei.
You are given the number of elements in the list n, the index of the first element start, and the integer x.
You can make up to 2000 queries of the following two types:
* ? i (1 ≤ i ≤ n) — ask the values valuei and nexti,
* ! ans — give the answer for the problem: the minimum integer, greater than or equal to x, or ! -1, if there are no such integers. Your program should terminate after this query.
Write a program that solves this problem.
Input
The first line contains three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109) — the number of elements in the list, the index of the first element and the integer x.
Output
To print the answer for the problem, print ! ans, where ans is the minimum integer in the list greater than or equal to x, or -1, if there is no such integer.
Interaction
To make a query of the first type, print ? i (1 ≤ i ≤ n), where i is the index of element you want to know information about.
After each query of type ? read two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
It is guaranteed that if nexti ≠ - 1, then valuenexti > valuei, and that the array values give a valid singly linked list with start being the first element.
Note that you can't ask more than 1999 queries of the type ?.
If nexti = - 1 and valuei = - 1, then it means that you asked more queries than allowed, or asked an invalid query. Your program should immediately terminate (for example, by calling exit(0)). You will receive "Wrong Answer", it means that you asked more queries than allowed, or asked an invalid query. If you ignore this, you can get other verdicts since your program will continue to read from a closed stream.
Your solution will get "Idleness Limit Exceeded", if you don't print anything or forget to flush the output, including the final answer.
To flush you can use (just after printing a query and line end):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* For other languages see documentation.
Hacks format
For hacks, use the following format:
In the first line print three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109).
In the next n lines print the description of the elements of the list: in the i-th line print two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
The printed structure should be a valid singly linked list. In particular, it should be possible to reach all elements from start by following links nexti, and the last element end should have -1 in the nextend.
Example
Input
5 3 80
97 -1
58 5
16 2
81 1
79 4
Output
? 1
? 2
? 3
? 4
? 5
! 81
Note
You can read more about singly linked list by the following link: <https://en.wikipedia.org/wiki/Linked_list#Singly_linked_list>
The illustration for the first sample case. Start and finish elements are marked dark. <image>
Submitted Solution:
```
import random
SIZE = 900
def R():
return map(int, input().split())
def ask(i):
print('?', i, flush=True)
v, nxt = R()
if v < 0:
exit()
return v, nxt
n, s, x = R()
q = range(1, n + 1)
if n > SIZE:
q = random.sample(q, SIZE)
v, nxt = max((t for t in map(ask, q) if t[0] < x), default=(-1, s))
while v < x and ~nxt:
v, nxt = ask(nxt)
print('!', v if v >= x else -1)
``` | instruction | 0 | 80,537 | 5 | 161,074 |
Yes | output | 1 | 80,537 | 5 | 161,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
You are given a sorted in increasing order singly linked list. You should find the minimum integer in the list which is greater than or equal to x.
More formally, there is a singly liked list built on an array of n elements. Element with index i contains two integers: valuei is the integer value in this element, and nexti that is the index of the next element of the singly linked list (or -1, if the current element is the last). The list is sorted, i.e. if nexti ≠ - 1, then valuenexti > valuei.
You are given the number of elements in the list n, the index of the first element start, and the integer x.
You can make up to 2000 queries of the following two types:
* ? i (1 ≤ i ≤ n) — ask the values valuei and nexti,
* ! ans — give the answer for the problem: the minimum integer, greater than or equal to x, or ! -1, if there are no such integers. Your program should terminate after this query.
Write a program that solves this problem.
Input
The first line contains three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109) — the number of elements in the list, the index of the first element and the integer x.
Output
To print the answer for the problem, print ! ans, where ans is the minimum integer in the list greater than or equal to x, or -1, if there is no such integer.
Interaction
To make a query of the first type, print ? i (1 ≤ i ≤ n), where i is the index of element you want to know information about.
After each query of type ? read two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
It is guaranteed that if nexti ≠ - 1, then valuenexti > valuei, and that the array values give a valid singly linked list with start being the first element.
Note that you can't ask more than 1999 queries of the type ?.
If nexti = - 1 and valuei = - 1, then it means that you asked more queries than allowed, or asked an invalid query. Your program should immediately terminate (for example, by calling exit(0)). You will receive "Wrong Answer", it means that you asked more queries than allowed, or asked an invalid query. If you ignore this, you can get other verdicts since your program will continue to read from a closed stream.
Your solution will get "Idleness Limit Exceeded", if you don't print anything or forget to flush the output, including the final answer.
To flush you can use (just after printing a query and line end):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* For other languages see documentation.
Hacks format
For hacks, use the following format:
In the first line print three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109).
In the next n lines print the description of the elements of the list: in the i-th line print two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
The printed structure should be a valid singly linked list. In particular, it should be possible to reach all elements from start by following links nexti, and the last element end should have -1 in the nextend.
Example
Input
5 3 80
97 -1
58 5
16 2
81 1
79 4
Output
? 1
? 2
? 3
? 4
? 5
! 81
Note
You can read more about singly linked list by the following link: <https://en.wikipedia.org/wiki/Linked_list#Singly_linked_list>
The illustration for the first sample case. Start and finish elements are marked dark. <image>
Submitted Solution:
```
import random, sys
from datetime import datetime
def main():
n, start, x = map(int, input().split())
indexes = list(map(lambda x: random.randint(1, n), range(1000)))
best, bestNext = 0, 0
for index in indexes:
sys.stdout.write('? %d\n' % index)
sys.stdout.flush()
value, nextId = map(int, input().split())
if value < x and best < x:
best, bestNext = value, nextId
while best < x and bestNext != -1:
sys.stdout.write('? %d\n' % bestNext)
sys.stdout.flush()
best, bestNext = map(int, input().split())
if best < x:
sys.stdout.write('! -1\n')
sys.stdout.flush()
else:
sys.stdout.write('! %d\n' % best)
sys.stdout.flush()
if __name__ == '__main__':
random.seed(datetime.now())
main()
``` | instruction | 0 | 80,538 | 5 | 161,076 |
No | output | 1 | 80,538 | 5 | 161,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
You are given a sorted in increasing order singly linked list. You should find the minimum integer in the list which is greater than or equal to x.
More formally, there is a singly liked list built on an array of n elements. Element with index i contains two integers: valuei is the integer value in this element, and nexti that is the index of the next element of the singly linked list (or -1, if the current element is the last). The list is sorted, i.e. if nexti ≠ - 1, then valuenexti > valuei.
You are given the number of elements in the list n, the index of the first element start, and the integer x.
You can make up to 2000 queries of the following two types:
* ? i (1 ≤ i ≤ n) — ask the values valuei and nexti,
* ! ans — give the answer for the problem: the minimum integer, greater than or equal to x, or ! -1, if there are no such integers. Your program should terminate after this query.
Write a program that solves this problem.
Input
The first line contains three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109) — the number of elements in the list, the index of the first element and the integer x.
Output
To print the answer for the problem, print ! ans, where ans is the minimum integer in the list greater than or equal to x, or -1, if there is no such integer.
Interaction
To make a query of the first type, print ? i (1 ≤ i ≤ n), where i is the index of element you want to know information about.
After each query of type ? read two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
It is guaranteed that if nexti ≠ - 1, then valuenexti > valuei, and that the array values give a valid singly linked list with start being the first element.
Note that you can't ask more than 1999 queries of the type ?.
If nexti = - 1 and valuei = - 1, then it means that you asked more queries than allowed, or asked an invalid query. Your program should immediately terminate (for example, by calling exit(0)). You will receive "Wrong Answer", it means that you asked more queries than allowed, or asked an invalid query. If you ignore this, you can get other verdicts since your program will continue to read from a closed stream.
Your solution will get "Idleness Limit Exceeded", if you don't print anything or forget to flush the output, including the final answer.
To flush you can use (just after printing a query and line end):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* For other languages see documentation.
Hacks format
For hacks, use the following format:
In the first line print three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109).
In the next n lines print the description of the elements of the list: in the i-th line print two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
The printed structure should be a valid singly linked list. In particular, it should be possible to reach all elements from start by following links nexti, and the last element end should have -1 in the nextend.
Example
Input
5 3 80
97 -1
58 5
16 2
81 1
79 4
Output
? 1
? 2
? 3
? 4
? 5
! 81
Note
You can read more about singly linked list by the following link: <https://en.wikipedia.org/wiki/Linked_list#Singly_linked_list>
The illustration for the first sample case. Start and finish elements are marked dark. <image>
Submitted Solution:
```
from sys import stdout
n, start, x = map(int, input().split())
not_vis = set(range(1, n+1))
print('? ' + str(start))
stdout.flush()
val, nxt = map(int, input().split())
vis = set([start])
while 1:
if nxt == -1 and val == -1:
break
if val >= x:
print('! ' + str(val))
stdout.flush()
break
if nxt == -1:
print('! -1')
break
print('? ' + str(nxt))
stdout.flush()
v1, n1 = map(int, input().split())
vis.add(nxt)
rnd = next(iter(not_vis))
not_vis.remove(rnd)
print('? ' + str(rnd))
stdout.flush()
v2, n2 = map(int, input().split())
vis.add(rnd)
if v2 > v1 and v2 <= x:
val = v2
nxt = n2
else:
val = v1
nxt = n1
``` | instruction | 0 | 80,539 | 5 | 161,078 |
No | output | 1 | 80,539 | 5 | 161,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
You are given a sorted in increasing order singly linked list. You should find the minimum integer in the list which is greater than or equal to x.
More formally, there is a singly liked list built on an array of n elements. Element with index i contains two integers: valuei is the integer value in this element, and nexti that is the index of the next element of the singly linked list (or -1, if the current element is the last). The list is sorted, i.e. if nexti ≠ - 1, then valuenexti > valuei.
You are given the number of elements in the list n, the index of the first element start, and the integer x.
You can make up to 2000 queries of the following two types:
* ? i (1 ≤ i ≤ n) — ask the values valuei and nexti,
* ! ans — give the answer for the problem: the minimum integer, greater than or equal to x, or ! -1, if there are no such integers. Your program should terminate after this query.
Write a program that solves this problem.
Input
The first line contains three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109) — the number of elements in the list, the index of the first element and the integer x.
Output
To print the answer for the problem, print ! ans, where ans is the minimum integer in the list greater than or equal to x, or -1, if there is no such integer.
Interaction
To make a query of the first type, print ? i (1 ≤ i ≤ n), where i is the index of element you want to know information about.
After each query of type ? read two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
It is guaranteed that if nexti ≠ - 1, then valuenexti > valuei, and that the array values give a valid singly linked list with start being the first element.
Note that you can't ask more than 1999 queries of the type ?.
If nexti = - 1 and valuei = - 1, then it means that you asked more queries than allowed, or asked an invalid query. Your program should immediately terminate (for example, by calling exit(0)). You will receive "Wrong Answer", it means that you asked more queries than allowed, or asked an invalid query. If you ignore this, you can get other verdicts since your program will continue to read from a closed stream.
Your solution will get "Idleness Limit Exceeded", if you don't print anything or forget to flush the output, including the final answer.
To flush you can use (just after printing a query and line end):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* For other languages see documentation.
Hacks format
For hacks, use the following format:
In the first line print three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109).
In the next n lines print the description of the elements of the list: in the i-th line print two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
The printed structure should be a valid singly linked list. In particular, it should be possible to reach all elements from start by following links nexti, and the last element end should have -1 in the nextend.
Example
Input
5 3 80
97 -1
58 5
16 2
81 1
79 4
Output
? 1
? 2
? 3
? 4
? 5
! 81
Note
You can read more about singly linked list by the following link: <https://en.wikipedia.org/wiki/Linked_list#Singly_linked_list>
The illustration for the first sample case. Start and finish elements are marked dark. <image>
Submitted Solution:
```
from random import randint
from sys import stdout
n, start, x = [int(t) for t in input().split()]
best_index = None
best_value = None
for i in range(10):
ind = randint(0, n - 1)
print('? {}'.format(ind + 1))
stdout.flush()
value, nxt = [int(t) for t in input().split()]
if value < x and (best_value is None or value > best_value):
best_value = value
best_index = ind
if best_value is None:
best_index = start
ind = best_index
while True:
print('? {}'.format(ind + 1))
stdout.flush()
value, nxt = [int(t) for t in input().split()]
if value >= x:
print('! {}'.format(value))
break
ind = nxt - 1
``` | instruction | 0 | 80,540 | 5 | 161,080 |
No | output | 1 | 80,540 | 5 | 161,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
You are given a sorted in increasing order singly linked list. You should find the minimum integer in the list which is greater than or equal to x.
More formally, there is a singly liked list built on an array of n elements. Element with index i contains two integers: valuei is the integer value in this element, and nexti that is the index of the next element of the singly linked list (or -1, if the current element is the last). The list is sorted, i.e. if nexti ≠ - 1, then valuenexti > valuei.
You are given the number of elements in the list n, the index of the first element start, and the integer x.
You can make up to 2000 queries of the following two types:
* ? i (1 ≤ i ≤ n) — ask the values valuei and nexti,
* ! ans — give the answer for the problem: the minimum integer, greater than or equal to x, or ! -1, if there are no such integers. Your program should terminate after this query.
Write a program that solves this problem.
Input
The first line contains three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109) — the number of elements in the list, the index of the first element and the integer x.
Output
To print the answer for the problem, print ! ans, where ans is the minimum integer in the list greater than or equal to x, or -1, if there is no such integer.
Interaction
To make a query of the first type, print ? i (1 ≤ i ≤ n), where i is the index of element you want to know information about.
After each query of type ? read two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
It is guaranteed that if nexti ≠ - 1, then valuenexti > valuei, and that the array values give a valid singly linked list with start being the first element.
Note that you can't ask more than 1999 queries of the type ?.
If nexti = - 1 and valuei = - 1, then it means that you asked more queries than allowed, or asked an invalid query. Your program should immediately terminate (for example, by calling exit(0)). You will receive "Wrong Answer", it means that you asked more queries than allowed, or asked an invalid query. If you ignore this, you can get other verdicts since your program will continue to read from a closed stream.
Your solution will get "Idleness Limit Exceeded", if you don't print anything or forget to flush the output, including the final answer.
To flush you can use (just after printing a query and line end):
* fflush(stdout) in C++;
* System.out.flush() in Java;
* stdout.flush() in Python;
* flush(output) in Pascal;
* For other languages see documentation.
Hacks format
For hacks, use the following format:
In the first line print three integers n, start, x (1 ≤ n ≤ 50000, 1 ≤ start ≤ n, 0 ≤ x ≤ 109).
In the next n lines print the description of the elements of the list: in the i-th line print two integers valuei and nexti (0 ≤ valuei ≤ 109, - 1 ≤ nexti ≤ n, nexti ≠ 0).
The printed structure should be a valid singly linked list. In particular, it should be possible to reach all elements from start by following links nexti, and the last element end should have -1 in the nextend.
Example
Input
5 3 80
97 -1
58 5
16 2
81 1
79 4
Output
? 1
? 2
? 3
? 4
? 5
! 81
Note
You can read more about singly linked list by the following link: <https://en.wikipedia.org/wiki/Linked_list#Singly_linked_list>
The illustration for the first sample case. Start and finish elements are marked dark. <image>
Submitted Solution:
```
import random
from sys import stdout
n,start,x=map(int,input().split())
ans=start
annnsss=-1
z=0
print("?",start)
stdout.flush()
anss,b=map(int,input().split())
if anss>=x:
print("!",anss)
stdout.flush()
else:
i = 0
while z != 10 and i!= 1750:
i=i+1
anns=random.randrange(1,n+1)
print("?",anns)
stdout.flush()
annss,b=map(int,input().split())
if (annss<x) and (anns<annss):
ans=anns
anss=annss
if annss == x:
print("!", x)
stdout.flush()
z = 10
if z!=10:
print("?",ans)
stdout.flush()
annss,b=map(int,input().split())
while (anss<x and b!=-1):
print("?",b)
stdout.flush()
annss,b=map(int,input().split())
annnsss=anss
anss=annss
if anss<x:
print("! -1")
stdout.flush()
else:
print("!", anss)
stdout.flush()
``` | instruction | 0 | 80,541 | 5 | 161,082 |
No | output | 1 | 80,541 | 5 | 161,083 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Mr. ukuku1333 is a little sloppy, so when I expanded the product of the linear expressions of x, I couldn't figure out the original linear expression.
Given the nth degree polynomial of x, factor it into the product of the original linear expressions of x.
The nth degree polynomial of x is given by the following BNF.
<Polynomial>: = <Term> | <Polynomial> & plus; <Polynomial> | <Polynomial> − <Term>
<Term>: = x ^ <exponent> | <coefficient> x ^ <index> | <coefficient> x | <constant>
<Index>: = [2-5]
<Coefficient>: = [1-9] [0-9] *
<Constant>: = [1-9] [0-9] *
If the exponent and coefficient are omitted, it is regarded as 1.
Constraints
The input satisfies the following conditions.
* 2 ≤ n ≤ 5
* For any set of i, j such that 1 ≤ i <j ≤ m, where m is the number of terms in the given expression,
The degree of the i-th term is guaranteed to be greater than the degree of the j-th term
* It is guaranteed that the nth degree polynomial of x given can be factored into the product form of the linear expression of x.
* Absolute values of coefficients and constants are 2 × 103 or less, respectively.
* The coefficient with the highest degree is 1, which is guaranteed to be omitted.
* The original constant term of each linear expression before expansion is guaranteed to be a non-zero integer
* It is guaranteed that the original constant terms of each linear expression before expansion are different.
Input
The input is given in the following format.
S
The string S representing the nth degree polynomial of x is given on one line.
Output
Factor S into the product of a linear expression of x, and output it in ascending order of the constant term.
Insert a line break at the end of the output.
Examples
Input
x^2+3x+2
Output
(x+1)(x+2)
Input
x^2-1
Output
(x-1)(x+1)
Input
x^5+15x^4+85x^3+225x^2+274x+120
Output
(x+1)(x+2)(x+3)(x+4)(x+5)
Input
x^3-81x^2-1882x-1800
Output
(x-100)(x+1)(x+18) | instruction | 0 | 80,771 | 5 | 161,542 |
"Correct Solution:
```
import copy
s = input()
k = [0] * 6
t = []
while True:
i = 0
if "+" in s[1:]:
i = 1 + s[1:].index("+")
if "-" in s[1:]:
if i > 0:i = min(i, 1 + s[1:].index("-"))
else:i = 1 + s[1:].index("-")
if i == 0:
k[0] = int(s)
break
j = s.index("x")
if j + 3 == i:
if j == 0:k[int(s[j + 2])] = 1
else:k[int(s[j + 2])] = int(s[:j])
else:
if j == 0:k[1] = 1
elif j == 1:k[1] = int(s[0] + "1")
else:k[1] = int(s[:j])
s = s[i:]
while True:
if k[-1] == 0:k.pop()
else:break
while True:
i = 1
while i <= abs(k[0]):
if k[0] % i == 0:
j = k[0] // i
kn = copy.deepcopy(k)
for l in range(len(kn) - 1):
if kn[l] % j == 0:
kn[l + 1] -= kn[l] // j
kn[l] //= j
else:
break
else:
if kn[-1] == 0:
kn.pop()
t.append(j)
k = copy.deepcopy(kn)
break
j *= -1
kn = copy.deepcopy(k)
for l in range(len(kn) - 1):
if kn[l] % j == 0:
kn[l + 1] -= kn[l] // j
kn[l] //= j
else:
break
else:
if kn[-1] == 0:
kn.pop()
t.append(j)
k = copy.deepcopy(kn)
break
i += 1
if len(k) == 2:
t.append(k[0])
break
t.sort()
s = ""
for i in t:s += "(x{})".format(i) if i < 0 else "(x+{})".format(i)
print(s)
``` | output | 1 | 80,771 | 5 | 161,543 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Mr. ukuku1333 is a little sloppy, so when I expanded the product of the linear expressions of x, I couldn't figure out the original linear expression.
Given the nth degree polynomial of x, factor it into the product of the original linear expressions of x.
The nth degree polynomial of x is given by the following BNF.
<Polynomial>: = <Term> | <Polynomial> & plus; <Polynomial> | <Polynomial> − <Term>
<Term>: = x ^ <exponent> | <coefficient> x ^ <index> | <coefficient> x | <constant>
<Index>: = [2-5]
<Coefficient>: = [1-9] [0-9] *
<Constant>: = [1-9] [0-9] *
If the exponent and coefficient are omitted, it is regarded as 1.
Constraints
The input satisfies the following conditions.
* 2 ≤ n ≤ 5
* For any set of i, j such that 1 ≤ i <j ≤ m, where m is the number of terms in the given expression,
The degree of the i-th term is guaranteed to be greater than the degree of the j-th term
* It is guaranteed that the nth degree polynomial of x given can be factored into the product form of the linear expression of x.
* Absolute values of coefficients and constants are 2 × 103 or less, respectively.
* The coefficient with the highest degree is 1, which is guaranteed to be omitted.
* The original constant term of each linear expression before expansion is guaranteed to be a non-zero integer
* It is guaranteed that the original constant terms of each linear expression before expansion are different.
Input
The input is given in the following format.
S
The string S representing the nth degree polynomial of x is given on one line.
Output
Factor S into the product of a linear expression of x, and output it in ascending order of the constant term.
Insert a line break at the end of the output.
Examples
Input
x^2+3x+2
Output
(x+1)(x+2)
Input
x^2-1
Output
(x-1)(x+1)
Input
x^5+15x^4+85x^3+225x^2+274x+120
Output
(x+1)(x+2)(x+3)(x+4)(x+5)
Input
x^3-81x^2-1882x-1800
Output
(x-100)(x+1)(x+18) | instruction | 0 | 80,772 | 5 | 161,544 |
"Correct Solution:
```
import re
S = input()
S = S.replace('x', '*({0})')
S = S.replace('^', '**')
S = re.sub(r'([+-])\*', r'\1', S)[1:]
for i in range(2000, -2001, -1):
if eval(S.format(i)) == 0:
print('(x{:+})'.format(-i), end='')
else:
print()
``` | output | 1 | 80,772 | 5 | 161,545 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Mr. ukuku1333 is a little sloppy, so when I expanded the product of the linear expressions of x, I couldn't figure out the original linear expression.
Given the nth degree polynomial of x, factor it into the product of the original linear expressions of x.
The nth degree polynomial of x is given by the following BNF.
<Polynomial>: = <Term> | <Polynomial> & plus; <Polynomial> | <Polynomial> − <Term>
<Term>: = x ^ <exponent> | <coefficient> x ^ <index> | <coefficient> x | <constant>
<Index>: = [2-5]
<Coefficient>: = [1-9] [0-9] *
<Constant>: = [1-9] [0-9] *
If the exponent and coefficient are omitted, it is regarded as 1.
Constraints
The input satisfies the following conditions.
* 2 ≤ n ≤ 5
* For any set of i, j such that 1 ≤ i <j ≤ m, where m is the number of terms in the given expression,
The degree of the i-th term is guaranteed to be greater than the degree of the j-th term
* It is guaranteed that the nth degree polynomial of x given can be factored into the product form of the linear expression of x.
* Absolute values of coefficients and constants are 2 × 103 or less, respectively.
* The coefficient with the highest degree is 1, which is guaranteed to be omitted.
* The original constant term of each linear expression before expansion is guaranteed to be a non-zero integer
* It is guaranteed that the original constant terms of each linear expression before expansion are different.
Input
The input is given in the following format.
S
The string S representing the nth degree polynomial of x is given on one line.
Output
Factor S into the product of a linear expression of x, and output it in ascending order of the constant term.
Insert a line break at the end of the output.
Examples
Input
x^2+3x+2
Output
(x+1)(x+2)
Input
x^2-1
Output
(x-1)(x+1)
Input
x^5+15x^4+85x^3+225x^2+274x+120
Output
(x+1)(x+2)(x+3)(x+4)(x+5)
Input
x^3-81x^2-1882x-1800
Output
(x-100)(x+1)(x+18) | instruction | 0 | 80,773 | 5 | 161,546 |
"Correct Solution:
```
S = input()
L = len(S)
cur = 0
P = [0]*6
def expr():
global cur
op = "+"
while 1:
term(op)
if cur == L:
break
op = S[cur]
cur += 1 # "+"/"-"
def num():
global cur
v = 0
while cur < L and S[cur] in "0123456789":
v = 10*v + int(S[cur])
cur += 1
return v
def term(op):
global cur
k = 1
if S[cur] != 'x':
k = num()
d = 0
if cur < L and S[cur] == "x":
cur += 1 # "x"
d = 1
if cur < L and S[cur] == '^':
cur += 1 # "^"
d = num()
P[d] = k if op == '+' else -k
expr()
def f(P, x):
res = 0
for i in range(6):
res += P[i]*(x**i)
return res
d = 5
while P[d] == 0:
d -= 1
R = []
for x in range(2000, -2001, -1):
if f(P, x) == 0:
R.append(x)
print("".join("(x%+d)" % -x for x in R))
``` | output | 1 | 80,773 | 5 | 161,547 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Mr. ukuku1333 is a little sloppy, so when I expanded the product of the linear expressions of x, I couldn't figure out the original linear expression.
Given the nth degree polynomial of x, factor it into the product of the original linear expressions of x.
The nth degree polynomial of x is given by the following BNF.
<Polynomial>: = <Term> | <Polynomial> & plus; <Polynomial> | <Polynomial> − <Term>
<Term>: = x ^ <exponent> | <coefficient> x ^ <index> | <coefficient> x | <constant>
<Index>: = [2-5]
<Coefficient>: = [1-9] [0-9] *
<Constant>: = [1-9] [0-9] *
If the exponent and coefficient are omitted, it is regarded as 1.
Constraints
The input satisfies the following conditions.
* 2 ≤ n ≤ 5
* For any set of i, j such that 1 ≤ i <j ≤ m, where m is the number of terms in the given expression,
The degree of the i-th term is guaranteed to be greater than the degree of the j-th term
* It is guaranteed that the nth degree polynomial of x given can be factored into the product form of the linear expression of x.
* Absolute values of coefficients and constants are 2 × 103 or less, respectively.
* The coefficient with the highest degree is 1, which is guaranteed to be omitted.
* The original constant term of each linear expression before expansion is guaranteed to be a non-zero integer
* It is guaranteed that the original constant terms of each linear expression before expansion are different.
Input
The input is given in the following format.
S
The string S representing the nth degree polynomial of x is given on one line.
Output
Factor S into the product of a linear expression of x, and output it in ascending order of the constant term.
Insert a line break at the end of the output.
Examples
Input
x^2+3x+2
Output
(x+1)(x+2)
Input
x^2-1
Output
(x-1)(x+1)
Input
x^5+15x^4+85x^3+225x^2+274x+120
Output
(x+1)(x+2)(x+3)(x+4)(x+5)
Input
x^3-81x^2-1882x-1800
Output
(x-100)(x+1)(x+18) | instruction | 0 | 80,774 | 5 | 161,548 |
"Correct Solution:
```
S = input().replace('+x', '+1x').replace('-x', '-1x').replace('^', '**').replace('x', '*1*(x)')[1:]
for i in range(2000, -2001, -1):
if i == 0: continue
T = S.replace('x', str(i))
#print(T, eval(T))
if eval(T) == 0:
print('(x{:+})'.format(-i), end='')
else:
print()
``` | output | 1 | 80,774 | 5 | 161,549 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Mr. ukuku1333 is a little sloppy, so when I expanded the product of the linear expressions of x, I couldn't figure out the original linear expression.
Given the nth degree polynomial of x, factor it into the product of the original linear expressions of x.
The nth degree polynomial of x is given by the following BNF.
<Polynomial>: = <Term> | <Polynomial> & plus; <Polynomial> | <Polynomial> − <Term>
<Term>: = x ^ <exponent> | <coefficient> x ^ <index> | <coefficient> x | <constant>
<Index>: = [2-5]
<Coefficient>: = [1-9] [0-9] *
<Constant>: = [1-9] [0-9] *
If the exponent and coefficient are omitted, it is regarded as 1.
Constraints
The input satisfies the following conditions.
* 2 ≤ n ≤ 5
* For any set of i, j such that 1 ≤ i <j ≤ m, where m is the number of terms in the given expression,
The degree of the i-th term is guaranteed to be greater than the degree of the j-th term
* It is guaranteed that the nth degree polynomial of x given can be factored into the product form of the linear expression of x.
* Absolute values of coefficients and constants are 2 × 103 or less, respectively.
* The coefficient with the highest degree is 1, which is guaranteed to be omitted.
* The original constant term of each linear expression before expansion is guaranteed to be a non-zero integer
* It is guaranteed that the original constant terms of each linear expression before expansion are different.
Input
The input is given in the following format.
S
The string S representing the nth degree polynomial of x is given on one line.
Output
Factor S into the product of a linear expression of x, and output it in ascending order of the constant term.
Insert a line break at the end of the output.
Examples
Input
x^2+3x+2
Output
(x+1)(x+2)
Input
x^2-1
Output
(x-1)(x+1)
Input
x^5+15x^4+85x^3+225x^2+274x+120
Output
(x+1)(x+2)(x+3)(x+4)(x+5)
Input
x^3-81x^2-1882x-1800
Output
(x-100)(x+1)(x+18) | instruction | 0 | 80,775 | 5 | 161,550 |
"Correct Solution:
```
def parse(S):
poly = []
t = []
for x in S.split('+'):
if '-' in x:
t = t + ['-' + a if i != 0 else a for i, a in enumerate(x.split('-'))]
else:
t.append(x)
for x in t:
if '^' in x:
t = x.split('x^')
if len(t[0]) == 0:
a = 1
else:
a = int(t[0])
b = int(t[1])
else:
if 'x' in x:
if x == 'x':
a = 1
elif x == '-x':
a = -1
else:
a = int(x[:-1])
b = 1
else:
a = int(x)
b = 0
poly.append((a, b))
return poly
def calc_yaku(n):
ret = []
for i in range(n + 1):
if i != 0 and n % i == 0:
ret.append(i)
return reversed(sorted(ret + [-x for x in ret]))
def calc(poly, x):
ret = 0
for p in poly:
ret += p[0] * x ** p[1]
return ret
def solve(S):
poly = parse(S)
n = abs(poly[-1][0])
yaku = calc_yaku(n)
ans = []
for x in yaku:
if calc(poly, x) == 0:
ans.append(-x)
for x in ans:
if x > 0:
print('(x+{})'.format(x), end='')
else:
print('(x{})'.format(x), end='')
print('')
S=input()
solve(S)
``` | output | 1 | 80,775 | 5 | 161,551 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Mr. ukuku1333 is a little sloppy, so when I expanded the product of the linear expressions of x, I couldn't figure out the original linear expression.
Given the nth degree polynomial of x, factor it into the product of the original linear expressions of x.
The nth degree polynomial of x is given by the following BNF.
<Polynomial>: = <Term> | <Polynomial> & plus; <Polynomial> | <Polynomial> − <Term>
<Term>: = x ^ <exponent> | <coefficient> x ^ <index> | <coefficient> x | <constant>
<Index>: = [2-5]
<Coefficient>: = [1-9] [0-9] *
<Constant>: = [1-9] [0-9] *
If the exponent and coefficient are omitted, it is regarded as 1.
Constraints
The input satisfies the following conditions.
* 2 ≤ n ≤ 5
* For any set of i, j such that 1 ≤ i <j ≤ m, where m is the number of terms in the given expression,
The degree of the i-th term is guaranteed to be greater than the degree of the j-th term
* It is guaranteed that the nth degree polynomial of x given can be factored into the product form of the linear expression of x.
* Absolute values of coefficients and constants are 2 × 103 or less, respectively.
* The coefficient with the highest degree is 1, which is guaranteed to be omitted.
* The original constant term of each linear expression before expansion is guaranteed to be a non-zero integer
* It is guaranteed that the original constant terms of each linear expression before expansion are different.
Input
The input is given in the following format.
S
The string S representing the nth degree polynomial of x is given on one line.
Output
Factor S into the product of a linear expression of x, and output it in ascending order of the constant term.
Insert a line break at the end of the output.
Examples
Input
x^2+3x+2
Output
(x+1)(x+2)
Input
x^2-1
Output
(x-1)(x+1)
Input
x^5+15x^4+85x^3+225x^2+274x+120
Output
(x+1)(x+2)(x+3)(x+4)(x+5)
Input
x^3-81x^2-1882x-1800
Output
(x-100)(x+1)(x+18) | instruction | 0 | 80,776 | 5 | 161,552 |
"Correct Solution:
```
#E
import re
s = "+"+input()
c = 0
k = [0 for i in range(6)]
while c<len(s):
find_1 = re.search(r'^([\+\-])x\^([2-5])', s[c:])
if find_1:
k[int(find_1.group(2))] = int(str(find_1.group(1))+"1")
c += find_1.end()
continue
find_2 = re.search(r'^([\+\-][1-9][0-9]*)x\^([2-5])', s[c:])
if find_2:
k[int(find_2.group(2))] = int(str(find_2.group(1)))
c += find_2.end()
continue
find_3 = re.search(r'^([\+\-][1-9][0-9]*)x', s[c:])
if find_3:
k[1] = int(str(find_3.group(1)))
c += find_3.end()
continue
find_4 = re.search(r'^([\+\-])x', s[c:])
if find_4:
k[1] = int(str(find_4.group(1))+"1")
c += find_4.end()
continue
find_5 = re.search(r'^([\+\-][1-9][0-9]*)', s[c:])
if find_5:
k[0] = int(str(find_5.group(1)))
c += find_5.end()
continue
#print(k)
#
ans = ""
que = list(range(abs(k[0]), -abs(k[0])-1, -1))
#print(que)
while True:
buf = [0 for i in range(6)]
n = que[0]
p = 5
while(k[p] == 0): p-=1
# print(p)
if p==0: break
buf[p] = k[p]
while p>0:
p -= 1
buf[p] = k[p] + n * buf[p+1]
# print(k)
# print(buf)
if buf[0] == 0:
k = buf[1:]+[0]
if n < 0:
ans += "(x+"+str(-n)+")"
else:
ans += "(x"+str(-n)+")"
else:
que.pop(0)
print(ans)
``` | output | 1 | 80,776 | 5 | 161,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Mr. ukuku1333 is a little sloppy, so when I expanded the product of the linear expressions of x, I couldn't figure out the original linear expression.
Given the nth degree polynomial of x, factor it into the product of the original linear expressions of x.
The nth degree polynomial of x is given by the following BNF.
<Polynomial>: = <Term> | <Polynomial> & plus; <Polynomial> | <Polynomial> − <Term>
<Term>: = x ^ <exponent> | <coefficient> x ^ <index> | <coefficient> x | <constant>
<Index>: = [2-5]
<Coefficient>: = [1-9] [0-9] *
<Constant>: = [1-9] [0-9] *
If the exponent and coefficient are omitted, it is regarded as 1.
Constraints
The input satisfies the following conditions.
* 2 ≤ n ≤ 5
* For any set of i, j such that 1 ≤ i <j ≤ m, where m is the number of terms in the given expression,
The degree of the i-th term is guaranteed to be greater than the degree of the j-th term
* It is guaranteed that the nth degree polynomial of x given can be factored into the product form of the linear expression of x.
* Absolute values of coefficients and constants are 2 × 103 or less, respectively.
* The coefficient with the highest degree is 1, which is guaranteed to be omitted.
* The original constant term of each linear expression before expansion is guaranteed to be a non-zero integer
* It is guaranteed that the original constant terms of each linear expression before expansion are different.
Input
The input is given in the following format.
S
The string S representing the nth degree polynomial of x is given on one line.
Output
Factor S into the product of a linear expression of x, and output it in ascending order of the constant term.
Insert a line break at the end of the output.
Examples
Input
x^2+3x+2
Output
(x+1)(x+2)
Input
x^2-1
Output
(x-1)(x+1)
Input
x^5+15x^4+85x^3+225x^2+274x+120
Output
(x+1)(x+2)(x+3)(x+4)(x+5)
Input
x^3-81x^2-1882x-1800
Output
(x-100)(x+1)(x+18)
Submitted Solution:
```
import copy
s = input()
k = [0] * 6
t = []
while True:
i = 0
if "+" in s[1:]:
i = 1 + s[1:].index("+")
if "-" in s[1:]:
if i > 0:i = min(i, 1 + s[1:].index("-"))
else:i = 1 + s[1:].index("-")
if i == 0:
k[0] = int(s)
break
j = s.index("x")
if j + 3 == i:
if j == 0:k[int(s[j + 2])] = 1
else:k[int(s[j + 2])] = int(s[:j])
else:
if j == 0:k[1] = 1
elif j == 1:k[1] = -1
else:k[1] = int(s[:j])
s = s[i:]
while True:
if k[-1] == 0:k.pop()
else:break
while True:
i = 1
while i <= abs(k[0]):
if k[0] % i == 0:
j = k[0] // i
kn = copy.deepcopy(k)
for l in range(len(kn) - 1):
if kn[l] % j == 0:
kn[l + 1] -= kn[l] // j
kn[l] //= j
else:
break
else:
if kn[-1] == 0:
kn.pop()
t.append(j)
k = copy.deepcopy(kn)
break
j *= -1
kn = copy.deepcopy(k)
for l in range(len(kn) - 1):
if kn[l] % j == 0:
kn[l + 1] -= kn[l] // j
kn[l] //= j
else:
break
else:
if kn[-1] == 0:
kn.pop()
t.append(j)
k = copy.deepcopy(kn)
break
i += 1
if len(k) == 2:
print(k)
t.append(k[0])
break
t.sort()
s = ""
for i in t:s += "(x{})".format(i) if i < 0 else "(x+{})".format(i)
print(s)
``` | instruction | 0 | 80,777 | 5 | 161,554 |
No | output | 1 | 80,777 | 5 | 161,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Mr. ukuku1333 is a little sloppy, so when I expanded the product of the linear expressions of x, I couldn't figure out the original linear expression.
Given the nth degree polynomial of x, factor it into the product of the original linear expressions of x.
The nth degree polynomial of x is given by the following BNF.
<Polynomial>: = <Term> | <Polynomial> & plus; <Polynomial> | <Polynomial> − <Term>
<Term>: = x ^ <exponent> | <coefficient> x ^ <index> | <coefficient> x | <constant>
<Index>: = [2-5]
<Coefficient>: = [1-9] [0-9] *
<Constant>: = [1-9] [0-9] *
If the exponent and coefficient are omitted, it is regarded as 1.
Constraints
The input satisfies the following conditions.
* 2 ≤ n ≤ 5
* For any set of i, j such that 1 ≤ i <j ≤ m, where m is the number of terms in the given expression,
The degree of the i-th term is guaranteed to be greater than the degree of the j-th term
* It is guaranteed that the nth degree polynomial of x given can be factored into the product form of the linear expression of x.
* Absolute values of coefficients and constants are 2 × 103 or less, respectively.
* The coefficient with the highest degree is 1, which is guaranteed to be omitted.
* The original constant term of each linear expression before expansion is guaranteed to be a non-zero integer
* It is guaranteed that the original constant terms of each linear expression before expansion are different.
Input
The input is given in the following format.
S
The string S representing the nth degree polynomial of x is given on one line.
Output
Factor S into the product of a linear expression of x, and output it in ascending order of the constant term.
Insert a line break at the end of the output.
Examples
Input
x^2+3x+2
Output
(x+1)(x+2)
Input
x^2-1
Output
(x-1)(x+1)
Input
x^5+15x^4+85x^3+225x^2+274x+120
Output
(x+1)(x+2)(x+3)(x+4)(x+5)
Input
x^3-81x^2-1882x-1800
Output
(x-100)(x+1)(x+18)
Submitted Solution:
```
import copy
s = input()
k = [0] * 6
t = []
while True:
i = 0
if "+" in s[1:]:
i = 1 + s[1:].index("+")
if "-" in s[1:]:
if i > 0:i = min(i, 1 + s[1:].index("-"))
else:i = 1 + s[1:].index("-")
if i == 0:
k[0] = int(s)
break
j = s.index("x")
if j + 3 == i:
if j == 0:k[int(s[j + 2])] = 1
else:k[int(s[j + 2])] = int(s[:j])
else:
if j == 0:k[1] = 1
else:k[1] = int(s[:j])
s = s[i:]
while True:
if k[-1] == 0:k.pop()
else:break
while True:
i = 1
while i <= abs(k[0]):
if k[0] % i == 0:
j = k[0] // i
kn = copy.deepcopy(k)
for l in range(len(kn) - 1):
if kn[l] % j == 0:
kn[l + 1] -= kn[l] // j
kn[l] //= j
else:
break
else:
if kn[-1] == 0:
kn.pop()
t.append(j)
k = copy.deepcopy(kn)
break
j *= -1
kn = copy.deepcopy(k)
for l in range(len(kn) - 1):
if kn[l] % j == 0:
kn[l + 1] -= kn[l] // j
kn[l] //= j
else:
break
else:
if kn[-1] == 0:
kn.pop()
t.append(j)
k = copy.deepcopy(kn)
break
i += 1
if len(k) == 2:
t.append(k[0])
break
t.sort()
s = ""
for i in t:s += "(x{})".format(i) if i < 0 else "(x+{})".format(i)
print(s)
``` | instruction | 0 | 80,778 | 5 | 161,556 |
No | output | 1 | 80,778 | 5 | 161,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Mr. ukuku1333 is a little sloppy, so when I expanded the product of the linear expressions of x, I couldn't figure out the original linear expression.
Given the nth degree polynomial of x, factor it into the product of the original linear expressions of x.
The nth degree polynomial of x is given by the following BNF.
<Polynomial>: = <Term> | <Polynomial> & plus; <Polynomial> | <Polynomial> − <Term>
<Term>: = x ^ <exponent> | <coefficient> x ^ <index> | <coefficient> x | <constant>
<Index>: = [2-5]
<Coefficient>: = [1-9] [0-9] *
<Constant>: = [1-9] [0-9] *
If the exponent and coefficient are omitted, it is regarded as 1.
Constraints
The input satisfies the following conditions.
* 2 ≤ n ≤ 5
* For any set of i, j such that 1 ≤ i <j ≤ m, where m is the number of terms in the given expression,
The degree of the i-th term is guaranteed to be greater than the degree of the j-th term
* It is guaranteed that the nth degree polynomial of x given can be factored into the product form of the linear expression of x.
* Absolute values of coefficients and constants are 2 × 103 or less, respectively.
* The coefficient with the highest degree is 1, which is guaranteed to be omitted.
* The original constant term of each linear expression before expansion is guaranteed to be a non-zero integer
* It is guaranteed that the original constant terms of each linear expression before expansion are different.
Input
The input is given in the following format.
S
The string S representing the nth degree polynomial of x is given on one line.
Output
Factor S into the product of a linear expression of x, and output it in ascending order of the constant term.
Insert a line break at the end of the output.
Examples
Input
x^2+3x+2
Output
(x+1)(x+2)
Input
x^2-1
Output
(x-1)(x+1)
Input
x^5+15x^4+85x^3+225x^2+274x+120
Output
(x+1)(x+2)(x+3)(x+4)(x+5)
Input
x^3-81x^2-1882x-1800
Output
(x-100)(x+1)(x+18)
Submitted Solution:
```
import copy
s = input()
k = [0] * 6
t = []
while True:
i = 0
if "+" in s[1:]:
i = 1 + s[1:].index("+")
if "-" in s[1:]:
if i > 0:i = min(i, 1 + s[1:].index("-"))
else:i = 1 + s[1:].index("-")
if i == 0:
k[0] = int(s)
break
j = s.index("x")
if j + 3 == i:
if j == 0:k[int(s[j + 2])] = 1
else:k[int(s[j + 2])] = int(s[:j])
else:
if j == 0:k[1] = 1
else:k[1] = int(s[:j])
s = s[i:]
while True:
if k[-1] == 0:k.pop()
else:break
while True:
i = 1
while i <= abs(k[0]):
if k[0] % i == 0:
j = k[0] // i
kn = copy.deepcopy(k)
for l in range(len(kn) - 1):
if kn[l] % j == 0:
kn[l + 1] -= kn[l] // j
kn[l] //= j
else:
break
else:
if kn[-1] == 0:
kn.pop()
t.append(j)
k = copy.deepcopy(kn)
break
j *= -1
kn = copy.deepcopy(k)
for l in range(len(kn) - 1):
if kn[l] % j == 0:
kn[l + 1] -= kn[l] // j
kn[l] //= j
else:
break
else:
if kn[-1] == 0:
kn.pop()
t.append(j)
k = copy.deepcopy(kn)
break
i += 1
if len(k) == 2:
t.append(k[1])
break
t.sort()
s = ""
for i in t:s += "(x{})".format(i) if i < 0 else "(x+{})".format(i)
print(s)
``` | instruction | 0 | 80,779 | 5 | 161,558 |
No | output | 1 | 80,779 | 5 | 161,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Mr. ukuku1333 is a little sloppy, so when I expanded the product of the linear expressions of x, I couldn't figure out the original linear expression.
Given the nth degree polynomial of x, factor it into the product of the original linear expressions of x.
The nth degree polynomial of x is given by the following BNF.
<Polynomial>: = <Term> | <Polynomial> & plus; <Polynomial> | <Polynomial> − <Term>
<Term>: = x ^ <exponent> | <coefficient> x ^ <index> | <coefficient> x | <constant>
<Index>: = [2-5]
<Coefficient>: = [1-9] [0-9] *
<Constant>: = [1-9] [0-9] *
If the exponent and coefficient are omitted, it is regarded as 1.
Constraints
The input satisfies the following conditions.
* 2 ≤ n ≤ 5
* For any set of i, j such that 1 ≤ i <j ≤ m, where m is the number of terms in the given expression,
The degree of the i-th term is guaranteed to be greater than the degree of the j-th term
* It is guaranteed that the nth degree polynomial of x given can be factored into the product form of the linear expression of x.
* Absolute values of coefficients and constants are 2 × 103 or less, respectively.
* The coefficient with the highest degree is 1, which is guaranteed to be omitted.
* The original constant term of each linear expression before expansion is guaranteed to be a non-zero integer
* It is guaranteed that the original constant terms of each linear expression before expansion are different.
Input
The input is given in the following format.
S
The string S representing the nth degree polynomial of x is given on one line.
Output
Factor S into the product of a linear expression of x, and output it in ascending order of the constant term.
Insert a line break at the end of the output.
Examples
Input
x^2+3x+2
Output
(x+1)(x+2)
Input
x^2-1
Output
(x-1)(x+1)
Input
x^5+15x^4+85x^3+225x^2+274x+120
Output
(x+1)(x+2)(x+3)(x+4)(x+5)
Input
x^3-81x^2-1882x-1800
Output
(x-100)(x+1)(x+18)
Submitted Solution:
```
S = input().replace('^', '**').replace('x', '*1*(x)')[1:]
for i in range(2000, -2001, -1):
if i == 0: continue
T = S.replace('x', str(i))
#print(T, eval(T))
if eval(T) == 0:
print('(x{:+})'.format(-i), end='')
else:
print()
``` | instruction | 0 | 80,780 | 5 | 161,560 |
No | output | 1 | 80,780 | 5 | 161,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2.
Submitted Solution:
```
def solve(x):
answer = 1
while x > 0:
if x & 1 == 1:
answer *= 2
x >>= 1
return answer
if __name__ == '__main__':
test_count = int(input())
for x in range(test_count):
print(solve(int(input())))
# print(x, "-->", "{0:b}".format(x), "-->", solve(x))
``` | instruction | 0 | 80,839 | 5 | 161,678 |
Yes | output | 1 | 80,839 | 5 | 161,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2.
Submitted Solution:
```
import sys
def minp():
return sys.stdin.readline().strip()
t = int(minp())
for i in range(t):
a = int(minp())
c = 0
while a > 0:
if a & 1 != 0:
c += 1
a >>= 1
print(2**c)
``` | instruction | 0 | 80,840 | 5 | 161,680 |
Yes | output | 1 | 80,840 | 5 | 161,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2.
Submitted Solution:
```
def func(n):
if n==0:
return 1
if n==1:
return 2
if n%2==1 :
return 2*func(n//2)
if n%2==0 :
return func(n//2)
t = int(input())
while t:
# L,v,l,r=map(int,input().split())
a = int(input())
c = func(a)
print(c)
t-=1
``` | instruction | 0 | 80,841 | 5 | 161,682 |
Yes | output | 1 | 80,841 | 5 | 161,683 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2.
Submitted Solution:
```
t=int(input())
for i in range(0,t):
print(1 << str(bin(int(input()))).count('1'))
``` | instruction | 0 | 80,842 | 5 | 161,684 |
Yes | output | 1 | 80,842 | 5 | 161,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2.
Submitted Solution:
```
t = int(input())
for i in range(t):
a = int(input())
if a >= 3:
print(a + 1)
else:
print(max(a, 1))
``` | instruction | 0 | 80,843 | 5 | 161,686 |
No | output | 1 | 80,843 | 5 | 161,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2.
Submitted Solution:
```
import math
c = int(input())
for i in range(c):
a = int(input())
b = 0
if(a != 0):
b = int(math.log2(a))
else:
b = -1
if(2 ** b == a):
b -= 1
print(2 ** (b+1))
``` | instruction | 0 | 80,844 | 5 | 161,688 |
No | output | 1 | 80,844 | 5 | 161,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2.
Submitted Solution:
```
n=int(input())
arr=[]
for i in range(n):
a=int(input())
if a==0:
arr.append(1)
elif a==1:
arr.append(2)
else:
arr.append(a)
for i in arr:
print(i)
``` | instruction | 0 | 80,845 | 5 | 161,690 |
No | output | 1 | 80,845 | 5 | 161,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.
Arkadi and Boris Strugatsky. Monday starts on Saturday
Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: a - (a ⊕ x) - x = 0 for some given a, where ⊕ stands for a bitwise exclusive or (XOR) of two integers (this operation is denoted as ^ or xor in many modern programming languages). Oira-Oira quickly found some x, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.
Input
Each test contains several possible values of a and your task is to find the number of equation's solution for each of them. The first line contains an integer t (1 ≤ t ≤ 1000) — the number of these values.
The following t lines contain the values of parameter a, each value is an integer from 0 to 2^{30} - 1 inclusive.
Output
For each value of a print exactly one integer — the number of non-negative solutions of the equation for the given value of the parameter. Print answers in the same order as values of a appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3
0
2
1073741823
Output
1
2
1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers x and y, consider their binary representations (possibly with leading zeroes): x_k ... x_2 x_1 x_0 and y_k ... y_2 y_1 y_0. Here, x_i is the i-th bit of the number x and y_i is the i-th bit of the number y. Let r = x ⊕ y be the result of the XOR operation of x and y. Then r is defined as r_k ... r_2 r_1 r_0 where:
$$$ r_i = \left\{ \begin{aligned} 1, ~ if ~ x_i ≠ y_i \\\ 0, ~ if ~ x_i = y_i \end{aligned} \right. $$$
For the first value of the parameter, only x = 0 is a solution of the equation.
For the second value of the parameter, solutions are x = 0 and x = 2.
Submitted Solution:
```
t = int(input())
for i in range(t):
a = int(input())
c=1
if a%2==1:
c = a+1
else:
c = a//2+1
print(c)
``` | instruction | 0 | 80,846 | 5 | 161,692 |
No | output | 1 | 80,846 | 5 | 161,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At first, there was a legend related to the name of the problem, but now it's just a formal statement.
You are given n points a_1, a_2, ..., a_n on the OX axis. Now you are asked to find such an integer point x on OX axis that f_k(x) is minimal possible.
The function f_k(x) can be described in the following way:
* form a list of distances d_1, d_2, ..., d_n where d_i = |a_i - x| (distance between a_i and x);
* sort list d in non-descending order;
* take d_{k + 1} as a result.
If there are multiple optimal answers you can print any of them.
Input
The first line contains single integer T ( 1 ≤ T ≤ 2 ⋅ 10^5) — number of queries. Next 2 ⋅ T lines contain descriptions of queries. All queries are independent.
The first line of each query contains two integers n, k (1 ≤ n ≤ 2 ⋅ 10^5, 0 ≤ k < n) — the number of points and constant k.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9) — points in ascending order.
It's guaranteed that ∑{n} doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — corresponding points x which have minimal possible value of f_k(x). If there are multiple answers you can print any of them.
Example
Input
3
3 2
1 2 5
2 1
1 1000000000
1 0
4
Output
3
500000000
4
Submitted Solution:
```
import sys
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
t,=I()
for _ in range(t):
n,k = map(int,input().split())
arr = list(map(int,input().split()))
mx = int(10e9)
ans = arr[0]
for i in range(n-k):
le = arr[i]
r = arr[i+k]
x = (le + r) // 2
if mx > r - x:
mx = r - x
ans = x
print(ans)
``` | instruction | 0 | 80,897 | 5 | 161,794 |
Yes | output | 1 | 80,897 | 5 | 161,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At first, there was a legend related to the name of the problem, but now it's just a formal statement.
You are given n points a_1, a_2, ..., a_n on the OX axis. Now you are asked to find such an integer point x on OX axis that f_k(x) is minimal possible.
The function f_k(x) can be described in the following way:
* form a list of distances d_1, d_2, ..., d_n where d_i = |a_i - x| (distance between a_i and x);
* sort list d in non-descending order;
* take d_{k + 1} as a result.
If there are multiple optimal answers you can print any of them.
Input
The first line contains single integer T ( 1 ≤ T ≤ 2 ⋅ 10^5) — number of queries. Next 2 ⋅ T lines contain descriptions of queries. All queries are independent.
The first line of each query contains two integers n, k (1 ≤ n ≤ 2 ⋅ 10^5, 0 ≤ k < n) — the number of points and constant k.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_1 < a_2 < ... < a_n ≤ 10^9) — points in ascending order.
It's guaranteed that ∑{n} doesn't exceed 2 ⋅ 10^5.
Output
Print T integers — corresponding points x which have minimal possible value of f_k(x). If there are multiple answers you can print any of them.
Example
Input
3
3 2
1 2 5
2 1
1 1000000000
1 0
4
Output
3
500000000
4
Submitted Solution:
```
t= int(input())
for _ in range(t):
n,k = map(int,input().split())
a = list(map(int,input().split()))
mn = float('inf')
d= float('inf')
for i in range(n-k):
tmp = abs(a[i]-a[i+k])
if d>tmp:
d=tmp
mn = (a[i]+a[k])//2
print(mn)
``` | instruction | 0 | 80,899 | 5 | 161,798 |
No | output | 1 | 80,899 | 5 | 161,799 |
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