message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
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 grid n× n, where n is odd. Rows are enumerated from 1 to n from up to down, columns are enumerated from 1 to n from left to right. Cell, standing on the intersection of row x and column y, is denoted by (x, y).
Every cell contains 0 or 1. It is known that the top-left cell contains 1, and the bottom-right cell contains 0.
We want to know numbers in all cells of the grid. To do so we can ask the following questions:
"? x_1 y_1 x_2 y_2", where 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. In other words, we output two different cells (x_1, y_1), (x_2, y_2) of the grid such that we can get from the first to the second by moving only to the right and down, and they aren't adjacent.
As a response to such question you will be told if there exists a path between (x_1, y_1) and (x_2, y_2), going only to the right or down, numbers in cells of which form a palindrome.
For example, paths, shown in green, are palindromic, so answer for "? 1 1 2 3" and "? 1 2 3 3" would be that there exists such path. However, there is no palindromic path between (1, 1) and (3, 1).
<image>
Determine all cells of the grid by asking not more than n^2 questions. It can be shown that the answer always exists.
Input
The first line contains odd integer (3 ≤ n < 50) — the side of the grid.
Interaction
You begin the interaction by reading n.
To ask a question about cells (x_1, y_1), (x_2, y_2), in a separate line output "? x_1 y_1 x_2 y_2".
Numbers in the query have to satisfy 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. Don't forget to 'flush', to get the answer.
In response, you will receive 1, if there exists a path going from (x_1, y_1) to (x_2, y_2) only to the right or down, numbers in cells of which form a palindrome, and 0 otherwise.
In case your query is invalid or you asked more than n^2 queries, program will print -1 and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.
When you determine numbers in all cells, output "!".
Then output n lines, the i-th of which is a string of length n, corresponding to numbers in the i-th row of the grid.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, use the following format.
The first line should contain a single odd integer n (side of your grid).
The i-th of n following lines should contain a string of length n corresponding to the i-th row of the grid. Top left element of the grid has to be equal to 1, bottom right has to be equal to 0.
Example
Input
3
0
1
0
1
1
1
1
Output
? 1 1 1 3
? 1 1 2 3
? 2 1 2 3
? 3 1 3 3
? 2 2 3 3
? 1 2 3 2
? 1 2 3 3
!
100
001
000
Submitted Solution:
```
from sys import stdout
def full(k, get, mass):
global l
# print(f"k = {k}")
if k%2 == 1:
k = k
else:
k = k-1
if k-4 >= 0:
for i in range(l)[k-4::-2]:
answer = ask(get(i), get(i+2))
if answer:
mass[i] = mass[i+2]
else:
mass[i] = not mass[i+2]
# print((k+2))
# print(2*n-1)
if (k+2) < (2*n-1):
# print("pass here")
for i in range(l)[k+2::2]:
# print(f"l = {l}, i = {i}")
answer = ask(get(i-2), get(i))
if answer:
mass[i] = mass[i-2]
else:
mass[i] = not mass[i-2]
def ask(x, y):
x = [x[1], x[0]]
y = [y[1], y[0]]
q = f"? {x[0]+1} {x[1]+1} {y[0]+1} {y[1]+1}"
print(q)
stdout.flush()
answer = bool(int(input()))
return answer
def get_top(idx):
if idx>=n:
i = n-1
j = idx%n+1
else:
i = idx%n
j = 0
return i,j
def get_bot(idx):
if idx>=n:
j = n-1
i = idx%n+1
else:
j = idx%n
i = 0
return i,j
def fun( mass, f):
l = len(mass)
get = f
for i in range(l)[2:l-1:2]:
answer = ask(get(i-2), get(i))
# print(f"answer = {answer}")
if answer:
mass[i] = mass[i-2]
else:
mass[i] = not mass[i-2]
# print(mass)
for i in range(l)[4::2]:
if (mass[i-4] == mass[i-2]) and (mass[i-2] != mass[i]):
answer = ask(get(i-3), get(i-1))
if answer:
answer = ask(get(i-4), get(i-1))
if answer:
mass[i-3] = mass[i-4]
mass[i-1] = mass[i-4]
else:
mass[i-3] = not mass[i-4]
mass[i-1] = not mass[i-4]
full(i-1, get, mass)
else:
answer = ask(get(i-3), get(i))
if answer:
mass[i-3] = mass[i]
mass[i-1] = mass[i-2]
else:
mass[i-3] = not mass[i]
mass[i-1] = not mass[i-2]
full(i, get, mass)
break
elif (mass[i-4] != mass[i-2]) and (mass[i-2] == mass[i]):
answer = ask(get(i-3), get(i-1))
if answer:
answer = ask(get(i-3), get(i))
if answer:
mass[i-3] = mass[i]
mass[i-1] = mass[i]
else:
mass[i-3] = not mass[i]
mass[i-1] = not mass[i]
full(i, get, mass)
else:
answer = ask(get(i-4), get(i-1))
if answer:
mass[i-3] = mass[i-2]
mass[i-1] = mass[i-4]
else:
mass[i-3] = not mass[i-2]
mass[i-1] = not mass[i-4]
full(i-1, get, mass)
break
if __name__ == "__main__":
n = int(input())
a_top = [None]*(2*n-1)
a_top[0] = True
a_top[len(a_top)-1] = False
a_bot = a_top.copy()
l = len(a_bot)
fun(a_top, get_top)
# print(a_top)
fun(a_bot, get_bot)
# print(a_bot)
matrix = []
for i in range(n):
matrix.append([None]*n)
for i in range(len(a_top)):
idx = get_top(i)[::-1]
# print("idx_top", idx)
matrix[idx[0]][idx[1]] = a_top[i]
idx = get_bot(i)[::-1]
# print("idx_bot", idx)
matrix[idx[0]][idx[1]] = a_bot[i]
for i in range(n)[1:n-1:1]:
for j in range(n)[1:n-1:1]:
answer = ask([j-1,i-1], [j,i])
if answer:
matrix[i][j] = matrix[i-1][j-1]
else:
matrix[i][j] = not matrix[i-1][j-1]
print("!")
for i in matrix:
for j in i:
print(int(j), end = "")
print()
stdout.flush()
``` | instruction | 0 | 75,963 | 23 | 151,926 |
Yes | output | 1 | 75,963 | 23 | 151,927 |
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 grid n× n, where n is odd. Rows are enumerated from 1 to n from up to down, columns are enumerated from 1 to n from left to right. Cell, standing on the intersection of row x and column y, is denoted by (x, y).
Every cell contains 0 or 1. It is known that the top-left cell contains 1, and the bottom-right cell contains 0.
We want to know numbers in all cells of the grid. To do so we can ask the following questions:
"? x_1 y_1 x_2 y_2", where 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. In other words, we output two different cells (x_1, y_1), (x_2, y_2) of the grid such that we can get from the first to the second by moving only to the right and down, and they aren't adjacent.
As a response to such question you will be told if there exists a path between (x_1, y_1) and (x_2, y_2), going only to the right or down, numbers in cells of which form a palindrome.
For example, paths, shown in green, are palindromic, so answer for "? 1 1 2 3" and "? 1 2 3 3" would be that there exists such path. However, there is no palindromic path between (1, 1) and (3, 1).
<image>
Determine all cells of the grid by asking not more than n^2 questions. It can be shown that the answer always exists.
Input
The first line contains odd integer (3 ≤ n < 50) — the side of the grid.
Interaction
You begin the interaction by reading n.
To ask a question about cells (x_1, y_1), (x_2, y_2), in a separate line output "? x_1 y_1 x_2 y_2".
Numbers in the query have to satisfy 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. Don't forget to 'flush', to get the answer.
In response, you will receive 1, if there exists a path going from (x_1, y_1) to (x_2, y_2) only to the right or down, numbers in cells of which form a palindrome, and 0 otherwise.
In case your query is invalid or you asked more than n^2 queries, program will print -1 and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.
When you determine numbers in all cells, output "!".
Then output n lines, the i-th of which is a string of length n, corresponding to numbers in the i-th row of the grid.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, use the following format.
The first line should contain a single odd integer n (side of your grid).
The i-th of n following lines should contain a string of length n corresponding to the i-th row of the grid. Top left element of the grid has to be equal to 1, bottom right has to be equal to 0.
Example
Input
3
0
1
0
1
1
1
1
Output
? 1 1 1 3
? 1 1 2 3
? 2 1 2 3
? 3 1 3 3
? 2 2 3 3
? 1 2 3 2
? 1 2 3 3
!
100
001
000
Submitted Solution:
```
from sys import stdout
# matrix="""
# 101110111011101110111
# 011101110111011101110
# 111011101110111011101
# 110111011101110111011
# 101110111011101110111
# 011101110111011101110
# 111011101110111011101
# 110111011101110111011
# 101110111011101110111
# 011101110111011101110
# 111011101110111011100
# 110111011101110111001
# 101110111011101110010
# 011101110111011100100
# 111011101110111001000
# 110111011101110010001
# 101110111011100100010
# 011101110111001000100
# 111011101110010001000
# 110111011100100010001
# 101110111001000100010"""
#m=matrix.split()
n=int(input())#len(m)#
def query(i,j,k,l):
print("? {} {} {} {}".format(i,j,k,l))
if (k+l-i-j)==2:
res=1 if m[i-1][j-1]==m[k-1][l-1] else 0
if (k+l-i-j)==3:#la forme doit etre rect
assert k>i and l>j
isok=(a[i-1][j-1]==a[k-1][l-1])
if isok:
lautre= (m[k-2][j-1]==m[k-1][j-1]) if (k-i==2) else (m[i-1][l-1]==m[i-1][l-2])
matchpal= (m[i][j-1]==m[i][j]) or (m[i-1][j]==m[i][j]) or lautre
isok&=matchpal
res=1 if isok else 0
print(res)
return res
def req(i,j,k,l):
print("? {} {} {} {}".format(i,j,k,l))
stdout.flush()
rep=int(input())
return rep
a=[[0 for i in range(n)] for j in range(n)]
a[0][0]=1
a[n-1][n-1]=-1
a[0][1]=2
#a[2][1]
prev=0,1
i=2
j=1
rep=req(prev[0]+1,prev[1]+1,i+1,j+1)
if rep:
a[i][j]=a[prev[0]][prev[1]]
else:
a[i][j]=-a[prev[0]][prev[1]]
#a[1][0]
prev=2,1
i=1
j=0
rep=req(i+1,j+1,prev[0]+1,prev[1]+1)
if rep:
a[i][j]=a[prev[0]][prev[1]]
else:
a[i][j]=-a[prev[0]][prev[1]]
#on fait le reste
for i in range(n):
for j in range(n):
if not a[i][j]:
prev=(i-1,j-1) if i and j else ((i-2,j) if i else (i,j-2))
rep=req(prev[0]+1,prev[1]+1,i+1,j+1)
if rep:
a[i][j]=a[prev[0]][prev[1]]
else:
a[i][j]=-a[prev[0]][prev[1]]
assert(a[i][j])
#on cherche une case paire tq case +2 +2 est différente
baserec=-1,-1
for i in range(n-2):
for j in range(n-2):
if (i+j+1)%2 and a[i][j]!=a[i+2][j+2]:
baserec=i,j
break
if baserec[0]!=-1:
break
assert baserec[0]!=-1
i,j=baserec
#print("baserec",i,j)
rv2=None
#print(a[i][j],a[i][j+1],a[i][j+2])
#print(a[i+1][j],a[i+1][j+1],a[i+1][j+2])
#print(a[i+2][j],a[i+2][j+1],a[i+2][j+2])
if a[i][j+1]==a[i+1][j+2]:
if a[i+1][j+1]==a[i][j]:
prev=i,j
curr=i+1,j+2
rep=req(prev[0]+1,prev[1]+1,curr[0]+1,curr[1]+1)
if rep:
rv2=a[i][j] if a[i][j+1]==2 else -a[i][j]
else:
rv2=-a[i][j] if a[i][j+1]==2 else a[i][j]
else:
prev=i,j+1
curr=i+2,j+2
rep=req(prev[0]+1,prev[1]+1,curr[0]+1,curr[1]+1)
if rep:
rv2=a[i+1][j+1] if a[i][j+1]==2 else -a[i+1][j+1]
else:
rv2=-a[i+1][j+1] if a[i][j+1]==2 else a[i+1][j+1]
else:
if a[i+1][j+1]!=a[i][j]:
prev=i,j
curr=i+1,j+2
rep=req(prev[0]+1,prev[1]+1,curr[0]+1,curr[1]+1)
if rep:
rv2=a[i+1][j+1] if a[i][j+1]==2 else -a[i+1][j+1]
else:
rv2=-a[i+1][j+1] if a[i][j+1]==2 else a[i+1][j+1]
else:
prev=i,j+1
curr=i+2,j+2
rep=req(prev[0]+1,prev[1]+1,curr[0]+1,curr[1]+1)
if rep:
rv2=a[i+1][j+1] if a[i+1][j+2]==2 else -a[i+1][j+1]
else:
rv2=-a[i+1][j+1] if a[i+1][j+2]==2 else a[i+1][j+1]
for i in range(n):
for j in range(n):
if a[i][j] in (-2,2):
a[i][j]= rv2 if a[i][j]==2 else -rv2
print("!")
for i in range(n):
r=""
for j in range(n):
r+="1" if a[i][j]>0 else "0"
print(r)
stdout.flush()
``` | instruction | 0 | 75,964 | 23 | 151,928 |
Yes | output | 1 | 75,964 | 23 | 151,929 |
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 grid n× n, where n is odd. Rows are enumerated from 1 to n from up to down, columns are enumerated from 1 to n from left to right. Cell, standing on the intersection of row x and column y, is denoted by (x, y).
Every cell contains 0 or 1. It is known that the top-left cell contains 1, and the bottom-right cell contains 0.
We want to know numbers in all cells of the grid. To do so we can ask the following questions:
"? x_1 y_1 x_2 y_2", where 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. In other words, we output two different cells (x_1, y_1), (x_2, y_2) of the grid such that we can get from the first to the second by moving only to the right and down, and they aren't adjacent.
As a response to such question you will be told if there exists a path between (x_1, y_1) and (x_2, y_2), going only to the right or down, numbers in cells of which form a palindrome.
For example, paths, shown in green, are palindromic, so answer for "? 1 1 2 3" and "? 1 2 3 3" would be that there exists such path. However, there is no palindromic path between (1, 1) and (3, 1).
<image>
Determine all cells of the grid by asking not more than n^2 questions. It can be shown that the answer always exists.
Input
The first line contains odd integer (3 ≤ n < 50) — the side of the grid.
Interaction
You begin the interaction by reading n.
To ask a question about cells (x_1, y_1), (x_2, y_2), in a separate line output "? x_1 y_1 x_2 y_2".
Numbers in the query have to satisfy 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. Don't forget to 'flush', to get the answer.
In response, you will receive 1, if there exists a path going from (x_1, y_1) to (x_2, y_2) only to the right or down, numbers in cells of which form a palindrome, and 0 otherwise.
In case your query is invalid or you asked more than n^2 queries, program will print -1 and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.
When you determine numbers in all cells, output "!".
Then output n lines, the i-th of which is a string of length n, corresponding to numbers in the i-th row of the grid.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, use the following format.
The first line should contain a single odd integer n (side of your grid).
The i-th of n following lines should contain a string of length n corresponding to the i-th row of the grid. Top left element of the grid has to be equal to 1, bottom right has to be equal to 0.
Example
Input
3
0
1
0
1
1
1
1
Output
? 1 1 1 3
? 1 1 2 3
? 2 1 2 3
? 3 1 3 3
? 2 2 3 3
? 1 2 3 2
? 1 2 3 3
!
100
001
000
Submitted Solution:
```
n=int(input())
mat=[[-1 for j in range(n)] for i in range(n)]
def call(x1,y1,x2,y2):
if (x2==n-1 and y2==n-1):
if (mat[x1][y1]==1):
return 0
else:
return 1
print ("?",x1+1,y1+1,x2+1,y2+1)
t=int(input())
return t
mat[0][0]=1
if (call(0,0,1,1)):
mat[1][1]=1
else:
mat[1][1]=1-mat[0][0]
for i in range(2,n,2):
if (call(0,i-2,0,i)):
mat[0][i]=mat[0][i-2]
else:
mat[0][i]=1-mat[0][i-2]
for i in range(2,n,2):
for j in range(0,n,2):
if (call(i-2,j,i,j)):
mat[i][j]=mat[i-2][j]
else:
mat[i][j]=1-mat[i-2][j]
for i in range(3,n,2):
if (call(1,i-2,1,i)):
mat[1][i]=mat[1][i-2]
else:
mat[1][i]=1-mat[1][i-2]
for i in range(3,n,2):
for j in range(1,n,2):
if (call(i-2,j,i,j)):
mat[i][j]=mat[i-2][j]
else:
mat[i][j]=1-mat[i-2][j]
mat[0][1]=1
if (call(0,1,1,0)):
mat[1][0]=mat[0][1]
else:
mat[1][0]=1-mat[0][1]
for i in range(3,n,2):
if (call(0,i-2,0,i)):
mat[0][i]=mat[0][i-2]
else:
mat[0][i]=1-mat[0][i-2]
for i in range(2,n,2):
for j in range(1,n,2):
if (call(i-2,j,i,j)):
mat[i][j]=mat[i-2][j]
else:
mat[i][j]=1-mat[i-2][j]
for i in range(2,n,2):
if (call(1,i-2,1,i)):
mat[1][i]=mat[1][i-2]
else:
mat[1][i]=1-mat[1][i-2]
for i in range(3,n,2):
for j in range(0,n,2):
if (call(i-2,j,i,j)):
mat[i][j]=mat[i-2][j]
else:
mat[i][j]=1-mat[i-2][j]
change=1
print ("!")
for i in range(n):
for j in range(n):
if ((i+j)%2==1 and change==1):
print (1-mat[i][j],end='')
else:
print (mat[i][j],end='')
print()
``` | instruction | 0 | 75,965 | 23 | 151,930 |
No | output | 1 | 75,965 | 23 | 151,931 |
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 grid n× n, where n is odd. Rows are enumerated from 1 to n from up to down, columns are enumerated from 1 to n from left to right. Cell, standing on the intersection of row x and column y, is denoted by (x, y).
Every cell contains 0 or 1. It is known that the top-left cell contains 1, and the bottom-right cell contains 0.
We want to know numbers in all cells of the grid. To do so we can ask the following questions:
"? x_1 y_1 x_2 y_2", where 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. In other words, we output two different cells (x_1, y_1), (x_2, y_2) of the grid such that we can get from the first to the second by moving only to the right and down, and they aren't adjacent.
As a response to such question you will be told if there exists a path between (x_1, y_1) and (x_2, y_2), going only to the right or down, numbers in cells of which form a palindrome.
For example, paths, shown in green, are palindromic, so answer for "? 1 1 2 3" and "? 1 2 3 3" would be that there exists such path. However, there is no palindromic path between (1, 1) and (3, 1).
<image>
Determine all cells of the grid by asking not more than n^2 questions. It can be shown that the answer always exists.
Input
The first line contains odd integer (3 ≤ n < 50) — the side of the grid.
Interaction
You begin the interaction by reading n.
To ask a question about cells (x_1, y_1), (x_2, y_2), in a separate line output "? x_1 y_1 x_2 y_2".
Numbers in the query have to satisfy 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. Don't forget to 'flush', to get the answer.
In response, you will receive 1, if there exists a path going from (x_1, y_1) to (x_2, y_2) only to the right or down, numbers in cells of which form a palindrome, and 0 otherwise.
In case your query is invalid or you asked more than n^2 queries, program will print -1 and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.
When you determine numbers in all cells, output "!".
Then output n lines, the i-th of which is a string of length n, corresponding to numbers in the i-th row of the grid.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, use the following format.
The first line should contain a single odd integer n (side of your grid).
The i-th of n following lines should contain a string of length n corresponding to the i-th row of the grid. Top left element of the grid has to be equal to 1, bottom right has to be equal to 0.
Example
Input
3
0
1
0
1
1
1
1
Output
? 1 1 1 3
? 1 1 2 3
? 2 1 2 3
? 3 1 3 3
? 2 2 3 3
? 1 2 3 2
? 1 2 3 3
!
100
001
000
Submitted Solution:
```
n=int(input())
board1=[[-1]*n for _ in range(n)]
board1[0][0]=1
board1[n-1][n-1]=0
board2=[[-1]*n for _ in range(n)]
board2[0][0]=1
board2[n-1][n-1]=0
print('? 1 1 2 3',flush=True)
flag=int(input())
if flag==1:
board1[2][1]=1
print('? 2 1 2 3',flush=True)
flag=int(input())
if flag==0:
board1[0][1]=0
else:
board1[0][1]=1
print('? 1 2 2 3',flush=True)
flag=int(input())
if flag==0:
board1[1][0]=0
else:
board1[1][0]=1
for x in range(1,n+1):
for y in range(1,n+1):
for dx in range(0,2+1):
if x+dx>n:
continue
for dy in range(0,2+1):
if dx+dy!=2:
continue
if y+dy>n:
continue
if board1[y+dy-1][x+dx-1]!=-1:
continue
arr=['?',x,y,x+dx,y+dy]
print(*arr,flush=True)
flag=int(input())
if flag==0:
if board1[y-1][x-1]==1:
board1[y+dy-1][x+dx-1]=0
else:
board1[y+dy-1][x+dx-1]=1
else:
if board1[y-1][x-1]==1:
board1[y+dy-1][x+dx-1]=1
else:
board1[y+dy-1][x+dx-1]=0
print('!')
for column in board1:
print(''.join(map(str,column)))
else:
board1[2][1]=1
board2[2][1]=0
print('? 2 1 2 3',flush=True)
flag=int(input())
if flag==0:
board1[0][1]=0
board2[0][1]=1
else:
board1[0][1]=1
board2[0][1]=0
print('? 1 2 2 3',flush=True)
flag=int(input())
if flag==0:
board1[1][0]=0
board2[1][0]=1
else:
board1[1][0]=1
board2[1][0]=0
for x in range(1,n+1):
for y in range(1,n+1):
for dx in range(0,2+1):
if x+dx>n:
continue
for dy in range(0,2+1):
if dx+dy!=2:
continue
if y+dy>n:
continue
if board1[y+dy-1][x+dx-1]!=-1:
continue
arr=['?',x,y,x+dx,y+dy]
print(*arr,flush=True)
flag=int(input())
if flag==0:
if board1[y-1][x-1]==1:
board1[y+dy-1][x+dx-1]=0
else:
board1[y+dy-1][x+dx-1]=1
if board2[y-1][x-1]==1:
board2[y+dy-1][x+dx-1]=0
else:
board2[y+dy-1][x+dx-1]=1
else:
if board1[y-1][x-1]==1:
board1[y+dy-1][x+dx-1]=1
else:
board1[y+dy-1][x+dx-1]=0
if board2[y-1][x-1]==1:
board2[y+dy-1][x+dx-1]=1
else:
board2[y+dy-1][x+dx-1]=0
print('? 1 1 3 2',flush=True)
flag=int(input())
b1flag=-1
b2flag=-1
if board1[1][2]!=1 or (board1[0][1]!=board1[0][2]) or (board1[0][1]!=board1[1][1]) or (board1[1][0]!=board1[1][1]):
b1flag=0
else:
b1flag=1
if board2[1][2]!=1 or (board2[0][1]!=board2[0][2]) or (board2[0][1]!=board2[1][1]) or (board2[1][0]!=board2[1][1]):
b2flag=0
else:
b2flag=1
if b1flag==flag:
print('b1')
print('!')
for column in board1:
print(''.join(map(str,column)))
else:
print('b2')
print('!')
for column in board2:
print(''.join(map(str,column)))
``` | instruction | 0 | 75,966 | 23 | 151,932 |
No | output | 1 | 75,966 | 23 | 151,933 |
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 grid n× n, where n is odd. Rows are enumerated from 1 to n from up to down, columns are enumerated from 1 to n from left to right. Cell, standing on the intersection of row x and column y, is denoted by (x, y).
Every cell contains 0 or 1. It is known that the top-left cell contains 1, and the bottom-right cell contains 0.
We want to know numbers in all cells of the grid. To do so we can ask the following questions:
"? x_1 y_1 x_2 y_2", where 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. In other words, we output two different cells (x_1, y_1), (x_2, y_2) of the grid such that we can get from the first to the second by moving only to the right and down, and they aren't adjacent.
As a response to such question you will be told if there exists a path between (x_1, y_1) and (x_2, y_2), going only to the right or down, numbers in cells of which form a palindrome.
For example, paths, shown in green, are palindromic, so answer for "? 1 1 2 3" and "? 1 2 3 3" would be that there exists such path. However, there is no palindromic path between (1, 1) and (3, 1).
<image>
Determine all cells of the grid by asking not more than n^2 questions. It can be shown that the answer always exists.
Input
The first line contains odd integer (3 ≤ n < 50) — the side of the grid.
Interaction
You begin the interaction by reading n.
To ask a question about cells (x_1, y_1), (x_2, y_2), in a separate line output "? x_1 y_1 x_2 y_2".
Numbers in the query have to satisfy 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. Don't forget to 'flush', to get the answer.
In response, you will receive 1, if there exists a path going from (x_1, y_1) to (x_2, y_2) only to the right or down, numbers in cells of which form a palindrome, and 0 otherwise.
In case your query is invalid or you asked more than n^2 queries, program will print -1 and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.
When you determine numbers in all cells, output "!".
Then output n lines, the i-th of which is a string of length n, corresponding to numbers in the i-th row of the grid.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, use the following format.
The first line should contain a single odd integer n (side of your grid).
The i-th of n following lines should contain a string of length n corresponding to the i-th row of the grid. Top left element of the grid has to be equal to 1, bottom right has to be equal to 0.
Example
Input
3
0
1
0
1
1
1
1
Output
? 1 1 1 3
? 1 1 2 3
? 2 1 2 3
? 3 1 3 3
? 2 2 3 3
? 1 2 3 2
? 1 2 3 3
!
100
001
000
Submitted Solution:
```
def gns():
return list(map(int,input().split()))
n=int(input())
dp=[[None]*n for i in range(n)]
ans=set()
to={}
to[0]=(0,1)
to[1]=(1,0)
to[2]=(1,1)
to[3]=(2,0)
to[4]=(2,1)
for i in range(32):
mp = [[0] * 2 for i in range(3)]
mp[0][0]=1
for j in range(5):
if (i>>j)&1>0:
x,y=to[j]
mp[x][y]=1
ans.add(tuple([tuple(x)for x in mp]))
def check(sx,sy,tx,ty):
print('? {} {} {} {}'.format(sx+1,sy+1,tx+1,ty+1))
a=int(input())
if a==1:
for t in set(ans):
if t[sx][sy]!=t[tx][ty]:
ans.remove(t)
else:
for t in set(ans):
if t[sx][sy]==t[tx][ty]:
ans.remove(t)
def con(ns,mp):
ans=''
for x,y in ns:
ans+=str(mp[x][y])
return ans
def p(ns,mp):
s=con(ns,mp)
for i in range(len(s)):
j=len(s)-i-1
if s[i]==s[j]:
continue
return False
return True
def check2():
print('? {} {} {} {}'.format(1, 1, 3, 2))
a=int(input())
nss=[]
nss.append([(0,0),(0,1),(1,1),(2,1)])
nss.append([(0,0),(1,0),(1,1),(2,1)])
nss.append([(0,0),(1,0),(2,0),(2,1)])
for t in set(ans):
f=p(nss[0],t)or p(nss[1],t) or p(nss[2],t)
if (f==True) != (a==1):
ans.remove(t)
for t in ans:
for i in range(3):
for j in range(2):
dp[i][j]=t[i][j]
return
def get(i,j):
if (i<3 and j<2) or dp[i][j]!=None:
return
if j<2:
print('? {} {} {} {}'.format(i-1, j+1, i+1, j+1))
a=int(input())
if a==1:
dp[i][j]=dp[i-2][j]
else:
dp[i][j] =1- dp[i - 2][j]
return
print('? {} {} {} {}'.format(i+1, j-1, i+1, j+1))
a = int(input())
if a == 1:
dp[i][j] = dp[i][j-2]
else:
dp[i][j] = 1 - dp[i][j-2]
for sx,sy,tx,ty in [(0,0,1,1),(1,0,2,1),(0,0,2,0),(0,1,2,1)]:
check(sx,sy,tx,ty)
check2()
for i in range(n):
get(i,0)
get(i,1)
for i in range(n):
for j in range(2,n):
get(i,j)
print('!')
for x in dp:
print(''.join(map(str,x)))
``` | instruction | 0 | 75,967 | 23 | 151,934 |
No | output | 1 | 75,967 | 23 | 151,935 |
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 grid n× n, where n is odd. Rows are enumerated from 1 to n from up to down, columns are enumerated from 1 to n from left to right. Cell, standing on the intersection of row x and column y, is denoted by (x, y).
Every cell contains 0 or 1. It is known that the top-left cell contains 1, and the bottom-right cell contains 0.
We want to know numbers in all cells of the grid. To do so we can ask the following questions:
"? x_1 y_1 x_2 y_2", where 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. In other words, we output two different cells (x_1, y_1), (x_2, y_2) of the grid such that we can get from the first to the second by moving only to the right and down, and they aren't adjacent.
As a response to such question you will be told if there exists a path between (x_1, y_1) and (x_2, y_2), going only to the right or down, numbers in cells of which form a palindrome.
For example, paths, shown in green, are palindromic, so answer for "? 1 1 2 3" and "? 1 2 3 3" would be that there exists such path. However, there is no palindromic path between (1, 1) and (3, 1).
<image>
Determine all cells of the grid by asking not more than n^2 questions. It can be shown that the answer always exists.
Input
The first line contains odd integer (3 ≤ n < 50) — the side of the grid.
Interaction
You begin the interaction by reading n.
To ask a question about cells (x_1, y_1), (x_2, y_2), in a separate line output "? x_1 y_1 x_2 y_2".
Numbers in the query have to satisfy 1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ n, and x_1 + y_1 + 2 ≤ x_2 + y_2. Don't forget to 'flush', to get the answer.
In response, you will receive 1, if there exists a path going from (x_1, y_1) to (x_2, y_2) only to the right or down, numbers in cells of which form a palindrome, and 0 otherwise.
In case your query is invalid or you asked more than n^2 queries, program will print -1 and will finish interaction. You will receive Wrong answer verdict. Make sure to exit immediately to avoid getting other verdicts.
When you determine numbers in all cells, output "!".
Then output n lines, the i-th of which is a string of length n, corresponding to numbers in the i-th row of the grid.
After printing a query do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hack Format
To hack, use the following format.
The first line should contain a single odd integer n (side of your grid).
The i-th of n following lines should contain a string of length n corresponding to the i-th row of the grid. Top left element of the grid has to be equal to 1, bottom right has to be equal to 0.
Example
Input
3
0
1
0
1
1
1
1
Output
? 1 1 1 3
? 1 1 2 3
? 2 1 2 3
? 3 1 3 3
? 2 2 3 3
? 1 2 3 2
? 1 2 3 3
!
100
001
000
Submitted Solution:
```
from sys import stdin, stdout
def ask(x1,y1,x2,y2):
stdout.write('? {} {} {} {}\n'.format(x1+1,y1+1,x2+1,y2+1))
stdout.flush()
return int(stdin.readline())
def hack(D, x1, y1, x,y):
assert x-x1 == 1 or y-y1 == 1
assert x-x1 == 0 or y-y1 == 0
v = ask(x1,y1,x,y)
if v == 1:
D[x][y] = D[x1][y1]
else:
D[x][y] = 1-D[x1][y1]
def main():
N = int(stdin.readline())
D = [N*[0] for x in range(N)]
D[0][0] = 1
# first row
for y in range(1, N):
hack(D, 0, y-1, 0, y)
# and next rows
for x in range(1, N):
for y in range(N):
hack(D, x-1, y, x, y)
stdout.write('!\n')
for a in D:
for x in a:
stdout.write('{}'.format(x))
stdout.write('\n')
main()
``` | instruction | 0 | 75,968 | 23 | 151,936 |
No | output | 1 | 75,968 | 23 | 151,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of n segments on a line [L_i; R_i]. All 2n segment endpoints are pairwise distinct integers.
The set is laminar — any two segments are either disjoint or one of them contains the other.
Choose a non-empty subsegment [l_i, r_i] with integer endpoints in each segment (L_i ≤ l_i < r_i ≤ R_i) in such a way that no two subsegments intersect (they are allowed to have common endpoints though) and the sum of their lengths (∑_{i=1}^n r_i - l_i) is maximized.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^3) — the number of segments.
The i-th of the next n lines contains two integers L_i and R_i (0 ≤ L_i < R_i ≤ 10^9) — the endpoints of the i-th segment.
All the given 2n segment endpoints are distinct. The set of segments is laminar.
Output
On the first line, output the maximum possible sum of subsegment lengths.
On the i-th of the next n lines, output two integers l_i and r_i (L_i ≤ l_i < r_i ≤ R_i), denoting the chosen subsegment of the i-th segment.
Example
Input
4
1 10
2 3
5 9
6 7
Output
7
3 6
2 3
7 9
6 7
Note
The example input and the example output are illustrated below. <image> <image>
Submitted Solution:
```
def is_contained (Ii,Ij):
li,ri = Ii
lj,rj = Ij
return lj <= li and ri <= rj
def setminus (Ii, list_Ij):
li,ri = Ii
if len(list_Ij)==0:
return Ii
else:
list_Ij.sort (key=lambda item: item[0])
min_lj = list_Ij[0][0]
max_rj = list_Ij[-1][1]
delta1 = min_lj - li
delta2 = ri - max_rj
ans = [li, min_lj] if delta1>= delta2 else [max_rj,ri]
delta = max (delta1,delta2)
m = len(list_Ij)
for k in range (m-1):
l_next_k = list_Ij[k+1][0]
r_k = list_Ij[k][1]
delta_k = l_next_k - r_k
if delta_k>delta:
delta = delta_k
ans = [ r_k , l_next_k ]
return ans
def solve (I,Tree):
indices = set(i for (i,j) in Tree)
if len (Tree)==0:
return []
elif len (Tree)==1: ## indices = [i] only one index
i = list(indices)[0]
return [I[i]]
else:
sub_trees = {} ## also connected components of the main graph
for i in indices:
if not any([Tree[(i,j)] for j in indices]):
sub_trees [i] = [j for j in indices if Tree[(j,i)] ]
## The subtree is rooted at Ii
#print (sub_trees)
ans = []
for i in sub_trees:
cc_i = sub_trees [i]
Tree_i = {(k,j):Tree[(k,j)] for k in cc_i for j in cc_i}
sub_intervals_i = solve (I,Tree_i)
interval_i = setminus (I[i], sub_intervals_i)
ans.append(interval_i)
ans += sub_intervals_i
return ans
def main():
n = int (input())
I = {i:[] for i in range(n)}
for i in range (n):
I[i] = list (map(int, input().split(' ')))
Tree = { (i,j):False for i in range(n) for j in range(n)}
for (i,j) in Tree:
Tree[(i,j)] = (i!=j and is_contained (I[i],I[j]))
'''
print (I)
indices = set(i for (i,j) in Tree)
print ('##########################################')
for i in indices:
print (i,'--',end =" ")
for j in indices:
print (Tree[(i,j)],end =" ")
print ('')
print ('##########################################')
'''
intervals = solve (I,Tree)
#print (intervals)
print (sum ([ri-li for [li,ri] in intervals]))
for interval in intervals:
print (*interval)
main ()
``` | instruction | 0 | 76,108 | 23 | 152,216 |
No | output | 1 | 76,108 | 23 | 152,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of n segments on a line [L_i; R_i]. All 2n segment endpoints are pairwise distinct integers.
The set is laminar — any two segments are either disjoint or one of them contains the other.
Choose a non-empty subsegment [l_i, r_i] with integer endpoints in each segment (L_i ≤ l_i < r_i ≤ R_i) in such a way that no two subsegments intersect (they are allowed to have common endpoints though) and the sum of their lengths (∑_{i=1}^n r_i - l_i) is maximized.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^3) — the number of segments.
The i-th of the next n lines contains two integers L_i and R_i (0 ≤ L_i < R_i ≤ 10^9) — the endpoints of the i-th segment.
All the given 2n segment endpoints are distinct. The set of segments is laminar.
Output
On the first line, output the maximum possible sum of subsegment lengths.
On the i-th of the next n lines, output two integers l_i and r_i (L_i ≤ l_i < r_i ≤ R_i), denoting the chosen subsegment of the i-th segment.
Example
Input
4
1 10
2 3
5 9
6 7
Output
7
3 6
2 3
7 9
6 7
Note
The example input and the example output are illustrated below. <image> <image>
Submitted Solution:
```
print("I love 2020-2021 ICPC, NERC, Northern Eurasia so much")
``` | instruction | 0 | 76,109 | 23 | 152,218 |
No | output | 1 | 76,109 | 23 | 152,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a set of n segments on a line [L_i; R_i]. All 2n segment endpoints are pairwise distinct integers.
The set is laminar — any two segments are either disjoint or one of them contains the other.
Choose a non-empty subsegment [l_i, r_i] with integer endpoints in each segment (L_i ≤ l_i < r_i ≤ R_i) in such a way that no two subsegments intersect (they are allowed to have common endpoints though) and the sum of their lengths (∑_{i=1}^n r_i - l_i) is maximized.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^3) — the number of segments.
The i-th of the next n lines contains two integers L_i and R_i (0 ≤ L_i < R_i ≤ 10^9) — the endpoints of the i-th segment.
All the given 2n segment endpoints are distinct. The set of segments is laminar.
Output
On the first line, output the maximum possible sum of subsegment lengths.
On the i-th of the next n lines, output two integers l_i and r_i (L_i ≤ l_i < r_i ≤ R_i), denoting the chosen subsegment of the i-th segment.
Example
Input
4
1 10
2 3
5 9
6 7
Output
7
3 6
2 3
7 9
6 7
Note
The example input and the example output are illustrated below. <image> <image>
Submitted Solution:
```
def is_contained (Ii,Ij):
li,ri = Ii
lj,rj = Ij
return lj <= li and ri <= rj
def setminus (Ii, list_Ij):
li,ri = Ii
if len(list_Ij)==0:
return Ii
else:
list_Ij.sort (key=lambda item: item[0])
min_lj = list_Ij[0][0]
max_rj = list_Ij[-1][1]
delta1 = min_lj - li
delta2 = ri - max_rj
ans = [li, min_lj] if delta1>= delta2 else [max_rj,ri]
delta = max (delta1,delta2)
m = len(list_Ij)
for k in range (m-1):
l_next_k = list_Ij[k+1][0]
r_k = list_Ij[k][1]
delta_k = l_next_k - r_k
if delta_k>delta:
delta = delta_k
ans = [ r_k , l_next_k ]
return ans
def solve (I,Tree):
indices = set(i for (i,j) in Tree)
if len (Tree)==0:
return {}
elif len (Tree)==1: ## indices = [i] only one index
i = list(indices)[0]
return {i:I[i]}
else:
sub_trees = {} ## also connected components of the main graph
for i in indices:
if not any([Tree[(i,j)] for j in indices]):
sub_trees [i] = [j for j in indices if Tree[(j,i)] ]
## The subtree is rooted at Ii
#print (sub_trees)
ans = {}
for i in sub_trees:
cc_i = sub_trees [i]
Tree_i = {(k,j):Tree[(k,j)] for k in cc_i for j in cc_i}
sub_intervals_i = solve (I,Tree_i)
interval_i = setminus (I[i], list(sub_intervals_i.values()))
ans[i] = interval_i
ans.update (sub_intervals_i)
return ans
def main():
n = int (input())
I = {i:[] for i in range(n)}
for i in range (n):
I[i] = list (map(int, input().split(' ')))
Tree = { (i,j):False for i in range(n) for j in range(n)}
for (i,j) in Tree:
Tree[(i,j)] = (i!=j and is_contained (I[i],I[j]))
'''
print (I)
indices = set(i for (i,j) in Tree)
print ('##########################################')
for i in indices:
print (i,'--',end =" ")
for j in indices:
print (Tree[(i,j)],end =" ")
print ('')
print ('##########################################')
'''
intervals = solve (I,Tree)
sum_ = 0
for i in range (n):
sum_ += intervals[i] [1] - intervals[i] [0]
print (sum_)
for i in range (n):
print (*intervals[i])
main ()
``` | instruction | 0 | 76,110 | 23 | 152,220 |
No | output | 1 | 76,110 | 23 | 152,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4. | instruction | 0 | 76,129 | 23 | 152,258 |
Tags: brute force, geometry
Correct Solution:
```
x,y,z=map(int,input().split())
x1,y1,z1=map(int,input().split())
arr=list(map(int,input().split()))
sums=0
if x1>0:
if x>x1:
sums+=arr[5]
elif x<0:
sums+=arr[4]
elif x1<0:
if x<x1:
sums+=arr[5]
elif x>0:
sums+=arr[4]
if y1>0:
if y>y1:
sums+=arr[1]
elif y<0:
sums+=arr[0]
elif y1<0:
if y<y1:
sums+=arr[1]
elif y>0:
sums+=arr[0]
if z1>0:
if z>z1:
sums+=arr[3]
elif z<0:
sums+=arr[2]
elif z1<0:
if z<z1:
sums+=arr[3]
elif z>0:
sums+=arr[2]
print(sums)
``` | output | 1 | 76,129 | 23 | 152,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4. | instruction | 0 | 76,130 | 23 | 152,260 |
Tags: brute force, geometry
Correct Solution:
```
x, y, z = map(int, input().split())
x1, y1, z1 = map(int, input().split())
a1, a2, a3, a4, a5, a6 = map(int, input().split())
ans = 0
if x > x1:
ans += a6
if x < 0:
ans += a5
if y > y1:
ans += a2
if y < 0:
ans += a1
if z > z1:
ans += a4
if z < 0:
ans += a3
print(ans)
``` | output | 1 | 76,130 | 23 | 152,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4. | instruction | 0 | 76,131 | 23 | 152,262 |
Tags: brute force, geometry
Correct Solution:
```
x,y,z=map(int,input().split())
a,b,c=map(int,input().split())
a1,a2,a3,a4,a5,a6=map(int,input().split())
sum=0
if x>a:
sum+=a6
if x<0:
sum+=a5
if y>b:
sum+=a2
if y<0:
sum+=a1
if z>c:
sum+=a4
if z<0:
sum+=a3
print(sum)
``` | output | 1 | 76,131 | 23 | 152,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4. | instruction | 0 | 76,132 | 23 | 152,264 |
Tags: brute force, geometry
Correct Solution:
```
x, y, z = map(int, input().split())
x1, y1, z1 = map(int, input().split())
a1, a2, a3, a4, a5, a6 = map(int, input().split())
s = 0
if x > x1: s += a6
elif x < 0: s += a5
if y > y1: s += a2
elif y < 0: s += a1
if z > z1: s += a4
elif z < 0: s += a3
print(s)
``` | output | 1 | 76,132 | 23 | 152,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4. | instruction | 0 | 76,133 | 23 | 152,266 |
Tags: brute force, geometry
Correct Solution:
```
from decimal import *
getcontext().prec=16
x,y,z=[int(element) for element in input().split(" ")]
a,b,c=[int(element) for element in input().split(" ")]
liste=[int(element) for element in input().split(" ")]
count=0
if x<0:
count+=liste[4]
if x>a:
count+=liste[5]
if y<0:
count+=liste[0]
if y>b:
count+=liste[1]
if z>c:
count+=liste[3]
if z<0:
count+=liste[2]
print(count)
``` | output | 1 | 76,133 | 23 | 152,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4. | instruction | 0 | 76,134 | 23 | 152,268 |
Tags: brute force, geometry
Correct Solution:
```
#input
x,y,z=map(int,input().strip().split(" "));
x1,y1,z1=map(int,input().strip().split(" "));
a1,a2,a3,a4,a5,a6=map(int,input().strip().split(" "));
#process
ans=0;
if(z>z1):
ans+=a4;
if(z<0):
ans+=a3;
if(x>x1):
ans+=a6;
if(x<0):
ans+=a5;
if(y>y1):
ans+=a2;
if(y<0):
ans+=a1;
print(ans);
``` | output | 1 | 76,134 | 23 | 152,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4. | instruction | 0 | 76,135 | 23 | 152,270 |
Tags: brute force, geometry
Correct Solution:
```
# import itertools
# import bisect
import math
from collections import defaultdict, Counter, deque
import os
import sys
from io import BytesIO, IOBase
# sys.setrecursionlimit(10 ** 5)
ii = lambda: int(input())
lmii = lambda: list(map(int, input().split()))
slmii = lambda: sorted(map(int, input().split()))
li = lambda: list(input())
mii = lambda: map(int, input().split())
msi = lambda: map(str, input().split())
def gcd(a, b):
if b == 0: return a
return gcd(b, a % b)
def lcm(a, b): return (a * b) // gcd(a, b)
def main():
# for _ in " " * int(input()):
x, y, z = map(int, input().split())
x1, y1, z1 = map(int, input().split())
xz,xzp,xy,xyp,yz,yzp = map(int, input().split())
print((yz if x<0 else 0)+(xz if y<0 else 0)+(xy if z<0 else 0)+(yzp if x>x1 else 0)+(xzp if y>y1 else 0)+(xyp if z>z1 else 0))
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")
if __name__ == "__main__":
main()
``` | output | 1 | 76,135 | 23 | 152,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4. | instruction | 0 | 76,136 | 23 | 152,272 |
Tags: brute force, geometry
Correct Solution:
```
x, y, z = map(int, input().split())
x1, y1, z1, s = map(int, input().split() + [0])
a = list(map(int, input().split()))
s += a[0] if y < 0 else a[1] if y > y1 else 0
s += a[2] if z < 0 else a[3] if z > z1 else 0
s += a[4] if x < 0 else a[5] if x > x1 else 0
print(s)
``` | output | 1 | 76,136 | 23 | 152,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4.
Submitted Solution:
```
x,y,z=list(map(int,input().split()))
x1,y1,z1=list(map(int,input().split()))
a=list(map(int,input().split()))
b=0
if y<0:
b+=a[0]
if y>y1:
b+=a[1]
if z<0:
b+=a[2]
if z>z1:
b+=a[3]
if x<0:
b+=a[4]
if x>x1:
b+=a[5]
print(b)
``` | instruction | 0 | 76,137 | 23 | 152,274 |
Yes | output | 1 | 76,137 | 23 | 152,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4.
Submitted Solution:
```
x, y, z = map(int, input().split())
x1, y1, z1 = map(int, input().split())
a = list(map(int, input().split()))
print(a[0] * (y < 0) + a[1] * (y > y1) + a[2] * (z < 0) + a[3] * (z > z1) + a[4] * (x < 0) + a[5] * (x > x1))
``` | instruction | 0 | 76,138 | 23 | 152,276 |
Yes | output | 1 | 76,138 | 23 | 152,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4.
Submitted Solution:
```
x=(input())
a=[int(i) for i in x.split()]
x=(input())
b=[int(i) for i in x.split()]
x=(input())
num=[int(i) for i in x.split()]
ans=0
if a[1]<0:
ans+=num[0]
if a[2]<0:
ans+=num[2]
if a[0]<0:
ans+=num[4]
if a[0]-b[0]>0:
ans+=num[5]
if a[1]-b[1]>0:
ans+=num[1]
if a[2]-b[2]>0:
ans+=num[3]
print(ans)
``` | instruction | 0 | 76,139 | 23 | 152,278 |
Yes | output | 1 | 76,139 | 23 | 152,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4.
Submitted Solution:
```
v=list(map(int, input().split()))
b=list(map(int, input().split()))
s=list(map(int, input().split()))
sol=0
if v[1]<0:
sol+=s[0]
if v[1]>b[1]:
sol+=s[1]
if v[2]<0:
sol+=s[2]
if v[2]>b[2]:
sol+=s[3]
if v[0]<0:
sol+=s[4]
if v[0]>b[0]:
sol+=s[5]
print(sol)
``` | instruction | 0 | 76,140 | 23 | 152,280 |
Yes | output | 1 | 76,140 | 23 | 152,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4.
Submitted Solution:
```
x,y,z=map(int,input().split())
x1,y1,z1=map(int,input().split())
arr=list(map(int,input().split()))
sums=0
if x>0:
if x>x1:
sums+=arr[5]
elif x<0:
sums+=arr[4]
elif x<0:
if x<x1:
sums+=arr[5]
elif x>0:
sums+=arr[4]
if y>0:
if y>y1:
sums+=arr[1]
elif y<0:
sums+=arr[0]
elif y<0:
if y<y1:
sums+=arr[1]
elif y>0:
sums+=arr[0]
if z>0:
if z>z1:
sums+=arr[3]
elif z<0:
sums+=arr[2]
elif z<0:
if z<z1:
sums+=arr[3]
elif z>0:
sums+=arr[2]
print(sums)
``` | instruction | 0 | 76,141 | 23 | 152,282 |
No | output | 1 | 76,141 | 23 | 152,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4.
Submitted Solution:
```
x,y,z=map(int,input().split())
x1,y1,z1=map(int,input().split())
arr=list(map(int,input().split()))
sums=0
if x<0:
if x>x1:
sums+=arr[5]
elif x<0:
sums+=arr[4]
else:
if x<x1:
sums+=arr[5]
elif x>0:
sums+=arr[4]
if y>0:
if y>y1:
sums+=arr[1]
elif y<0:
sums+=arr[0]
else:
if y<y1:
sums+=arr[1]
elif y>0:
sums+=arr[0]
if z>0:
if z>z1:
sums+=arr[3]
elif z<0:
sums+=arr[2]
else:
if z<z1:
sums+=arr[3]
elif z>0:
sums+=arr[2]
print(sums)
``` | instruction | 0 | 76,142 | 23 | 152,284 |
No | output | 1 | 76,142 | 23 | 152,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4.
Submitted Solution:
```
from math import sqrt,gcd
def primeFactors(n):
# Print the number of two's that divide n
sa = []
while n % 2 == 0:
sa.append(2)
n = n / 2
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used
for i in range(3,int(sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
sa.append(int(i))
n = n / i
# Condition if n is a prime
# number greater than 2
if n > 2:
sa.append(int(n))
return list(set(sa))
from math import sqrt
from collections import defaultdict
from bisect import bisect_right
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def cal(n):
seti = set()
for i in range(2,int(sqrt(n))+1):
if n%i == 0:
seti.add(n//i)
seti.add(i)
return seti
def pre(s):
n = len(s)
lps = [0]*(n)
for i in range(1,n):
j = lps[i-1]
while j>0 and s[i]!=s[j]:
j = lps[j-1]
if s[i] == s[j]:
j+=1
lps[i] = j
for i in range(1,n):
if lps[i] == i+1:
lps[i]-=1
return lps
from collections import defaultdict
def solve():
x,y,z = map(int,input().split())
x1,y1,z1 = map(int,input().split())
m = sqrt(x1**2 + y1**2 + z1**2)
m1 = sqrt((x1-x)**2 + (y1-y)**2 + (z1-z)**2)
a1,a2,a3,a4,a5,a6 = map(int,input().split())
ans = 0
if y<0:
z5 = abs(y/m)
if 0.5<=z5<=1:
ans+=a1
if y>y1:
z5 = abs((y-y1)/m1)
# print(z)
if 0.5<=z5<=1:
ans+=a2
if z<0:
z5 = abs(z/m)
if 0.5<=z5<=1:
ans+=a3
if z>z1:
z5 = abs((z-z1)/m1)
if 0.5<=z5<=1:
ans+=a4
if x<0:
z5 = abs(x/m)
if 0.5<=z5<=1:
ans+=a5
if x>x1:
z5 = abs((x-x1)/m)
if 0.5<=z5<=1:
ans+=a6
print(ans)
# t = int(stdin.readline())
# for _ in range(t):
solve()
``` | instruction | 0 | 76,143 | 23 | 152,286 |
No | output | 1 | 76,143 | 23 | 152,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.
<image>
The numbers are located on the box like that:
* number a1 is written on the face that lies on the ZOX plane;
* a2 is written on the face, parallel to the plane from the previous point;
* a3 is written on the face that lies on the XOY plane;
* a4 is written on the face, parallel to the plane from the previous point;
* a5 is written on the face that lies on the YOZ plane;
* a6 is written on the face, parallel to the plane from the previous point.
At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).
Input
The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.
It is guaranteed that point (x, y, z) is located strictly outside the box.
Output
Print a single integer — the sum of all numbers on the box faces that Vasya sees.
Examples
Input
2 2 2
1 1 1
1 2 3 4 5 6
Output
12
Input
0 0 10
3 2 3
1 2 3 4 5 6
Output
4
Note
The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6 (on the right face that is the lightest) and a4 (on the left visible face).
In the second sample Vasya can only see number a4.
Submitted Solution:
```
import sys,os.path
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")
if __name__ == '__main__':
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
# for _ in range(int(input())):
x,y,z = map(int,input().split())
x1,y1,z1 = map(int,input().split())
a = list(map(int,input().split()))
s = 0
if x>=x1:
s+=a[5]
elif x<0:
s+=a[4]
if y>=y1:
s+=a[1]
elif y<0:
s+=a[0]
if z>=z1:
s+=a[3]
elif z<0:
s+=a[2]
print(s)
``` | instruction | 0 | 76,144 | 23 | 152,288 |
No | output | 1 | 76,144 | 23 | 152,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image> | instruction | 0 | 76,145 | 23 | 152,290 |
Tags: brute force, geometry, math
Correct Solution:
```
def angle(v):
x, y = v
return atan2(x, y)
def main():
n = int(input())
import sys
if n == 1:
input()
print(0)
else:
l = sorted(atan2(*(int(i)for i in line.split())) for line in sys.stdin)
print(360 - 180 * max(max(j - i for i, j in zip(l[:-1], l[1:])), 2 *pi +l[0] - l[-1]) / pi)
if __name__ == "__main__":
from math import pi, atan2
import doctest
main()
``` | output | 1 | 76,145 | 23 | 152,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image> | instruction | 0 | 76,146 | 23 | 152,292 |
Tags: brute force, geometry, math
Correct Solution:
```
from math import atan2, pi
n = int(input())
a = [map(int, input().split()) for i in range(n)]
a = sorted(atan2(y, x) for x, y in a)
d = [a[i + 1] - a[i] for i in range(n - 1)]
d.append(2 * pi - a[n - 1] + a[0])
print(360 - 180 * max(d) / pi)
``` | output | 1 | 76,146 | 23 | 152,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image> | instruction | 0 | 76,147 | 23 | 152,294 |
Tags: brute force, geometry, math
Correct Solution:
```
import math
n = int(input())
coors = [[int(item) for item in input().split(' ')] for i in range(n)]
def get_pos(item):
(x, y) = item
deg = math.degrees(math.atan(abs(math.inf if x == 0 else y / x)))
if y < 0:
if x >= 0:
return 360 - deg
else:
return 180 + deg
else:
if x >= 0:
return deg
else:
return 180 - deg
angles = [get_pos(item) for item in coors]
angles.sort()
t = [angles[0] + 360 - angles[-1]]
for i in range(1, n):
t.append(angles[i] - angles[i - 1])
print(360 - max(t))
``` | output | 1 | 76,147 | 23 | 152,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image> | instruction | 0 | 76,148 | 23 | 152,296 |
Tags: brute force, geometry, math
Correct Solution:
```
import math
n=int(input())
angles=[]
for i in range(n):
a,b=map(int,input().split())
if a==0 and b>0:
angle=90
elif a==0 and b<0:
angle=270
elif b==0 and a>0:
angle=0
elif b==0 and a<0:
angle=180
elif a<0:
angle=(math.atan(b/a)/math.pi)*180+180
else:
angle=(math.atan(b/a)/math.pi)*180
if angle<0:
angle+=360
angles.append(angle)
angles.sort()
minn=angles[-1]-angles[0]
for i in range(len(angles)-1):
if minn>360-(angles[i+1]-angles[i]):
minn=360-(angles[i+1]-angles[i])
print(f"{minn:.10f}")
``` | output | 1 | 76,148 | 23 | 152,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image> | instruction | 0 | 76,149 | 23 | 152,298 |
Tags: brute force, geometry, math
Correct Solution:
```
import math
n, max_diff = int(input()), 0
values = [[int(i) for i in input().split()] for j in range(n)]
values = sorted([math.atan2(values[i][1], values[i][0]) for i in range(n)])
for i in range(n - 1):
max_diff = max(max_diff, values[i + 1] - values[i])
max_diff = max(max_diff, 2 * math.pi - (values[-1] - values[0]))
print(360 - max_diff * 180 / math.pi)
``` | output | 1 | 76,149 | 23 | 152,299 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image> | instruction | 0 | 76,150 | 23 | 152,300 |
Tags: brute force, geometry, math
Correct Solution:
```
import math
a=-400
m=400
n=int(input())
ang=[]
for _ in ' '*n:
x,y=map(int,input().split())
theta =(180*math.atan2(y, x)/math.pi)
ang.append(theta)
ang.sort()
ans=ang[-1]-ang[0]
for i in range(n-1):
ans=min(ans,360-(ang[i+1]-ang[i]))
print(ans)
``` | output | 1 | 76,150 | 23 | 152,301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image> | instruction | 0 | 76,151 | 23 | 152,302 |
Tags: brute force, geometry, math
Correct Solution:
```
from math import pi, atan2
n = int(input())
if n == 1:
input()
print(0)
else:
l = sorted([atan2(*(int(x) for x in input().split())) for i in range(n)])
print(360 - 180 * max(max(j - i for i, j in zip(l[:-1], l[1:])), 2 *pi +l[0] - l[-1]) / pi)
``` | output | 1 | 76,151 | 23 | 152,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image> | instruction | 0 | 76,152 | 23 | 152,304 |
Tags: brute force, geometry, math
Correct Solution:
```
from math import atan2
from math import pi
from math import degrees
n = int(input())
if n == 1:
print(0)
exit()
arr = []
for _ in range(n):
x, y = map(int, input().split())
arr.append(degrees(atan2(y, x)))
arr.sort()
arr.append(arr[0]+360)
print(360-max([arr[i+1] - arr[i] for i in range(n)]))
``` | output | 1 | 76,152 | 23 | 152,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image>
Submitted Solution:
```
import math
n = int(input())
arr = []
pnts = list(tuple(map(int, input().split()) for _ in range(n)))
for x, y in pnts:
arr.append (math.atan2(y, x) % (2*math.pi))
arr.sort()
if abs(arr[0] - arr[-1]) < 1e-9:
print (0)
exit()
res = 0
for i in range(n):
res = max(res, arr[i] - arr[i-1])
res = max(res, 2*math.pi - (arr[-1] - arr[0]))
print(360 - res / math.pi * 180)
``` | instruction | 0 | 76,153 | 23 | 152,306 |
Yes | output | 1 | 76,153 | 23 | 152,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image>
Submitted Solution:
```
def main():
from sys import stdin, stdout
from math import atan2, pi
n = int(stdin.readline().strip())
angles = []
for i in range(n):
(x, y) = map(float, stdin.readline().strip().split(' '))
angles.append((atan2(y, x) * 180 / pi) % 360)
angles = sorted(angles)
m = 0
for i in range(n - 1):
m = max(m, angles[i + 1] - angles[i])
m = max(m, (angles[0] - angles[-1]) % 360)
stdout.write('{:.15f}\n'.format((360 - m) % 360))
main()
``` | instruction | 0 | 76,154 | 23 | 152,308 |
Yes | output | 1 | 76,154 | 23 | 152,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image>
Submitted Solution:
```
from math import atan2, pi
def solve():
n = int(input().rstrip())
angles = []
for _ in range(n):
a, b = map(int, input().rstrip().split())
angles.append(atan2(b, a))
angles.sort()
maxangle = 2 * pi - (angles[n - 1] - angles[0])
for i in range(n - 1):
maxangle = max(angles[i + 1] - angles[i], maxangle)
return 360 - maxangle * 180 / pi
if __name__ == '__main__':
print(solve())
``` | instruction | 0 | 76,155 | 23 | 152,310 |
Yes | output | 1 | 76,155 | 23 | 152,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image>
Submitted Solution:
```
t = []
from math import pi
from math import atan
def f(x,y):
if y == 0:
if x>0:
return 0
else:
return pi
if x == 0:
if y>0:
return pi/2
else:
return 3*pi/2
if x > 0 and y > 0:
return atan(y/x)
if x < 0 and y < 0:
return pi + atan(y/x)
if x > 0 and y < 0:
return 2*pi+atan(y/x)
if x < 0 and y > 0:
return pi+atan(y/x)
def g(x,y):
return 180*f(x,y)/pi
for i in range(int(input())):
x, y = list(map(int, input().split(' ')))
t.append(g(x,y))
t.sort()
m = t[-1] - t[0]
for i in range(1, len(t)):
m = min(m, 360 - (t[i]-t[i-1]))
print(m)
``` | instruction | 0 | 76,156 | 23 | 152,312 |
Yes | output | 1 | 76,156 | 23 | 152,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image>
Submitted Solution:
```
from math import *
n,angs = int(input()), []
for _ in range(n):
x,y = input().split()
t = atan2(float(y),float(x))
if t < 0:
t += 2*pi
angs.append(t)
#print(list(map(degrees,angs)))
angs.sort()
dif = []
for i in range(len(angs)):
al = angs[i]-angs[i-1]
if al < 0:
al += 2*pi
dif.append(al)
res = 0
#print(list(map(degrees,dif)))
if len(angs)>2:
res = degrees(2*pi-max(dif))
elif len(angs)==2 :
res = degrees(min(dif))
print('%.10f' % res)
# C:\Users\Usuario\HOME2\Programacion\ACM
``` | instruction | 0 | 76,157 | 23 | 152,314 |
No | output | 1 | 76,157 | 23 | 152,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image>
Submitted Solution:
```
from sys import stdin
import math
N = int(stdin.readline())
angles = []
for n in range(N):
x, y = map(int, stdin.readline().split())
angles.append(math.atan2(y, x))
angles.sort()
print(360 - 180 * max((angles[i - 1] - angles[i] + 2 * math.pi) % (2 * math.pi) for i in range(len(angles))) / math.pi)
``` | instruction | 0 | 76,158 | 23 | 152,316 |
No | output | 1 | 76,158 | 23 | 152,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image>
Submitted Solution:
```
from math import atan2
from math import pi
def mooc():
n=int(input())
liss=[]
y1=0
for i in range(n):
l=list(map(int,input().split()))
liss.append(atan2(l[0],l[1]))
liss.sort()
for i in range(n-1):
y=liss[i+1]-liss[i]
if(y>y1):
y1=y
y=liss[n-1]-liss[0]
if(y>y1):
y1=y
print((y1/pi)*180)
mooc()
``` | instruction | 0 | 76,159 | 23 | 152,318 |
No | output | 1 | 76,159 | 23 | 152,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.
As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.
Input
The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.
Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.
Output
Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.
Examples
Input
2
2 0
0 2
Output
90.0000000000
Input
3
2 0
0 2
-2 2
Output
135.0000000000
Input
4
2 0
0 2
-2 0
0 -2
Output
270.0000000000
Input
2
2 1
1 2
Output
36.8698976458
Note
Solution for the first sample test is shown below:
<image>
Solution for the second sample test is shown below:
<image>
Solution for the third sample test is shown below:
<image>
Solution for the fourth sample test is shown below:
<image>
Submitted Solution:
```
from math import atan2
def mooc():
pi=atan2(1,1)*4
n=int(input())
liss=[]
y1=0
for i in range(n):
l=list(map(int,input().split()))
liss.append(atan2(l[0],l[1]))
liss.sort()
for i in range(n-1):
y=liss[i+1]-liss[i]
if(y>y1):
y1=y
y=liss[n-1]-liss[0]
if(y>y1):
y1=y
print(y1*180/pi)
mooc()
``` | instruction | 0 | 76,160 | 23 | 152,320 |
No | output | 1 | 76,160 | 23 | 152,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangle grid. That grid's size is n × m. Let's denote the coordinate system on the grid. So, each point on the grid will have coordinates — a pair of integers (x, y) (0 ≤ x ≤ n, 0 ≤ y ≤ m).
Your task is to find a maximum sub-rectangle on the grid (x1, y1, x2, y2) so that it contains the given point (x, y), and its length-width ratio is exactly (a, b). In other words the following conditions must hold: 0 ≤ x1 ≤ x ≤ x2 ≤ n, 0 ≤ y1 ≤ y ≤ y2 ≤ m, <image>.
The sides of this sub-rectangle should be parallel to the axes. And values x1, y1, x2, y2 should be integers.
<image>
If there are multiple solutions, find the rectangle which is closest to (x, y). Here "closest" means the Euclid distance between (x, y) and the center of the rectangle is as small as possible. If there are still multiple solutions, find the lexicographically minimum one. Here "lexicographically minimum" means that we should consider the sub-rectangle as sequence of integers (x1, y1, x2, y2), so we can choose the lexicographically minimum one.
Input
The first line contains six integers n, m, x, y, a, b (1 ≤ n, m ≤ 109, 0 ≤ x ≤ n, 0 ≤ y ≤ m, 1 ≤ a ≤ n, 1 ≤ b ≤ m).
Output
Print four integers x1, y1, x2, y2, which represent the founded sub-rectangle whose left-bottom point is (x1, y1) and right-up point is (x2, y2).
Examples
Input
9 9 5 5 2 1
Output
1 3 9 7
Input
100 100 52 50 46 56
Output
17 8 86 92 | instruction | 0 | 76,163 | 23 | 152,326 |
Tags: implementation, math
Correct Solution:
```
import sys
from fractions import gcd
with sys.stdin as fin, sys.stdout as fout:
n, m, x, y, a, b = map(int, next(fin).split())
d = gcd(a, b)
a //= d
b //= d
k = min(n // a, m // b)
w = k * a
h = k * b
x1 = x - (w + 1) // 2
y1 = y - (h + 1) // 2
x1 = min(x1, n - w)
y1 = min(y1, m - h)
x1 = max(x1, 0)
y1 = max(y1, 0)
print(x1, y1, x1 + w, y1 + h, file=fout)
``` | output | 1 | 76,163 | 23 | 152,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangle grid. That grid's size is n × m. Let's denote the coordinate system on the grid. So, each point on the grid will have coordinates — a pair of integers (x, y) (0 ≤ x ≤ n, 0 ≤ y ≤ m).
Your task is to find a maximum sub-rectangle on the grid (x1, y1, x2, y2) so that it contains the given point (x, y), and its length-width ratio is exactly (a, b). In other words the following conditions must hold: 0 ≤ x1 ≤ x ≤ x2 ≤ n, 0 ≤ y1 ≤ y ≤ y2 ≤ m, <image>.
The sides of this sub-rectangle should be parallel to the axes. And values x1, y1, x2, y2 should be integers.
<image>
If there are multiple solutions, find the rectangle which is closest to (x, y). Here "closest" means the Euclid distance between (x, y) and the center of the rectangle is as small as possible. If there are still multiple solutions, find the lexicographically minimum one. Here "lexicographically minimum" means that we should consider the sub-rectangle as sequence of integers (x1, y1, x2, y2), so we can choose the lexicographically minimum one.
Input
The first line contains six integers n, m, x, y, a, b (1 ≤ n, m ≤ 109, 0 ≤ x ≤ n, 0 ≤ y ≤ m, 1 ≤ a ≤ n, 1 ≤ b ≤ m).
Output
Print four integers x1, y1, x2, y2, which represent the founded sub-rectangle whose left-bottom point is (x1, y1) and right-up point is (x2, y2).
Examples
Input
9 9 5 5 2 1
Output
1 3 9 7
Input
100 100 52 50 46 56
Output
17 8 86 92 | instruction | 0 | 76,164 | 23 | 152,328 |
Tags: implementation, math
Correct Solution:
```
import math
n, m, x, y, a, b = map(int, input().split())
gcd = math.gcd(a, b)
a //= gcd
b //= gcd
max_ratio = min(n // a, m // b)
a *= max_ratio
b *= max_ratio
x1 = max(0, min(x - (a + 1) // 2, n - a))
y1 = max(0, min(y - (b + 1) // 2, m - b))
print(x1, y1, x1 + a, y1 + b)
``` | output | 1 | 76,164 | 23 | 152,329 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangle grid. That grid's size is n × m. Let's denote the coordinate system on the grid. So, each point on the grid will have coordinates — a pair of integers (x, y) (0 ≤ x ≤ n, 0 ≤ y ≤ m).
Your task is to find a maximum sub-rectangle on the grid (x1, y1, x2, y2) so that it contains the given point (x, y), and its length-width ratio is exactly (a, b). In other words the following conditions must hold: 0 ≤ x1 ≤ x ≤ x2 ≤ n, 0 ≤ y1 ≤ y ≤ y2 ≤ m, <image>.
The sides of this sub-rectangle should be parallel to the axes. And values x1, y1, x2, y2 should be integers.
<image>
If there are multiple solutions, find the rectangle which is closest to (x, y). Here "closest" means the Euclid distance between (x, y) and the center of the rectangle is as small as possible. If there are still multiple solutions, find the lexicographically minimum one. Here "lexicographically minimum" means that we should consider the sub-rectangle as sequence of integers (x1, y1, x2, y2), so we can choose the lexicographically minimum one.
Input
The first line contains six integers n, m, x, y, a, b (1 ≤ n, m ≤ 109, 0 ≤ x ≤ n, 0 ≤ y ≤ m, 1 ≤ a ≤ n, 1 ≤ b ≤ m).
Output
Print four integers x1, y1, x2, y2, which represent the founded sub-rectangle whose left-bottom point is (x1, y1) and right-up point is (x2, y2).
Examples
Input
9 9 5 5 2 1
Output
1 3 9 7
Input
100 100 52 50 46 56
Output
17 8 86 92 | instruction | 0 | 76,165 | 23 | 152,330 |
Tags: implementation, math
Correct Solution:
```
from fractions import gcd
n, m, x, y, a, b = map(int, input().split())
r = gcd(a, b)
a, b = a // r, b // r
r = min(n // a, m // b)
a, b = a * r, b * r
cx, cy = (a + 1) // 2, (b + 1) // 2
dx, dy = min(n - a, max(cx, x) - cx), min(m - b, max(cy, y) - cy)
print(dx, dy, a + dx, b + dy)
``` | output | 1 | 76,165 | 23 | 152,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangle grid. That grid's size is n × m. Let's denote the coordinate system on the grid. So, each point on the grid will have coordinates — a pair of integers (x, y) (0 ≤ x ≤ n, 0 ≤ y ≤ m).
Your task is to find a maximum sub-rectangle on the grid (x1, y1, x2, y2) so that it contains the given point (x, y), and its length-width ratio is exactly (a, b). In other words the following conditions must hold: 0 ≤ x1 ≤ x ≤ x2 ≤ n, 0 ≤ y1 ≤ y ≤ y2 ≤ m, <image>.
The sides of this sub-rectangle should be parallel to the axes. And values x1, y1, x2, y2 should be integers.
<image>
If there are multiple solutions, find the rectangle which is closest to (x, y). Here "closest" means the Euclid distance between (x, y) and the center of the rectangle is as small as possible. If there are still multiple solutions, find the lexicographically minimum one. Here "lexicographically minimum" means that we should consider the sub-rectangle as sequence of integers (x1, y1, x2, y2), so we can choose the lexicographically minimum one.
Input
The first line contains six integers n, m, x, y, a, b (1 ≤ n, m ≤ 109, 0 ≤ x ≤ n, 0 ≤ y ≤ m, 1 ≤ a ≤ n, 1 ≤ b ≤ m).
Output
Print four integers x1, y1, x2, y2, which represent the founded sub-rectangle whose left-bottom point is (x1, y1) and right-up point is (x2, y2).
Examples
Input
9 9 5 5 2 1
Output
1 3 9 7
Input
100 100 52 50 46 56
Output
17 8 86 92 | instruction | 0 | 76,166 | 23 | 152,332 |
Tags: implementation, math
Correct Solution:
```
from fractions import gcd
n, m, x, y, a, b = map(int, input().split())
r = gcd(a, b)
a, b = a // r, b // r
r = min(n // a, m // b)
a, b = a * r, b * r
cx, cy = (a + 1) // 2, (b + 1) // 2
dx, dy = min(n - a, max(cx, x) - cx), min(m - b, max(cy, y) - cy)
print(dx, dy, a + dx, b + dy)
# Made By Mostafa_Khaled
``` | output | 1 | 76,166 | 23 | 152,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangle grid. That grid's size is n × m. Let's denote the coordinate system on the grid. So, each point on the grid will have coordinates — a pair of integers (x, y) (0 ≤ x ≤ n, 0 ≤ y ≤ m).
Your task is to find a maximum sub-rectangle on the grid (x1, y1, x2, y2) so that it contains the given point (x, y), and its length-width ratio is exactly (a, b). In other words the following conditions must hold: 0 ≤ x1 ≤ x ≤ x2 ≤ n, 0 ≤ y1 ≤ y ≤ y2 ≤ m, <image>.
The sides of this sub-rectangle should be parallel to the axes. And values x1, y1, x2, y2 should be integers.
<image>
If there are multiple solutions, find the rectangle which is closest to (x, y). Here "closest" means the Euclid distance between (x, y) and the center of the rectangle is as small as possible. If there are still multiple solutions, find the lexicographically minimum one. Here "lexicographically minimum" means that we should consider the sub-rectangle as sequence of integers (x1, y1, x2, y2), so we can choose the lexicographically minimum one.
Input
The first line contains six integers n, m, x, y, a, b (1 ≤ n, m ≤ 109, 0 ≤ x ≤ n, 0 ≤ y ≤ m, 1 ≤ a ≤ n, 1 ≤ b ≤ m).
Output
Print four integers x1, y1, x2, y2, which represent the founded sub-rectangle whose left-bottom point is (x1, y1) and right-up point is (x2, y2).
Examples
Input
9 9 5 5 2 1
Output
1 3 9 7
Input
100 100 52 50 46 56
Output
17 8 86 92 | instruction | 0 | 76,167 | 23 | 152,334 |
Tags: implementation, math
Correct Solution:
```
import sys
from fractions import gcd
with sys.stdin as fin, sys.stdout as fout:
n, m, x, y, a, b = map(int, next(fin).split())
d = gcd(a, b)
a //= d
b //= d
k = min(n // a, m // b) # >_<
w = k * a
h = k * b
best = tuple([float('inf')] * 3)
for add1 in 0, 1:
for add2 in 0, 1:
x1 = x - w // 2 - add1
y1 = y - h // 2 - add2
cur = ((2 * x1 + w - 2 * x) ** 2 + (2 * y1 + h - 2 * y) ** 2, x1, y1)
if cur < best:
best = cur
x1, y1 = best[1:]
x1 = min(x1, n - w)
y1 = min(y1, m - h)
x1 = max(x1, 0)
y1 = max(y1, 0)
print(x1, y1, x1 + w, y1 + h)
``` | output | 1 | 76,167 | 23 | 152,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangle grid. That grid's size is n × m. Let's denote the coordinate system on the grid. So, each point on the grid will have coordinates — a pair of integers (x, y) (0 ≤ x ≤ n, 0 ≤ y ≤ m).
Your task is to find a maximum sub-rectangle on the grid (x1, y1, x2, y2) so that it contains the given point (x, y), and its length-width ratio is exactly (a, b). In other words the following conditions must hold: 0 ≤ x1 ≤ x ≤ x2 ≤ n, 0 ≤ y1 ≤ y ≤ y2 ≤ m, <image>.
The sides of this sub-rectangle should be parallel to the axes. And values x1, y1, x2, y2 should be integers.
<image>
If there are multiple solutions, find the rectangle which is closest to (x, y). Here "closest" means the Euclid distance between (x, y) and the center of the rectangle is as small as possible. If there are still multiple solutions, find the lexicographically minimum one. Here "lexicographically minimum" means that we should consider the sub-rectangle as sequence of integers (x1, y1, x2, y2), so we can choose the lexicographically minimum one.
Input
The first line contains six integers n, m, x, y, a, b (1 ≤ n, m ≤ 109, 0 ≤ x ≤ n, 0 ≤ y ≤ m, 1 ≤ a ≤ n, 1 ≤ b ≤ m).
Output
Print four integers x1, y1, x2, y2, which represent the founded sub-rectangle whose left-bottom point is (x1, y1) and right-up point is (x2, y2).
Examples
Input
9 9 5 5 2 1
Output
1 3 9 7
Input
100 100 52 50 46 56
Output
17 8 86 92 | instruction | 0 | 76,168 | 23 | 152,336 |
Tags: implementation, math
Correct Solution:
```
from fractions import gcd
n, m, x, y, a, b = map(int, input().split())
g = gcd(a, b)
a, b = a // g, b // g
k = min(n // a, m // b)
a, b = k * a, k * b
x1, x2 = x - (a - a // 2), x + a // 2
y1, y2 = y - (b - b // 2), y + b // 2
d = max(0, 0 - x1)
x1, x2 = x1 + d, x2 + d
d = max(0, x2 - n)
x1, x2 = x1 - d, x2 - d
d = max(0, 0 - y1)
y1, y2 = y1 + d, y2 + d
d = max(0, y2 - m)
y1, y2 = y1 - d, y2 - d
print(" ".join(map(str, [x1, y1, x2, y2])));
``` | output | 1 | 76,168 | 23 | 152,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rectangle grid. That grid's size is n × m. Let's denote the coordinate system on the grid. So, each point on the grid will have coordinates — a pair of integers (x, y) (0 ≤ x ≤ n, 0 ≤ y ≤ m).
Your task is to find a maximum sub-rectangle on the grid (x1, y1, x2, y2) so that it contains the given point (x, y), and its length-width ratio is exactly (a, b). In other words the following conditions must hold: 0 ≤ x1 ≤ x ≤ x2 ≤ n, 0 ≤ y1 ≤ y ≤ y2 ≤ m, <image>.
The sides of this sub-rectangle should be parallel to the axes. And values x1, y1, x2, y2 should be integers.
<image>
If there are multiple solutions, find the rectangle which is closest to (x, y). Here "closest" means the Euclid distance between (x, y) and the center of the rectangle is as small as possible. If there are still multiple solutions, find the lexicographically minimum one. Here "lexicographically minimum" means that we should consider the sub-rectangle as sequence of integers (x1, y1, x2, y2), so we can choose the lexicographically minimum one.
Input
The first line contains six integers n, m, x, y, a, b (1 ≤ n, m ≤ 109, 0 ≤ x ≤ n, 0 ≤ y ≤ m, 1 ≤ a ≤ n, 1 ≤ b ≤ m).
Output
Print four integers x1, y1, x2, y2, which represent the founded sub-rectangle whose left-bottom point is (x1, y1) and right-up point is (x2, y2).
Examples
Input
9 9 5 5 2 1
Output
1 3 9 7
Input
100 100 52 50 46 56
Output
17 8 86 92
Submitted Solution:
```
import sys
from fractions import gcd
with sys.stdin as fin, sys.stdout as fout:
n, m, x, y, a, b = map(int, next(fin).split())
d = gcd(a, b)
a //= d
b //= d
k = min(n // a, m // b) # >_<
w = k * a
h = k * b
best = tuple([float('inf')] * 3)
for add1 in 0, 1:
for add2 in 0, 1:
x1 = x - w // 2 - add1
y1 = y - h // 2 - add2
cur = ((x1 + w / 2 - x) ** 2 + (y1 + h / 2 - y) ** 2, x1, y1)
if cur < best:
best = cur
print(best[1], best[2], best[1] + w, best[2] + h)
``` | instruction | 0 | 76,169 | 23 | 152,338 |
No | output | 1 | 76,169 | 23 | 152,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rectangle grid. That grid's size is n × m. Let's denote the coordinate system on the grid. So, each point on the grid will have coordinates — a pair of integers (x, y) (0 ≤ x ≤ n, 0 ≤ y ≤ m).
Your task is to find a maximum sub-rectangle on the grid (x1, y1, x2, y2) so that it contains the given point (x, y), and its length-width ratio is exactly (a, b). In other words the following conditions must hold: 0 ≤ x1 ≤ x ≤ x2 ≤ n, 0 ≤ y1 ≤ y ≤ y2 ≤ m, <image>.
The sides of this sub-rectangle should be parallel to the axes. And values x1, y1, x2, y2 should be integers.
<image>
If there are multiple solutions, find the rectangle which is closest to (x, y). Here "closest" means the Euclid distance between (x, y) and the center of the rectangle is as small as possible. If there are still multiple solutions, find the lexicographically minimum one. Here "lexicographically minimum" means that we should consider the sub-rectangle as sequence of integers (x1, y1, x2, y2), so we can choose the lexicographically minimum one.
Input
The first line contains six integers n, m, x, y, a, b (1 ≤ n, m ≤ 109, 0 ≤ x ≤ n, 0 ≤ y ≤ m, 1 ≤ a ≤ n, 1 ≤ b ≤ m).
Output
Print four integers x1, y1, x2, y2, which represent the founded sub-rectangle whose left-bottom point is (x1, y1) and right-up point is (x2, y2).
Examples
Input
9 9 5 5 2 1
Output
1 3 9 7
Input
100 100 52 50 46 56
Output
17 8 86 92
Submitted Solution:
```
import sys
from fractions import gcd
with sys.stdin as fin, sys.stdout as fout:
n, m, x, y, a, b = map(int, next(fin).split())
d = gcd(a, b)
a //= d
b //= d
k = min(n // a, m // b) # >_<
w = k * a
h = k * b
x1 = x - (w + 1) // 2
y1 = y - (h + 1) // 2
print(x1, y1, x1 + w, y1 + h)
``` | instruction | 0 | 76,170 | 23 | 152,340 |
No | output | 1 | 76,170 | 23 | 152,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rectangle grid. That grid's size is n × m. Let's denote the coordinate system on the grid. So, each point on the grid will have coordinates — a pair of integers (x, y) (0 ≤ x ≤ n, 0 ≤ y ≤ m).
Your task is to find a maximum sub-rectangle on the grid (x1, y1, x2, y2) so that it contains the given point (x, y), and its length-width ratio is exactly (a, b). In other words the following conditions must hold: 0 ≤ x1 ≤ x ≤ x2 ≤ n, 0 ≤ y1 ≤ y ≤ y2 ≤ m, <image>.
The sides of this sub-rectangle should be parallel to the axes. And values x1, y1, x2, y2 should be integers.
<image>
If there are multiple solutions, find the rectangle which is closest to (x, y). Here "closest" means the Euclid distance between (x, y) and the center of the rectangle is as small as possible. If there are still multiple solutions, find the lexicographically minimum one. Here "lexicographically minimum" means that we should consider the sub-rectangle as sequence of integers (x1, y1, x2, y2), so we can choose the lexicographically minimum one.
Input
The first line contains six integers n, m, x, y, a, b (1 ≤ n, m ≤ 109, 0 ≤ x ≤ n, 0 ≤ y ≤ m, 1 ≤ a ≤ n, 1 ≤ b ≤ m).
Output
Print four integers x1, y1, x2, y2, which represent the founded sub-rectangle whose left-bottom point is (x1, y1) and right-up point is (x2, y2).
Examples
Input
9 9 5 5 2 1
Output
1 3 9 7
Input
100 100 52 50 46 56
Output
17 8 86 92
Submitted Solution:
```
import math
n, m, x, y, a, b = map(int, input().split())
gcd = math.gcd(a, b)
a //= gcd
b //= gcd
max_ratio = min(n // a, m // b)
#if (a * max_ratio) % 2 == 1 or (b * max_ratio) % 2 == 1:
# max_ratio -= 1
a *= max_ratio
b *= max_ratio
print(x - (a + 1) // 2, y - (b + 1) // 2, x + a // 2, y + b // 2)
``` | instruction | 0 | 76,171 | 23 | 152,342 |
No | output | 1 | 76,171 | 23 | 152,343 |
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