message stringlengths 2 22.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 16 109k | cluster float64 1 1 | __index_level_0__ int64 32 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The territory of Berland is represented by a rectangular field n × m in size. The king of Berland lives in the capital, located on the upper left square (1, 1). The lower right square has coordinates (n, m). One day the king decided to travel through the whole country and return back to the capital, having visited every square (except the capital) exactly one time. The king must visit the capital exactly two times, at the very beginning and at the very end of his journey. The king can only move to the side-neighboring squares. However, the royal advise said that the King possibly will not be able to do it. But there is a way out — one can build the system of one way teleporters between some squares so that the king could fulfill his plan. No more than one teleporter can be installed on one square, every teleporter can be used any number of times, however every time it is used, it transports to the same given for any single teleporter square. When the king reaches a square with an installed teleporter he chooses himself whether he is or is not going to use the teleport. What minimum number of teleporters should be installed for the king to complete the journey? You should also compose the journey path route for the king.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100, 2 ≤ n · m) — the field size. The upper left square has coordinates (1, 1), and the lower right square has coordinates of (n, m).
Output
On the first line output integer k — the minimum number of teleporters. Then output k lines each containing 4 integers x1 y1 x2 y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — the coordinates of the square where the teleporter is installed (x1, y1), and the coordinates of the square where the teleporter leads (x2, y2).
Then print nm + 1 lines containing 2 numbers each — the coordinates of the squares in the order in which they are visited by the king. The travel path must start and end at (1, 1). The king can move to side-neighboring squares and to the squares where a teleporter leads. Besides, he also should visit the capital exactly two times and he should visit other squares exactly one time.
Examples
Input
2 2
Output
0
1 1
1 2
2 2
2 1
1 1
Input
3 3
Output
1
3 3 1 1
1 1
1 2
1 3
2 3
2 2
2 1
3 1
3 2
3 3
1 1
Submitted Solution:
```
n, m = map(int,input().split())
if n == 1 or m == 1:
print(1)
print(n,m,1,1)
for i in range(1,n+1):
for j in range(1,m+1):
print(i,j)
elif (n * m)%2 == 0:
print(0)
row = 1
column = 1
print(row,column)
while row <= n:
if row % 2 == 1:
column = 2
while column <= m:
print(row,column)
column += 1
else:
column = m
while column >= 2:
print(row,column)
column -= 1
row += 1
for x in range(n,1,-1):
print(x,1)
else:
print(1)
print(n,m,1,1)
row = 1
column = 1
while row <= n:
if row % 2 == 1:
column = 1
while column <= m:
print(row, column)
column += 1
else:
column = m
while column >= 1:
print(row, column)
column -= 1
row += 1
print(1,1)
``` | instruction | 0 | 39,961 | 1 | 79,922 |
No | output | 1 | 39,961 | 1 | 79,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The territory of Berland is represented by a rectangular field n × m in size. The king of Berland lives in the capital, located on the upper left square (1, 1). The lower right square has coordinates (n, m). One day the king decided to travel through the whole country and return back to the capital, having visited every square (except the capital) exactly one time. The king must visit the capital exactly two times, at the very beginning and at the very end of his journey. The king can only move to the side-neighboring squares. However, the royal advise said that the King possibly will not be able to do it. But there is a way out — one can build the system of one way teleporters between some squares so that the king could fulfill his plan. No more than one teleporter can be installed on one square, every teleporter can be used any number of times, however every time it is used, it transports to the same given for any single teleporter square. When the king reaches a square with an installed teleporter he chooses himself whether he is or is not going to use the teleport. What minimum number of teleporters should be installed for the king to complete the journey? You should also compose the journey path route for the king.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100, 2 ≤ n · m) — the field size. The upper left square has coordinates (1, 1), and the lower right square has coordinates of (n, m).
Output
On the first line output integer k — the minimum number of teleporters. Then output k lines each containing 4 integers x1 y1 x2 y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — the coordinates of the square where the teleporter is installed (x1, y1), and the coordinates of the square where the teleporter leads (x2, y2).
Then print nm + 1 lines containing 2 numbers each — the coordinates of the squares in the order in which they are visited by the king. The travel path must start and end at (1, 1). The king can move to side-neighboring squares and to the squares where a teleporter leads. Besides, he also should visit the capital exactly two times and he should visit other squares exactly one time.
Examples
Input
2 2
Output
0
1 1
1 2
2 2
2 1
1 1
Input
3 3
Output
1
3 3 1 1
1 1
1 2
1 3
2 3
2 2
2 1
3 1
3 2
3 3
1 1
Submitted Solution:
```
#!/usr/bin/env python3
from sys import stdin
input_tokens = stdin.readline()
tokens = [int(x) for x in input_tokens.split()]
assert len(tokens) == 2
n, m = tokens[0], tokens[1]
# if they are not both odd
if not (n % 2 != 0 and m % 2 != 0):
print(0)
# 1 2 3 4
# 6 1 0 5
# 5 2 9 6
# 4 3 8 7
# 1 2 3
# 2 1 4
# 9 0 5
# 8 7 6
# 1 2 3 4
# 2 9 8 5
# 1 0 7 6
if(m % 2 == 0):
for i in range(1,m):
print(1,i)
for i in range(1,n+1):
print(i,m)
for i in range(m-1,0,-1):
for j in range(n,1,-1):
if(i%2==1):
print(j,i)
else:
print(n-j+2,i)
else:
for i in range(1,m):
print(1,i)
for i in range(1,n+1):
print(i,m)
for i in range(n,1,-1):
for j in range(m-1,0,-1):
if(i%2==0):
print(i,j)
else:
print(i,m-j)
else:
print(1)
# 1 2 3
# 6 5 4
# 7 8 9
print(n,m,1,1)
for i in range(1,n+1):
for j in range(1,m+1):
if(i%2==1):
print(i,j)
else:
print(i,m-j+1)
print(1,1)
``` | instruction | 0 | 39,962 | 1 | 79,924 |
No | output | 1 | 39,962 | 1 | 79,925 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The territory of Berland is represented by a rectangular field n × m in size. The king of Berland lives in the capital, located on the upper left square (1, 1). The lower right square has coordinates (n, m). One day the king decided to travel through the whole country and return back to the capital, having visited every square (except the capital) exactly one time. The king must visit the capital exactly two times, at the very beginning and at the very end of his journey. The king can only move to the side-neighboring squares. However, the royal advise said that the King possibly will not be able to do it. But there is a way out — one can build the system of one way teleporters between some squares so that the king could fulfill his plan. No more than one teleporter can be installed on one square, every teleporter can be used any number of times, however every time it is used, it transports to the same given for any single teleporter square. When the king reaches a square with an installed teleporter he chooses himself whether he is or is not going to use the teleport. What minimum number of teleporters should be installed for the king to complete the journey? You should also compose the journey path route for the king.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100, 2 ≤ n · m) — the field size. The upper left square has coordinates (1, 1), and the lower right square has coordinates of (n, m).
Output
On the first line output integer k — the minimum number of teleporters. Then output k lines each containing 4 integers x1 y1 x2 y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — the coordinates of the square where the teleporter is installed (x1, y1), and the coordinates of the square where the teleporter leads (x2, y2).
Then print nm + 1 lines containing 2 numbers each — the coordinates of the squares in the order in which they are visited by the king. The travel path must start and end at (1, 1). The king can move to side-neighboring squares and to the squares where a teleporter leads. Besides, he also should visit the capital exactly two times and he should visit other squares exactly one time.
Examples
Input
2 2
Output
0
1 1
1 2
2 2
2 1
1 1
Input
3 3
Output
1
3 3 1 1
1 1
1 2
1 3
2 3
2 2
2 1
3 1
3 2
3 3
1 1
Submitted Solution:
```
def inRows():
print(0)
row = 1
column = 1
print(row, column)
while row <= n:
if row % 2 == 1:
column = 2
while column <= m:
print(row, column)
column += 1
else:
column = m
while column >= 2:
print(row, column)
column -= 1
row += 1
for x in range(n, 1, -1):
print(x, 1)
def inCol():
print(0)
row = 1
column = 1
print(row, column)
while column <= m:
if column % 2 == 1:
row = 2
while row <= n:
print(row, column)
row += 1
else:
row = n
while row >= 2:
print(row, column)
row -= 1
column += 1
for x in range(m, 1, -1):
print(1, x)
n, m = map(int,input().split())
if n == 1 or m == 1:
print(1)
print(n,m,1,1)
for i in range(1,n+1):
for j in range(1,m+1):
print(i,j)
elif (n * m)%2 == 0:
if n % 2 == 1:
inCol()
else:
inRows()
else:
print(1)
print(n,m,1,1)
row = 1
column = 1
while row <= n:
if row % 2 == 1:
column = 1
while column <= m:
print(row, column)
column += 1
else:
column = m
while column >= 1:
print(row, column)
column -= 1
row += 1
print(1,1)
``` | instruction | 0 | 39,963 | 1 | 79,926 |
No | output | 1 | 39,963 | 1 | 79,927 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The territory of Berland is represented by a rectangular field n × m in size. The king of Berland lives in the capital, located on the upper left square (1, 1). The lower right square has coordinates (n, m). One day the king decided to travel through the whole country and return back to the capital, having visited every square (except the capital) exactly one time. The king must visit the capital exactly two times, at the very beginning and at the very end of his journey. The king can only move to the side-neighboring squares. However, the royal advise said that the King possibly will not be able to do it. But there is a way out — one can build the system of one way teleporters between some squares so that the king could fulfill his plan. No more than one teleporter can be installed on one square, every teleporter can be used any number of times, however every time it is used, it transports to the same given for any single teleporter square. When the king reaches a square with an installed teleporter he chooses himself whether he is or is not going to use the teleport. What minimum number of teleporters should be installed for the king to complete the journey? You should also compose the journey path route for the king.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100, 2 ≤ n · m) — the field size. The upper left square has coordinates (1, 1), and the lower right square has coordinates of (n, m).
Output
On the first line output integer k — the minimum number of teleporters. Then output k lines each containing 4 integers x1 y1 x2 y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — the coordinates of the square where the teleporter is installed (x1, y1), and the coordinates of the square where the teleporter leads (x2, y2).
Then print nm + 1 lines containing 2 numbers each — the coordinates of the squares in the order in which they are visited by the king. The travel path must start and end at (1, 1). The king can move to side-neighboring squares and to the squares where a teleporter leads. Besides, he also should visit the capital exactly two times and he should visit other squares exactly one time.
Examples
Input
2 2
Output
0
1 1
1 2
2 2
2 1
1 1
Input
3 3
Output
1
3 3 1 1
1 1
1 2
1 3
2 3
2 2
2 1
3 1
3 2
3 3
1 1
Submitted Solution:
```
def p(a, b, swap):
if swap:
a, b = b, a
print(a, b)
n, m = [int(x) for x in input().split()]
if n * m % 2 == 0:
print(0)
print(1, 1)
swap = m % 2 != 0
if swap:
m, n = n, m
for c in range(m):
for r in range(n - 1):
if c % 2 == 0:
p(r + 2, c + 1, swap)
else:
p(n - r, c + 1, swap)
for c in range(m):
p(1, m - c, swap)
else:
print(1)
print(n, m, 1, 1)
for c in range(m):
for r in range(n):
if c % 2 == 0:
print(r + 1, c + 1)
else:
print(n - r, c + 1)
print(1, 1)
``` | instruction | 0 | 39,964 | 1 | 79,928 |
No | output | 1 | 39,964 | 1 | 79,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur. | instruction | 0 | 40,094 | 1 | 80,188 |
Tags: implementation
Correct Solution:
```
from sys import *
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
l3=list(map(int,input().split()))
l4=list(map(int,input().split()))
if l1[-1]:
if sum(l1[:-1]) or l4[2] or l2[0] or l3[1]: print("YES"); exit()
if l2[-1]:
if sum(l2[:-1]) or l1[2] or l4[1] or l3[0]: print("YES"); exit()
if l3[-1]:
if sum(l3[:-1]) or l2[2] or l1[1] or l4[0]: print("YES"); exit()
if l4[-1]:
if sum(l4[:-1]) or l3[2] or l2[1] or l1[0]: print("YES"); exit()
print("NO")
``` | output | 1 | 40,094 | 1 | 80,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur. | instruction | 0 | 40,095 | 1 | 80,190 |
Tags: implementation
Correct Solution:
```
arr = [list(map(int, input().split())) for _ in range(4)]
flag = False
for i in range(4):
f = False
if arr[i][3] == 1:
s = sum([arr[i][j] for j in range(3)]) + arr[(i+2)%4][1] + arr[(i+1)%4][0] + arr[(i+3)%4][2]
if s > 0:
f = True
if f:
flag = True
break
if flag:
print("YES")
else:
print("NO")
``` | output | 1 | 40,095 | 1 | 80,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur. | instruction | 0 | 40,096 | 1 | 80,192 |
Tags: implementation
Correct Solution:
```
a = []
for i in range(4):
a.append(list(map(int, input().split())))
state = 0
for i in range(4):
if a[i][3] == 1:
if a[i-1][2] or a[i-2][1] or a[i-3][0]:
state = 1
if a[i][2] or a[i][1] or a[i][0]:
state = 1
if state:
print("YES")
else:
print("NO")
``` | output | 1 | 40,096 | 1 | 80,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur. | instruction | 0 | 40,097 | 1 | 80,194 |
Tags: implementation
Correct Solution:
```
conf = [[int(num) for num in input().split()] for i in range(4)]
bad = False
for ind in range(4):
bad |= conf[ind][3] and (1 in conf[ind][:3] or
conf[(ind + 1) % 4][0] or
conf[(ind + 2) % 4][1] or
conf[(ind + 3) % 4][2])
print('YES' if bad else 'NO')
``` | output | 1 | 40,097 | 1 | 80,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur. | instruction | 0 | 40,098 | 1 | 80,196 |
Tags: implementation
Correct Solution:
```
a=[int(i)for i in input().split()]
b=[int(i)for i in input().split()]
c=[int(i)for i in input().split()]
d=[int(i)for i in input().split()]
ans=False
if a[3]:ans = c[1]or b[0] or d[2] or a[0] or a[1] or a[2]
if not ans and b[3]: ans = d[1] or c[0] or a[2] or b[0] or b[1] or b[2]
if not ans and c[3]: ans = a[1] or d[0] or b[2] or c[0] or c[1] or c[2]
if not ans and d[3]: ans = b[1] or a[0] or c[2] or d[0] or d[1] or d[2]
print("YES"if ans else"NO")
``` | output | 1 | 40,098 | 1 | 80,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur. | instruction | 0 | 40,099 | 1 | 80,198 |
Tags: implementation
Correct Solution:
```
import sys
A = []
for i in range(4):
s = list(map(int, input().split()))
A.append(s)
B = [A[i][3] for i in range(4)]
for i in range(4):
for j in range(3):
if A[i][j] == 1:
if j == 0:
if B[i] or B[(i - 1)%4]:
print("YES")
sys.exit()
elif j == 2:
if B[i] or B[(i + 1)%4]:
print("YES")
sys.exit()
else:
if i < 2:
if B[i] or B[i + 2]:
print("YES")
sys.exit()
else:
if B[i] or B[i - 2]:
print("YES")
sys.exit()
print("NO")
``` | output | 1 | 40,099 | 1 | 80,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur. | instruction | 0 | 40,100 | 1 | 80,200 |
Tags: implementation
Correct Solution:
```
a = []
n = 4
for i in range(4):
temp = list(map(int,input().split()))
a.append(temp)
cnt = True
if a[0][3] == 1:
if a[0][0] == 1 or a[0][1] == 1 or a[0][2] == 1:
cnt = False
elif a[1][0] == 1 or a[2][1] == 1 or a[3][2] == 1:
cnt = False
if a[1][3] == 1:
if a[1][0] == 1 or a[1][1] == 1 or a[1][2] == 1:
cnt = False
elif a[0][2] == 1 or a[2][0] == 1 or a[3][1] == 1:
cnt = False
if a[2][3] == 1:
if a[2][0] == 1 or a[2][1] == 1 or a[2][2] == 1:
cnt = False
elif a[0][1] == 1 or a[1][2] == 1 or a[3][0] == 1:
cnt = False
if a[3][3] == 1:
if a[3][0] == 1 or a[3][1] == 1 or a[3][2] == 1:
cnt = False
elif a[0][0] == 1 or a[1][1] == 1 or a[2][2] == 1:
cnt = False
if cnt == False:
print("YES")
else:
print("NO")
``` | output | 1 | 40,100 | 1 | 80,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur. | instruction | 0 | 40,101 | 1 | 80,202 |
Tags: implementation
Correct Solution:
```
l = []
for i in range(4):
l.append(list(map(int, input().split())))
flag = False
for i in range(4):
if l[i][3] == 1:
if l[i][0] == 1 or l[i][1] == 1 or l[i][2] == 1 or l[(i+1)%4][0] == 1 or l[(i+2)%4][1] == 1 or l[(i+3)%4][2] == 1:
print("YES")
flag = True
break
if not flag:
print("NO")
``` | output | 1 | 40,101 | 1 | 80,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
Submitted Solution:
```
#Winners never quit, quiters never win............................................................................
from collections import deque as de
import math
from collections import Counter as cnt
from functools import reduce
from typing import MutableMapping
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
class My_stack():
def __init__(self):
self.data = []
def my_push(self, x):
return (self.data.append(x))
def my_pop(self):
return (self.data.pop())
def my_peak(self):
return (self.data[-1])
def my_contains(self, x):
return (self.data.count(x))
def my_show_all(self):
return (self.data)
def isEmpty(self):
return len(self.data)==0
arrStack = My_stack()
def decimalToBinary(n):
return bin(n).replace("0b", "")
def isPrime(n) :
if (n <= 1) :
return False
if (n <= 3) :
return True
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def get_prime_factors(number):
prime_factors = []
while number % 2 == 0:
prime_factors.append(2)
number = number / 2
for i in range(3, int(math.sqrt(number)) + 1, 2):
while number % i == 0:
prime_factors.append(int(i))
number = number / i
if number > 2:
prime_factors.append(int(number))
return prime_factors
def get_frequency(list):
dic={}
for ele in list:
if ele in dic:
dic[ele] += 1
else:
dic[ele] = 1
return dic
def Log2(x):
return (math.log10(x) /
math.log10(2));
def isPowerOfTwo(n):
return (math.ceil(Log2(n)) == math.floor(Log2(n)));
#here we go......................
#winners never quit, quitters never win
n=4
dic={}
while n:
n-=1
l=list(map(int,input().split()))
dic[4-n]=l
#print(dic)
ch=1
if dic[1][3]==1:
kk=dic[1]
if kk.count(1)!=1:
ch=0
if dic[3][1]==1:
ch=0
if dic[4][2]==1: #left
ch=0
if dic[2][0]==1:
ch=0
if ch:
if dic[2][3]==1:
kk=dic[2]
if kk.count(1)!=1:
ch=0
if dic[4][1]==1:
ch=0
if dic[1][2]==1:
ch=0
if dic[3][0]==1:
ch=0
if ch:
if dic[3][3]==1:
kk=dic[3]
if kk.count(1)!=1:
ch=0
if dic[1][1]==1:
ch=0
if dic[2][2]==1:
ch=0
if dic[4][0]==1:
ch=0
if ch:
if dic[4][3]==1:
kk=dic[4]
if kk.count(1)!=1:
ch=0
if dic[2][1]==1:
ch=0
if dic[3][2]==1:
ch=0
if dic[1][0]==1:
ch=0
if ch:
print("NO")
else:
print("YES")
else:
print("YES")
else:
print("YES")
else:
print("YES")
``` | instruction | 0 | 40,102 | 1 | 80,204 |
Yes | output | 1 | 40,102 | 1 | 80,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
Submitted Solution:
```
sz = 4
r = []
for i in range(sz):
r.append(input().split(' '))
for i in range(sz):
for j in range(sz):
r[i][j] = int(r[i][j])
ans = True
for i in range(sz):
if (r[i][0] == 1 or r[i][1] == 1 or r[i][2] == 1) and r[i][3] == 1 : ans = False
if r[i][0] == 1 and r[(i+3)%4][3] == 1 : ans = False
if r[i][1] == 1 and r[(i+2)%4][3] == 1 : ans = False
if r[i][2] == 1 and r[(i+1)%4][3] == 1 : ans = False
if ans : print("NO")
else : print("YES")
``` | instruction | 0 | 40,103 | 1 | 80,206 |
Yes | output | 1 | 40,103 | 1 | 80,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
Submitted Solution:
```
def editor():
import sys
sys.stdin=open("input.txt",'r')
sys.stdout=open("output.txt",'w')
def solve():
m=[[int(i) for i in input().split()] for j in range(4)]
acc=False
for i in range(4):
if m[i][3]==1:
for j in range(3):
if m[i][j]==1: acc=True
if m[(i+2)%4][1]==1 or m[(i+3)%4][2]==1 or m[(i+1)%4][0]==1:acc=True
if acc is True: print("YES")
else: print("NO")
if __name__ == "__main__":
# editor()
solve()
``` | instruction | 0 | 40,104 | 1 | 80,208 |
Yes | output | 1 | 40,104 | 1 | 80,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
Submitted Solution:
```
# left, strait, right, pedastrian
# 3
#4 2
# 1
def check_if_safe(curr_part, left_part, right_part, opposite_part):
curr_left, curr_straight, curr_right, curr_pedastrian = curr_part
if curr_pedastrian == 0:
return True
elif (curr_left + curr_right + curr_straight) > 0:
return False
left_left, left_straight, left_right, left_pedastrian = left_part
if left_right == 1:
return False
right_left, right_straight, right_right, right_pedastrian = right_part
if right_left == 1:
return False
opposite_left, opposite_straight, opposite_right, opposite_pedastrian = opposite_part
if opposite_straight == 1:
return False
return True
part_one = [int(x) for x in input().split()]
part_two = [int(x) for x in input().split()]
part_three = [int(x) for x in input().split()]
part_four = [int(x) for x in input().split()]
ans = ( check_if_safe(part_one, part_four, part_two, part_three)
+ check_if_safe(part_two, part_one, part_three, part_four)
+ check_if_safe(part_three, part_two, part_four, part_one)
+ check_if_safe(part_four, part_three, part_one, part_two)
)
if ans != 4:
print('YES')
else:
print('NO')
``` | instruction | 0 | 40,105 | 1 | 80,210 |
Yes | output | 1 | 40,105 | 1 | 80,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
Submitted Solution:
```
counter = 0
for i in range(4):
l,s,r,p = input().split()
l = int(l)
s = int(s)
r = int(r)
p = int(p)
if(p == 1):
if(r == 1 or s == 1 or l == 1):
counter = 1
else:
continue
if(counter == 1):
print("YES")
else:
print("NO")
``` | instruction | 0 | 40,106 | 1 | 80,212 |
No | output | 1 | 40,106 | 1 | 80,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
Submitted Solution:
```
for _ in range(4):
l1, s1, r1, p1 = map(int, input().split())
if (l1 or s1 or r1) and p1:
print('YES')
exit(0)
print('NO')
``` | instruction | 0 | 40,107 | 1 | 80,214 |
No | output | 1 | 40,107 | 1 | 80,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
Submitted Solution:
```
import sys
firstline = True
secondline = True
thirdline = True
fourthline = True
numlines = 4
for line in sys.stdin:
if firstline:
line = line[:7]
list1 = line.split(" ")
numlines -= 1
firstline = False
continue
if secondline:
line = line[:7]
list2 = line.split(" ")
numlines -= 1
secondline = False
continue
if thirdline:
line = line[:7]
list3 = line.split(" ")
numlines -= 1
thirdline = False
continue
if fourthline:
line = line[:7]
list4 = line.split(" ")
numlines -= 1
fourthline = False
continue
if(numlines == 0):
break
if(list1[0] == list4[3] == 1 or list1[1] == list3[3] == 1 or list1[2] == list2[3] == 1 or list1[0] == list1[3] == 1):
print("YES")
elif(list2[0] == list1[3] == 1 or list2[1] == list4[3] == 1 or list2[2] == list3[3] == 1 or list2[0] == list2[3] == 1):
print("YES")
elif(list3[0] == list2[3] == 1 or list3[1] == list1[3] == 1 or list3[2] == list4[3] == 1 or list3[0] == list3[3] == 1):
print("YES")
elif(list4[0] == list3[3] == 1 or list4[1] == list2[3] == 1 or list4[2] == list1[3] == 1 or list4[0] == list4[3] == 1):
print("YES")
else:
print("NO")
``` | instruction | 0 | 40,108 | 1 | 80,216 |
No | output | 1 | 40,108 | 1 | 80,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (l — left, s — straight, r — right) and a light p for a pedestrian crossing.
<image>
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers l, s, r, p — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Examples
Input
1 0 0 1
0 1 0 0
0 0 1 0
0 0 0 1
Output
YES
Input
0 1 1 0
1 0 1 0
1 1 0 0
0 0 0 1
Output
NO
Input
1 0 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Output
NO
Note
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
Submitted Solution:
```
from sys import *
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
l3=list(map(int,input().split()))
l4=list(map(int,input().split()))
if l1[-1]:
if sum(l1[:-1]) or l4[2] or l2[0] or l3[1]: print("YES"); exit()
if l2[-1]:
if sum(l2[:-1]) or l1[2] or l4[1] or l3[0]: print("YES"); exit()
if l3[-1]:
if sum(l3[:-1]) or l2[2] or l1[1] or l2[0]: print("YES"); exit()
if l4[-1]:
if sum(l4[:-1]) or l3[2] or l2[1] or l1[0]: print("YES"); exit()
print("NO")
``` | instruction | 0 | 40,109 | 1 | 80,218 |
No | output | 1 | 40,109 | 1 | 80,219 |
Provide a correct Python 3 solution for this coding contest problem.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5 | instruction | 0 | 40,247 | 1 | 80,494 |
"Correct Solution:
```
n,t=map(int,input().split())
ans=chk=10000
for i in range(n):
x,y=map(int,input().split())
if y<=t: ans=min(ans,x)
print(ans if ans<10000 else "TLE")
``` | output | 1 | 40,247 | 1 | 80,495 |
Provide a correct Python 3 solution for this coding contest problem.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5 | instruction | 0 | 40,248 | 1 | 80,496 |
"Correct Solution:
```
n,t=map(int, input().split())
m=10**9
ans=m
for _ in range(n):
ci,ti=map(int, input().split())
if ti<=t:
ans=min(ans,ci)
print(ans if ans<m else "TLE")
``` | output | 1 | 40,248 | 1 | 80,497 |
Provide a correct Python 3 solution for this coding contest problem.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5 | instruction | 0 | 40,249 | 1 | 80,498 |
"Correct Solution:
```
N,T=map(int,input().split())
ans=2000
for i in range(N):
c,t=map(int,input().split())
if t<=T:
ans=min(ans,c)
print(ans if ans!=2000 else "TLE")
``` | output | 1 | 40,249 | 1 | 80,499 |
Provide a correct Python 3 solution for this coding contest problem.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5 | instruction | 0 | 40,250 | 1 | 80,500 |
"Correct Solution:
```
N,T=map(int, input().split())
R=[999999999]
for i in range(N):
c,t=map(int, input().split())
if T>=t:
R.append(c)
print('TLE' if len(R)==1 else min(R))
``` | output | 1 | 40,250 | 1 | 80,501 |
Provide a correct Python 3 solution for this coding contest problem.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5 | instruction | 0 | 40,251 | 1 | 80,502 |
"Correct Solution:
```
N,T=map(int,input().split())
ctl=[list(map(int,input().split())) for _ in range(N)]
cl =[ct[0] for ct in ctl if ct[1]<=T]
print(min(cl) if len(cl)>0 else 'TLE')
``` | output | 1 | 40,251 | 1 | 80,503 |
Provide a correct Python 3 solution for this coding contest problem.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5 | instruction | 0 | 40,252 | 1 | 80,504 |
"Correct Solution:
```
n, t = map(int,input().split())
ct = [list(map(int,input().split())) for _ in range(n)]
for c, ti in sorted(ct):
if int(ti) <= t:
print(c)
exit()
print("TLE")
``` | output | 1 | 40,252 | 1 | 80,505 |
Provide a correct Python 3 solution for this coding contest problem.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5 | instruction | 0 | 40,253 | 1 | 80,506 |
"Correct Solution:
```
N,T=map(int,input().split())
lst=[]
for i in range(N):
c,t=map(int,input().split())
if t<=T:
lst.append(c)
print("TLE" if len(lst)==0 else min(lst))
``` | output | 1 | 40,253 | 1 | 80,507 |
Provide a correct Python 3 solution for this coding contest problem.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5 | instruction | 0 | 40,254 | 1 | 80,508 |
"Correct Solution:
```
N, T = map(int, input().split())
mn = 1001
for i in range(N):
c,t = map(int, input().split())
if t <= T:
mn = min(c,mn)
print(mn if mn < 1001 else 'TLE')
``` | output | 1 | 40,254 | 1 | 80,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5
Submitted Solution:
```
n,T=map(int,input().split())
c=2000
for _ in range(n):
a,b=map(int,input().split())
if b<=T and a<c:c=a
print(c if c<2000 else "TLE")
``` | instruction | 0 | 40,255 | 1 | 80,510 |
Yes | output | 1 | 40,255 | 1 | 80,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5
Submitted Solution:
```
N,T=map(int,input().split())
ans=10**10
for i in range(N):
c,t=map(int,input().split())
if t<=T:
ans=min(ans,c)
if ans==10**10:
print('TLE')
else:
print(ans)
``` | instruction | 0 | 40,256 | 1 | 80,512 |
Yes | output | 1 | 40,256 | 1 | 80,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5
Submitted Solution:
```
n,tmax=map(int,input().split())
m=1001
for i in range(n):
c,t=map(int,input().split())
if t<= tmax:
m = min(m,c)
print("TLE" if m == 1001 else m)
``` | instruction | 0 | 40,257 | 1 | 80,514 |
Yes | output | 1 | 40,257 | 1 | 80,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5
Submitted Solution:
```
N,T = map(int,input().split())
L = list(tuple(map(int,input().split())) for _ in range(N))
print(min((c for c,t in L if t <= T), default='TLE'))
``` | instruction | 0 | 40,258 | 1 | 80,516 |
Yes | output | 1 | 40,258 | 1 | 80,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5
Submitted Solution:
```
a,b = map(int, input().split())
cl=[]
tl=[]
for i in range(a):
c,t = map(int, input().split())
cl.append(c)
tl.append(t)
ind=0
resut_list=[]
for p in range(a):
if tl[ind]<=b:
resut_list.append(cl[ind])
ind+=1
print(min(resut_list))
``` | instruction | 0 | 40,259 | 1 | 80,518 |
No | output | 1 | 40,259 | 1 | 80,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5
Submitted Solution:
```
N, T = map(int, input().split())
c = [0 for i in range(N)]
t = [0 for i in range(N)]
for i in range(N):
c[i], t[i] = map(int, input().split())
for i in range(N):
if t[i] > T:
del c[i]
c.sort()
print(c[0])
``` | instruction | 0 | 40,260 | 1 | 80,520 |
No | output | 1 | 40,260 | 1 | 80,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5
Submitted Solution:
```
N,T = map(int,input().split())
comp = []
for i in range(N):
nt = [int(i) for i in input().split()]
comp.append(nt)
cost = []
for i in range(N):
if comp[i][1] <= T:
temp = [comp[i][0],comp[i][1]]
cost.append(temp)
cost = sorted(cost, key = lambda x:x[1])
if cost:
print(cost[0][0])
else:
print("TLE")
``` | instruction | 0 | 40,261 | 1 | 80,522 |
No | output | 1 | 40,261 | 1 | 80,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.
You, the smartwatch, has found N routes to his home.
If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.
Find the smallest cost of a route that takes not longer than time T.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq T \leq 1000
* 1 \leq c_i \leq 1000
* 1 \leq t_i \leq 1000
* The pairs (c_i, t_i) are distinct.
Input
Input is given from Standard Input in the following format:
N T
c_1 t_1
c_2 t_2
:
c_N t_N
Output
Print the smallest cost of a route that takes not longer than time T.
If there is no route that takes not longer than time T, print `TLE` instead.
Examples
Input
3 70
7 60
1 80
4 50
Output
4
Input
4 3
1 1000
2 4
3 1000
4 500
Output
TLE
Input
5 9
25 8
5 9
4 10
1000 1000
6 1
Output
5
Submitted Solution:
```
a,b,k=map(int, input().split())
cnt=0
for i in range(min(a,b)+1,0,-1):
if a%i==0 and b%i==0:
cnt+=1
if cnt==k:
print(i)
``` | instruction | 0 | 40,262 | 1 | 80,524 |
No | output | 1 | 40,262 | 1 | 80,525 |
Provide a correct Python 3 solution for this coding contest problem.
E: Jam
problem
There are N cities in a country, numbered 1, \ 2, \ ..., \ N. These cities are connected in both directions by M roads, and the i-th road allows you to travel between the cities u_i and v_i in time t_i. Also, any two cities can be reached by using several roads.
Bread is sold in each town, and the deliciousness of bread sold in town i is P_i.
In addition, there are K flavors of jam in this country, and in town i you can buy delicious J_i jams at taste c_i.
Homu-chan, who lives in Town 1, decided to go out to buy bread and jam one by one. Homu-chan travels through several cities, buys bread and jam, and returns to city 1. More precisely, go to city u to buy bread, select city v to buy jam, and return to city 1. At this time, u = v, u = 1, and v = 1 can be used.
The happiness of Homu-chan after shopping is "the total of the deliciousness of the bread and jam I bought"-"the time it took to move". For each of the K types of jam, calculate the maximum possible happiness when going to buy jam and bread of that taste.
Input format
NMK
P_1 P_2 ... P_N
c_1 c_2 ... c_N
J_1 J_2 ... J_N
u_1 v_1 t_1
u_2 v_2 t_2
::
u_M v_M t_M
Constraint
* 1 \ leq N \ leq 10 ^ 5
* 1 \ leq M \ leq 2 \ times 10 ^ 5
* 1 \ leq K \ leq N
* 1 \ leq P_i, J_i \ leq 10 ^ 9
* 1 \ leq c_i \ leq K
* 1 \ leq u_i, v_i \ leq N
* 1 \ leq t_i \ leq 10 ^ 9
* For all i \ (1 \ leq i \ leq K), there exists j \ (1 \ leq j \ leq N) such that c_j = i.
* Given graphs are concatenated graphs that do not include multiple edges or self-loops.
* All inputs are given as integers
Output format
Please output K line. On line i, print the maximum happiness when you go to buy bread and taste i jam.
Input example 1
4 4 3
3 6 1 6
1 1 2 3
6 1 5 5
1 2 1
2 3 1
1 3 1
1 4 1
Output example 1
Ten
8
9
* When buying taste 1 jam
* If you buy jam in city 1, move to city 2, buy bread, and come back to city 1, your happiness will be 6 + 6-2 = 10, which is optimal.
* When buying taste 2 jam
* If you move to City 2 to buy bread, move to City 3 to buy jam, and return to City 1, your happiness will be 6 + 5-3 = 8.
* When buying taste 3 jam
* If you go to City 4 and buy both bread and jam and come back to City 1, your happiness will be 6 + 5-2 = 9.
Input example 2
2 1 2
1 1
1 2
1 1
1 2 1000000000
Output example 2
2
-1999999998
* Homu-chan seems to be dissatisfied because the distance is too far compared to the deliciousness of bread and jam.
Input example 3
6 8 3
31 41 59 26 53 58
1 2 3 1 2 3
27 18 28 18 28 46
one two Three
2 3 7
2 4 8
3 6 9
3 5 3
4 5 3
1 5 15
4 6 7
Output example 3
66
61
68
Example
Input
4 4 3
3 6 1 6
1 1 2 3
6 1 5 5
1 2 1
2 3 1
1 3 1
1 4 1
Output
10
8
9 | instruction | 0 | 40,402 | 1 | 80,804 |
"Correct Solution:
```
import sys
from heapq import heappop as hpp, heappush as hp
def dijkstra(N, s, Edge):
inf = 10**18
dist = [inf] * N
Q = [(0, s)]
decided = set()
for _ in range(N):
while True:
dn, vn = hpp(Q)
if vn not in decided:
decided.add(vn)
dist[vn] = dn
break
for vf, df in Edge[vn]:
if vf not in decided:
hp(Q, (dn + df, vf))
return dist
N, M ,K = map(int, input().split())
P = list(map(int, input().split()))
C = list(map(int, input().split()))
J = list(map(int, input().split()))
Edge1 = [[] for _ in range(N)]
L = []
for _ in range(M):
u, v, t = map(int, sys.stdin.readline().split())
u -= 1
v -= 1
Edge1[u].append((v, t))
Edge1[v].append((u, t))
L.append((u, v, t))
dist = dijkstra(N, 0, Edge1)
Edge2 = [[] for _ in range(3*N)]
for u, v, t in L:
Edge2[u].append((v, t))
Edge2[v].append((u, t))
Edge2[N+u].append((N+v, t))
Edge2[N+v].append((N+u, t))
inf = 10**17
for i in range(N):
Edge2[i].append((N+i, inf - P[i]))
Edge2[N+i].append((2*N+i, inf - J[i] + dist[i]))
dist2 = dijkstra(3*N, 0, Edge2)
Ans = [-10**20]*K
for i in range(N):
Ans[C[i]-1] = max(Ans[C[i]-1], 2*inf - dist2[2*N+i])
for a in Ans:
print(a)
``` | output | 1 | 40,402 | 1 | 80,805 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Codefortia is a small island country located somewhere in the West Pacific. It consists of n settlements connected by m bidirectional gravel roads. Curiously enough, the beliefs of the inhabitants require the time needed to pass each road to be equal either to a or b seconds. It's guaranteed that one can go between any pair of settlements by following a sequence of roads.
Codefortia was recently struck by the financial crisis. Therefore, the king decided to abandon some of the roads so that:
* it will be possible to travel between each pair of cities using the remaining roads only,
* the sum of times required to pass each remaining road will be minimum possible (in other words, remaining roads must form minimum spanning tree, using the time to pass the road as its weight),
* among all the plans minimizing the sum of times above, the time required to travel between the king's residence (in settlement 1) and the parliament house (in settlement p) using the remaining roads only will be minimum possible.
The king, however, forgot where the parliament house was. For each settlement p = 1, 2, ..., n, can you tell what is the minimum time required to travel between the king's residence and the parliament house (located in settlement p) after some roads are abandoned?
Input
The first line of the input contains four integers n, m, a and b (2 ≤ n ≤ 70, n - 1 ≤ m ≤ 200, 1 ≤ a < b ≤ 10^7) — the number of settlements and gravel roads in Codefortia, and two possible travel times. Each of the following lines contains three integers u, v, c (1 ≤ u, v ≤ n, u ≠ v, c ∈ \\{a, b\}) denoting a single gravel road between the settlements u and v, which requires c minutes to travel.
You can assume that the road network is connected and has no loops or multiedges.
Output
Output a single line containing n integers. The p-th of them should denote the minimum possible time required to travel from 1 to p after the selected roads are abandoned. Note that for each p you can abandon a different set of roads.
Examples
Input
5 5 20 25
1 2 25
2 3 25
3 4 20
4 5 20
5 1 20
Output
0 25 60 40 20
Input
6 7 13 22
1 2 13
2 3 13
1 4 22
3 4 13
4 5 13
5 6 13
6 1 13
Output
0 13 26 39 26 13
Note
The minimum possible sum of times required to pass each road in the first example is 85 — exactly one of the roads with passing time 25 must be abandoned. Note that after one of these roads is abandoned, it's now impossible to travel between settlements 1 and 3 in time 50. | instruction | 0 | 40,458 | 1 | 80,916 |
Tags: brute force, dp, graphs, greedy
Correct Solution:
```
import heapq
n,m,a,b=map(int,input().split())
graph={i:[] for i in range(n)}
for i in range(m):
u,v,w=map(int,input().split())
graph[u-1].append((v-1,w))
graph[v-1].append((u-1,w))
components=[-1]*n
comp=-1
for i in range(n):
if components[i]==-1:
comp+=1
components[i]=comp
prev=[]
layer=[i]
while layer!=[]:
newlayer=[]
for guy in layer:
for guy1 in graph[guy]:
if guy1[1]==a and components[guy1[0]]==-1:
newlayer.append(guy1[0])
components[guy1[0]]=comp
prev=layer[:]
layer=newlayer[:]
useless=[]
for guy in graph:
for neigh in graph[guy]:
if components[guy]==components[neigh[0]] and neigh[1]==b:
useless.append((guy,neigh))
for guy in useless:
graph[guy[0]].remove(guy[1])
counts=[0]*(comp+1)
for i in range(n):
counts[components[i]]+=1
bad=[]
for i in range(comp+1):
if counts[i]<=3:
bad.append(i)
for j in range(n):
if components[j]==i:
components[j]=-1
for guy in bad[::-1]:
for i in range(n):
if components[i]>guy:
components[i]-=1
comp-=len(bad)
comp+=1
dists=[[float("inf") for i in range(2**comp)] for j in range(n)]
dists[0][0]=0
pq=[]
heapq.heappush(pq,[0,0,0])
remaining=n
visited=[0]*n
while len(pq)>0 and remaining>0:
dist,vert,mask=heapq.heappop(pq)
if visited[vert]==0:
visited[vert]=1
remaining-=1
for neigh in graph[vert]:
if neigh[1]==b:
if components[vert]==components[neigh[0]] and components[vert]!=-1:
continue
if components[neigh[0]]!=-1:
if mask & (2**components[neigh[0]])>0:
continue
if components[vert]!=-1:
maskn=mask+2**(components[vert])
else:
maskn=mask
else:
maskn=mask
if dist+neigh[1]<dists[neigh[0]][maskn]:
dists[neigh[0]][maskn]=dist+neigh[1]
heapq.heappush(pq,[dist+neigh[1],neigh[0],maskn])
optimal=[str(min(dists[i])) for i in range(n)]
print(" ".join(optimal))
``` | output | 1 | 40,458 | 1 | 80,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Codefortia is a small island country located somewhere in the West Pacific. It consists of n settlements connected by m bidirectional gravel roads. Curiously enough, the beliefs of the inhabitants require the time needed to pass each road to be equal either to a or b seconds. It's guaranteed that one can go between any pair of settlements by following a sequence of roads.
Codefortia was recently struck by the financial crisis. Therefore, the king decided to abandon some of the roads so that:
* it will be possible to travel between each pair of cities using the remaining roads only,
* the sum of times required to pass each remaining road will be minimum possible (in other words, remaining roads must form minimum spanning tree, using the time to pass the road as its weight),
* among all the plans minimizing the sum of times above, the time required to travel between the king's residence (in settlement 1) and the parliament house (in settlement p) using the remaining roads only will be minimum possible.
The king, however, forgot where the parliament house was. For each settlement p = 1, 2, ..., n, can you tell what is the minimum time required to travel between the king's residence and the parliament house (located in settlement p) after some roads are abandoned?
Input
The first line of the input contains four integers n, m, a and b (2 ≤ n ≤ 70, n - 1 ≤ m ≤ 200, 1 ≤ a < b ≤ 10^7) — the number of settlements and gravel roads in Codefortia, and two possible travel times. Each of the following lines contains three integers u, v, c (1 ≤ u, v ≤ n, u ≠ v, c ∈ \\{a, b\}) denoting a single gravel road between the settlements u and v, which requires c minutes to travel.
You can assume that the road network is connected and has no loops or multiedges.
Output
Output a single line containing n integers. The p-th of them should denote the minimum possible time required to travel from 1 to p after the selected roads are abandoned. Note that for each p you can abandon a different set of roads.
Examples
Input
5 5 20 25
1 2 25
2 3 25
3 4 20
4 5 20
5 1 20
Output
0 25 60 40 20
Input
6 7 13 22
1 2 13
2 3 13
1 4 22
3 4 13
4 5 13
5 6 13
6 1 13
Output
0 13 26 39 26 13
Note
The minimum possible sum of times required to pass each road in the first example is 85 — exactly one of the roads with passing time 25 must be abandoned. Note that after one of these roads is abandoned, it's now impossible to travel between settlements 1 and 3 in time 50.
Submitted Solution:
```
print ( "JOPA")
``` | instruction | 0 | 40,459 | 1 | 80,918 |
No | output | 1 | 40,459 | 1 | 80,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Codefortia is a small island country located somewhere in the West Pacific. It consists of n settlements connected by m bidirectional gravel roads. Curiously enough, the beliefs of the inhabitants require the time needed to pass each road to be equal either to a or b seconds. It's guaranteed that one can go between any pair of settlements by following a sequence of roads.
Codefortia was recently struck by the financial crisis. Therefore, the king decided to abandon some of the roads so that:
* it will be possible to travel between each pair of cities using the remaining roads only,
* the sum of times required to pass each remaining road will be minimum possible (in other words, remaining roads must form minimum spanning tree, using the time to pass the road as its weight),
* among all the plans minimizing the sum of times above, the time required to travel between the king's residence (in settlement 1) and the parliament house (in settlement p) using the remaining roads only will be minimum possible.
The king, however, forgot where the parliament house was. For each settlement p = 1, 2, ..., n, can you tell what is the minimum time required to travel between the king's residence and the parliament house (located in settlement p) after some roads are abandoned?
Input
The first line of the input contains four integers n, m, a and b (2 ≤ n ≤ 70, n - 1 ≤ m ≤ 200, 1 ≤ a < b ≤ 10^7) — the number of settlements and gravel roads in Codefortia, and two possible travel times. Each of the following lines contains three integers u, v, c (1 ≤ u, v ≤ n, u ≠ v, c ∈ \\{a, b\}) denoting a single gravel road between the settlements u and v, which requires c minutes to travel.
You can assume that the road network is connected and has no loops or multiedges.
Output
Output a single line containing n integers. The p-th of them should denote the minimum possible time required to travel from 1 to p after the selected roads are abandoned. Note that for each p you can abandon a different set of roads.
Examples
Input
5 5 20 25
1 2 25
2 3 25
3 4 20
4 5 20
5 1 20
Output
0 25 60 40 20
Input
6 7 13 22
1 2 13
2 3 13
1 4 22
3 4 13
4 5 13
5 6 13
6 1 13
Output
0 13 26 39 26 13
Note
The minimum possible sum of times required to pass each road in the first example is 85 — exactly one of the roads with passing time 25 must be abandoned. Note that after one of these roads is abandoned, it's now impossible to travel between settlements 1 and 3 in time 50.
Submitted Solution:
```
n,m,a,b=map(int,input().split())
edges={}
for i in range(n):
edges[i+1]=[]
for i in range(m):
u,v,w=map(int,input().split())
edges[u].append((v,w))
edges[v].append((u,w))
dists={}
for i in range(n):
dists[i+1]=(1000000000000,[])
group={}
group[1]=0
groups=0
dists[1]=(0,[])
reachable={1}
layer=[1]
its=0
while layer!=[]:
its+=1
newlayer=[]
for guy in layer:
for edge in edges[guy]:
v=edge[0]
w=edge[1]
if dists[v][0]==1000000000000 and w==a:
dists[v]=(its*a,[])
group[v]=0
reachable.add(v)
newlayer.append(v)
layer=newlayer
for vert in edges:
for edge in edges[vert]:
if edge[1]==a and vert not in group:
groups+=1
reachable={vert}
group[vert]=groups
layer=[vert]
while layer!=[]:
newlayer=[]
for guy in layer:
for edge in edges[guy]:
v=edge[0]
w=edge[1]
if v not in group and w==a:
group[v]=groups
reachable.add(v)
newlayer.append(v)
layer=newlayer
for i in range(n):
if i+1 not in group:
group[i+1]=-1
def dfs(edges, reachable, dists, path, visited, start, left):
g=group[start]
for edge in edges[start]:
v=edge[0]
w=edge[1]
if group[v] not in left and v not in visited:
if not (g!=-1 and g==group[v] and w==b):
if g!=-1 and g!=group[v]:
newleft=left[:]+[g]
else:
newleft=left[:]
dist=dists[v][0]
if path+w<dist:
dists[v]=(path+w, newleft[:])
dfs(edges, reachable, dists, path+w, visited[:]+[v], v, newleft)
elif not set(dists[v][1]).issubset(set(newleft)):
dfs(edges, reachable, dists, path+w, visited[:]+[v], v, newleft)
dfs(edges, reachable, dists, 0, [], 1, [])
out=""
for i in range(n):
out+=str(dists[i+1][0])+" "
print(out)
``` | instruction | 0 | 40,460 | 1 | 80,920 |
No | output | 1 | 40,460 | 1 | 80,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Codefortia is a small island country located somewhere in the West Pacific. It consists of n settlements connected by m bidirectional gravel roads. Curiously enough, the beliefs of the inhabitants require the time needed to pass each road to be equal either to a or b seconds. It's guaranteed that one can go between any pair of settlements by following a sequence of roads.
Codefortia was recently struck by the financial crisis. Therefore, the king decided to abandon some of the roads so that:
* it will be possible to travel between each pair of cities using the remaining roads only,
* the sum of times required to pass each remaining road will be minimum possible (in other words, remaining roads must form minimum spanning tree, using the time to pass the road as its weight),
* among all the plans minimizing the sum of times above, the time required to travel between the king's residence (in settlement 1) and the parliament house (in settlement p) using the remaining roads only will be minimum possible.
The king, however, forgot where the parliament house was. For each settlement p = 1, 2, ..., n, can you tell what is the minimum time required to travel between the king's residence and the parliament house (located in settlement p) after some roads are abandoned?
Input
The first line of the input contains four integers n, m, a and b (2 ≤ n ≤ 70, n - 1 ≤ m ≤ 200, 1 ≤ a < b ≤ 10^7) — the number of settlements and gravel roads in Codefortia, and two possible travel times. Each of the following lines contains three integers u, v, c (1 ≤ u, v ≤ n, u ≠ v, c ∈ \\{a, b\}) denoting a single gravel road between the settlements u and v, which requires c minutes to travel.
You can assume that the road network is connected and has no loops or multiedges.
Output
Output a single line containing n integers. The p-th of them should denote the minimum possible time required to travel from 1 to p after the selected roads are abandoned. Note that for each p you can abandon a different set of roads.
Examples
Input
5 5 20 25
1 2 25
2 3 25
3 4 20
4 5 20
5 1 20
Output
0 25 60 40 20
Input
6 7 13 22
1 2 13
2 3 13
1 4 22
3 4 13
4 5 13
5 6 13
6 1 13
Output
0 13 26 39 26 13
Note
The minimum possible sum of times required to pass each road in the first example is 85 — exactly one of the roads with passing time 25 must be abandoned. Note that after one of these roads is abandoned, it's now impossible to travel between settlements 1 and 3 in time 50.
Submitted Solution:
```
n,m,a,b=map(int,input().split())
edges={}
for i in range(n):
edges[i+1]=[]
for i in range(m):
u,v,w=map(int,input().split())
edges[u].append((v,w))
edges[v].append((u,w))
dists={}
for i in range(n):
dists[i+1]=(1000000000000,[])
group={}
group[1]=0
groups=0
dists[1]=(0,[])
reachable={1}
layer=[1]
its=0
if n==70:
print(edges)
while layer!=[]:
its+=1
newlayer=[]
for guy in layer:
for edge in edges[guy]:
v=edge[0]
w=edge[1]
if dists[v][0]==1000000000000 and w==a:
dists[v]=(its*a,[])
group[v]=0
reachable.add(v)
newlayer.append(v)
layer=newlayer
for vert in edges:
for edge in edges[vert]:
if edge[1]==a and vert not in group:
groups+=1
reachable={vert}
group[vert]=groups
layer=[vert]
while layer!=[]:
newlayer=[]
for guy in layer:
for edge in edges[guy]:
v=edge[0]
w=edge[1]
if v not in group and w==a:
group[v]=groups
reachable.add(v)
newlayer.append(v)
layer=newlayer
for i in range(n):
if i+1 not in group:
group[i+1]=-1
def dfs(edges, reachable, dists, path, visited, start, left):
g=group[start]
for edge in edges[start]:
v=edge[0]
w=edge[1]
if group[v] not in left and v not in visited:
if not (g!=-1 and g==group[v] and w==b):
if g!=-1 and g!=group[v]:
newleft=left[:]+[g]
else:
newleft=left[:]
dist=dists[v][0]
if path+w<dist:
dists[v]=(path+w, newleft[:])
dfs(edges, reachable, dists, path+w, visited[:]+[v], v, newleft)
elif not set(dists[v][1]).issubset(set(newleft)):
dfs(edges, reachable, dists, path+w, visited[:]+[v], v, newleft)
dfs(edges, reachable, dists, 0, [], 1, [])
out=""
for i in range(n):
out+=str(dists[i+1][0])+" "
print(out)
``` | instruction | 0 | 40,461 | 1 | 80,922 |
No | output | 1 | 40,461 | 1 | 80,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Codefortia is a small island country located somewhere in the West Pacific. It consists of n settlements connected by m bidirectional gravel roads. Curiously enough, the beliefs of the inhabitants require the time needed to pass each road to be equal either to a or b seconds. It's guaranteed that one can go between any pair of settlements by following a sequence of roads.
Codefortia was recently struck by the financial crisis. Therefore, the king decided to abandon some of the roads so that:
* it will be possible to travel between each pair of cities using the remaining roads only,
* the sum of times required to pass each remaining road will be minimum possible (in other words, remaining roads must form minimum spanning tree, using the time to pass the road as its weight),
* among all the plans minimizing the sum of times above, the time required to travel between the king's residence (in settlement 1) and the parliament house (in settlement p) using the remaining roads only will be minimum possible.
The king, however, forgot where the parliament house was. For each settlement p = 1, 2, ..., n, can you tell what is the minimum time required to travel between the king's residence and the parliament house (located in settlement p) after some roads are abandoned?
Input
The first line of the input contains four integers n, m, a and b (2 ≤ n ≤ 70, n - 1 ≤ m ≤ 200, 1 ≤ a < b ≤ 10^7) — the number of settlements and gravel roads in Codefortia, and two possible travel times. Each of the following lines contains three integers u, v, c (1 ≤ u, v ≤ n, u ≠ v, c ∈ \\{a, b\}) denoting a single gravel road between the settlements u and v, which requires c minutes to travel.
You can assume that the road network is connected and has no loops or multiedges.
Output
Output a single line containing n integers. The p-th of them should denote the minimum possible time required to travel from 1 to p after the selected roads are abandoned. Note that for each p you can abandon a different set of roads.
Examples
Input
5 5 20 25
1 2 25
2 3 25
3 4 20
4 5 20
5 1 20
Output
0 25 60 40 20
Input
6 7 13 22
1 2 13
2 3 13
1 4 22
3 4 13
4 5 13
5 6 13
6 1 13
Output
0 13 26 39 26 13
Note
The minimum possible sum of times required to pass each road in the first example is 85 — exactly one of the roads with passing time 25 must be abandoned. Note that after one of these roads is abandoned, it's now impossible to travel between settlements 1 and 3 in time 50.
Submitted Solution:
```
import heapq
n,m,a,b=map(int,input().split())
graph={i:[] for i in range(n)}
for i in range(m):
u,v,w=map(int,input().split())
graph[u-1].append((v-1,w))
graph[v-1].append((u-1,w))
components=[-1]*n
comp=-1
for i in range(n):
if components[i]==-1:
comp+=1
components[i]=comp
prev=[]
layer=[i]
while layer!=[]:
newlayer=[]
for guy in layer:
for guy1 in graph[guy]:
if guy1[1]==a and components[guy1[0]]==-1:
newlayer.append(guy1[0])
components[guy1[0]]=comp
prev=layer[:]
layer=newlayer[:]
useless=[]
for guy in graph:
for neigh in graph[guy]:
if components[guy]==components[neigh[0]] and neigh[1]==b:
useless.append((guy,neigh))
for guy in useless:
graph[guy[0]].remove(guy[1])
counts=[0]*(comp+1)
for i in range(n):
counts[components[i]]+=1
bad=[]
for i in range(comp+1):
if counts[i]<=3:
bad.append(i)
for j in range(n):
if components[j]==i:
components[j]=-1
for guy in bad[::-1]:
for i in range(n):
if components[i]>guy:
components[i]-=1
comp-=len(bad)
comp+=1
dists=[[float("inf") for i in range(2**comp)] for j in range(n)]
dists[0][0]=0
pq=[]
heapq.heappush(pq,[0,0,0])
if n==70:
print("fuck")
exit()
while len(pq)>0:
dist,vert,mask=heapq.heappop(pq)
for neigh in graph[vert]:
if neigh[1]==b:
if components[vert]==components[neigh[0]] and components[vert]!=-1:
continue
if components[neigh[0]]!=-1:
if bin(mask)[-1-components[neigh[0]]]=="1":
continue
if components[vert]!=-1:
maskn=mask+2**(components[vert])
else:
maskn=mask
else:
maskn=mask
if dist+neigh[1]<dists[neigh[0]][maskn]:
dists[neigh[0]][maskn]=dist+neigh[1]
heapq.heappush(pq,[dist+neigh[1],neigh[0],maskn])
optimal=[str(min(dists[i])) for i in range(n)]
print(" ".join(optimal))
``` | instruction | 0 | 40,462 | 1 | 80,924 |
No | output | 1 | 40,462 | 1 | 80,925 |
Provide a correct Python 3 solution for this coding contest problem.
There is a double-track line (up and down are separate lines and pass each other everywhere). There are 11 stations on this line, including the terminal station, and each station is called by the section number shown in the figure.
<image>
Trains depart from both terminal stations on this line at the same time and run without stopping along the way. Create a program that reads the length of each section and the speed of the two trains and outputs the number of the section where the trains pass each other in each case. However, if you pass each other just at the station, the smaller number of the section numbers on both sides will be output. In addition, the length of the train and the length of the station cannot be ignored.
Input
Multiple datasets are given. Each dataset is given in the following format.
l1, l2, l3, l4, l5, l6, l7, l8, l9, l10, v1, v2
li (1 ≤ li ≤ 2,000) is an integer representing the length (km) of the interval i. v1 is the speed of the train departing from the terminal station on the section 1 side (km / h), and v2 is the speed of the train departing from the terminal station on the section 10 side (km / h) (1 ≤ v1, v2) ≤ 2,000).
The number of datasets does not exceed 50.
Output
For each data set, the number of the section where the train passes is output on one line.
Example
Input
1,1,1,1,1,1,1,1,1,1,40,60
1,1,1,1,1,3,3,3,3,3,50,50
10,10,10,10,10,10,10,10,10,10,50,49
Output
4
7
6 | instruction | 0 | 41,103 | 1 | 82,206 |
"Correct Solution:
```
while 1:
try:
tmplist = list(map(int, input().split(",")))
except:
break
lengthlist = tmplist[:10]
v1 = tmplist[10]
v2 = tmplist[11]
lsum = sum(lengthlist)
time = lsum / (v1 + v2)
spot = time * v1
le = 0
for i, l in enumerate(lengthlist):
le += l
if le >= spot:
print(i+1)
break
``` | output | 1 | 41,103 | 1 | 82,207 |
Provide a correct Python 3 solution for this coding contest problem.
There is a double-track line (up and down are separate lines and pass each other everywhere). There are 11 stations on this line, including the terminal station, and each station is called by the section number shown in the figure.
<image>
Trains depart from both terminal stations on this line at the same time and run without stopping along the way. Create a program that reads the length of each section and the speed of the two trains and outputs the number of the section where the trains pass each other in each case. However, if you pass each other just at the station, the smaller number of the section numbers on both sides will be output. In addition, the length of the train and the length of the station cannot be ignored.
Input
Multiple datasets are given. Each dataset is given in the following format.
l1, l2, l3, l4, l5, l6, l7, l8, l9, l10, v1, v2
li (1 ≤ li ≤ 2,000) is an integer representing the length (km) of the interval i. v1 is the speed of the train departing from the terminal station on the section 1 side (km / h), and v2 is the speed of the train departing from the terminal station on the section 10 side (km / h) (1 ≤ v1, v2) ≤ 2,000).
The number of datasets does not exceed 50.
Output
For each data set, the number of the section where the train passes is output on one line.
Example
Input
1,1,1,1,1,1,1,1,1,1,40,60
1,1,1,1,1,3,3,3,3,3,50,50
10,10,10,10,10,10,10,10,10,10,50,49
Output
4
7
6 | instruction | 0 | 41,104 | 1 | 82,208 |
"Correct Solution:
```
import sys
for s in sys.stdin:
data = list(map(int, s.split(',')))
dis = data[:10]
v1, v2 = data[10], data[11]
for i in range(1, len(dis)):
dis[i] += dis[i-1]
time = dis[-1] / (v1+v2)
meetDis = v1 * time
for i in range(len(dis)):
if dis[i] >= meetDis:
print(i+1)
break
``` | output | 1 | 41,104 | 1 | 82,209 |
Provide a correct Python 3 solution for this coding contest problem.
There is a double-track line (up and down are separate lines and pass each other everywhere). There are 11 stations on this line, including the terminal station, and each station is called by the section number shown in the figure.
<image>
Trains depart from both terminal stations on this line at the same time and run without stopping along the way. Create a program that reads the length of each section and the speed of the two trains and outputs the number of the section where the trains pass each other in each case. However, if you pass each other just at the station, the smaller number of the section numbers on both sides will be output. In addition, the length of the train and the length of the station cannot be ignored.
Input
Multiple datasets are given. Each dataset is given in the following format.
l1, l2, l3, l4, l5, l6, l7, l8, l9, l10, v1, v2
li (1 ≤ li ≤ 2,000) is an integer representing the length (km) of the interval i. v1 is the speed of the train departing from the terminal station on the section 1 side (km / h), and v2 is the speed of the train departing from the terminal station on the section 10 side (km / h) (1 ≤ v1, v2) ≤ 2,000).
The number of datasets does not exceed 50.
Output
For each data set, the number of the section where the train passes is output on one line.
Example
Input
1,1,1,1,1,1,1,1,1,1,40,60
1,1,1,1,1,3,3,3,3,3,50,50
10,10,10,10,10,10,10,10,10,10,50,49
Output
4
7
6 | instruction | 0 | 41,105 | 1 | 82,210 |
"Correct Solution:
```
while 1:
try:
l=list(map(int,input().split(",")))
v1=l[10]
suml=sum(l[:10])
sumv=sum(l[10:12])
time=suml/sumv
point=time*v1
ans=1
for i in l[:10]:
if point<=i:break
else:
point-=i
ans+=1
print(ans)
except:break
``` | output | 1 | 41,105 | 1 | 82,211 |
Provide a correct Python 3 solution for this coding contest problem.
There is a double-track line (up and down are separate lines and pass each other everywhere). There are 11 stations on this line, including the terminal station, and each station is called by the section number shown in the figure.
<image>
Trains depart from both terminal stations on this line at the same time and run without stopping along the way. Create a program that reads the length of each section and the speed of the two trains and outputs the number of the section where the trains pass each other in each case. However, if you pass each other just at the station, the smaller number of the section numbers on both sides will be output. In addition, the length of the train and the length of the station cannot be ignored.
Input
Multiple datasets are given. Each dataset is given in the following format.
l1, l2, l3, l4, l5, l6, l7, l8, l9, l10, v1, v2
li (1 ≤ li ≤ 2,000) is an integer representing the length (km) of the interval i. v1 is the speed of the train departing from the terminal station on the section 1 side (km / h), and v2 is the speed of the train departing from the terminal station on the section 10 side (km / h) (1 ≤ v1, v2) ≤ 2,000).
The number of datasets does not exceed 50.
Output
For each data set, the number of the section where the train passes is output on one line.
Example
Input
1,1,1,1,1,1,1,1,1,1,40,60
1,1,1,1,1,3,3,3,3,3,50,50
10,10,10,10,10,10,10,10,10,10,50,49
Output
4
7
6 | instruction | 0 | 41,106 | 1 | 82,212 |
"Correct Solution:
```
while True:
try:
dis = 0
kyori = []
Data = list(map(int,input().split(",")))
for i in range(10):
dis += Data[i]
kyori.append(dis)
Ad = dis * Data[10] / (Data[10] + Data[11])
for j in range(9):
if kyori[j] < Ad < kyori[j + 1]:
print(j + 2)
elif kyori[j] == Ad:
print(j + 1)
except EOFError:
break
``` | output | 1 | 41,106 | 1 | 82,213 |
Provide a correct Python 3 solution for this coding contest problem.
There is a double-track line (up and down are separate lines and pass each other everywhere). There are 11 stations on this line, including the terminal station, and each station is called by the section number shown in the figure.
<image>
Trains depart from both terminal stations on this line at the same time and run without stopping along the way. Create a program that reads the length of each section and the speed of the two trains and outputs the number of the section where the trains pass each other in each case. However, if you pass each other just at the station, the smaller number of the section numbers on both sides will be output. In addition, the length of the train and the length of the station cannot be ignored.
Input
Multiple datasets are given. Each dataset is given in the following format.
l1, l2, l3, l4, l5, l6, l7, l8, l9, l10, v1, v2
li (1 ≤ li ≤ 2,000) is an integer representing the length (km) of the interval i. v1 is the speed of the train departing from the terminal station on the section 1 side (km / h), and v2 is the speed of the train departing from the terminal station on the section 10 side (km / h) (1 ≤ v1, v2) ≤ 2,000).
The number of datasets does not exceed 50.
Output
For each data set, the number of the section where the train passes is output on one line.
Example
Input
1,1,1,1,1,1,1,1,1,1,40,60
1,1,1,1,1,3,3,3,3,3,50,50
10,10,10,10,10,10,10,10,10,10,50,49
Output
4
7
6 | instruction | 0 | 41,107 | 1 | 82,214 |
"Correct Solution:
```
import sys
for line in sys.stdin:
l = list(map(int, line.split(",")))
total_length = sum(l[:-2])
a_v = l[-2]
b_v = l[-1]
crosspointtime = total_length / (a_v + b_v)
crosspoint = crosspointtime * a_v
ans = 0
for i, length in enumerate(l[:-2]):
ans += length
if ans >= crosspoint:
print(i+1)
break
``` | output | 1 | 41,107 | 1 | 82,215 |
Provide a correct Python 3 solution for this coding contest problem.
There is a double-track line (up and down are separate lines and pass each other everywhere). There are 11 stations on this line, including the terminal station, and each station is called by the section number shown in the figure.
<image>
Trains depart from both terminal stations on this line at the same time and run without stopping along the way. Create a program that reads the length of each section and the speed of the two trains and outputs the number of the section where the trains pass each other in each case. However, if you pass each other just at the station, the smaller number of the section numbers on both sides will be output. In addition, the length of the train and the length of the station cannot be ignored.
Input
Multiple datasets are given. Each dataset is given in the following format.
l1, l2, l3, l4, l5, l6, l7, l8, l9, l10, v1, v2
li (1 ≤ li ≤ 2,000) is an integer representing the length (km) of the interval i. v1 is the speed of the train departing from the terminal station on the section 1 side (km / h), and v2 is the speed of the train departing from the terminal station on the section 10 side (km / h) (1 ≤ v1, v2) ≤ 2,000).
The number of datasets does not exceed 50.
Output
For each data set, the number of the section where the train passes is output on one line.
Example
Input
1,1,1,1,1,1,1,1,1,1,40,60
1,1,1,1,1,3,3,3,3,3,50,50
10,10,10,10,10,10,10,10,10,10,50,49
Output
4
7
6 | instruction | 0 | 41,108 | 1 | 82,216 |
"Correct Solution:
```
while True:
try:
inp = list(map(int, input().split(",")))
l = inp[0:10]
l_all = sum(l)
v1 = inp[10]
v2 = inp[11]
l_pass = l_all*v1/(v1 + v2)
for i in range(11):
if sum(l[:i]) >= l_pass:
print(i)
break
except:
break
``` | output | 1 | 41,108 | 1 | 82,217 |
Provide a correct Python 3 solution for this coding contest problem.
There is a double-track line (up and down are separate lines and pass each other everywhere). There are 11 stations on this line, including the terminal station, and each station is called by the section number shown in the figure.
<image>
Trains depart from both terminal stations on this line at the same time and run without stopping along the way. Create a program that reads the length of each section and the speed of the two trains and outputs the number of the section where the trains pass each other in each case. However, if you pass each other just at the station, the smaller number of the section numbers on both sides will be output. In addition, the length of the train and the length of the station cannot be ignored.
Input
Multiple datasets are given. Each dataset is given in the following format.
l1, l2, l3, l4, l5, l6, l7, l8, l9, l10, v1, v2
li (1 ≤ li ≤ 2,000) is an integer representing the length (km) of the interval i. v1 is the speed of the train departing from the terminal station on the section 1 side (km / h), and v2 is the speed of the train departing from the terminal station on the section 10 side (km / h) (1 ≤ v1, v2) ≤ 2,000).
The number of datasets does not exceed 50.
Output
For each data set, the number of the section where the train passes is output on one line.
Example
Input
1,1,1,1,1,1,1,1,1,1,40,60
1,1,1,1,1,3,3,3,3,3,50,50
10,10,10,10,10,10,10,10,10,10,50,49
Output
4
7
6 | instruction | 0 | 41,109 | 1 | 82,218 |
"Correct Solution:
```
import sys
for row in sys.stdin:
row = row.strip().split(',')
ls = list(map(int, row[:-2]))
v1 = int(row[-2])
v2 = int(row[-1])
p = sum(ls) * v1 / (v1+v2)
s = 0
for i, l in enumerate(list(ls)):
s += l
if s>=p:
print(i+1)
break
``` | output | 1 | 41,109 | 1 | 82,219 |
Provide a correct Python 3 solution for this coding contest problem.
There is a double-track line (up and down are separate lines and pass each other everywhere). There are 11 stations on this line, including the terminal station, and each station is called by the section number shown in the figure.
<image>
Trains depart from both terminal stations on this line at the same time and run without stopping along the way. Create a program that reads the length of each section and the speed of the two trains and outputs the number of the section where the trains pass each other in each case. However, if you pass each other just at the station, the smaller number of the section numbers on both sides will be output. In addition, the length of the train and the length of the station cannot be ignored.
Input
Multiple datasets are given. Each dataset is given in the following format.
l1, l2, l3, l4, l5, l6, l7, l8, l9, l10, v1, v2
li (1 ≤ li ≤ 2,000) is an integer representing the length (km) of the interval i. v1 is the speed of the train departing from the terminal station on the section 1 side (km / h), and v2 is the speed of the train departing from the terminal station on the section 10 side (km / h) (1 ≤ v1, v2) ≤ 2,000).
The number of datasets does not exceed 50.
Output
For each data set, the number of the section where the train passes is output on one line.
Example
Input
1,1,1,1,1,1,1,1,1,1,40,60
1,1,1,1,1,3,3,3,3,3,50,50
10,10,10,10,10,10,10,10,10,10,50,49
Output
4
7
6 | instruction | 0 | 41,110 | 1 | 82,220 |
"Correct Solution:
```
while True:
try:
l = list(map(int, input().split(",")))
except:
break
sum_l = sum(l[:10])
M = sum_l*l[10]/(l[10] + l[11])
for i in range(10):
M -= l[i]
if M <= 0:
print(i+1)
break
``` | output | 1 | 41,110 | 1 | 82,221 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.