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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.
Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?
Input
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
Output
If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Examples
Input
5 5 11
Output
No
Input
10 15 25
Output
Yes
Input
0 5 1
Output
No
Input
0 0 2
Output
Yes
Note
In fourth sample case one possible route is: <image>. | instruction | 0 | 32,061 | 1 | 64,122 |
Tags: math
Correct Solution:
```
a, b, s = map(lambda x: abs(int(x)), input().split())
print("Yes" if a + b == s or (a + b < s and (s - (a + b)) % 2 == 0) else "No")
``` | output | 1 | 32,061 | 1 | 64,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.
Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?
Input
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
Output
If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Examples
Input
5 5 11
Output
No
Input
10 15 25
Output
Yes
Input
0 5 1
Output
No
Input
0 0 2
Output
Yes
Note
In fourth sample case one possible route is: <image>. | instruction | 0 | 32,062 | 1 | 64,124 |
Tags: math
Correct Solution:
```
a, b, s = map(int, input().split())
if(abs(a)+abs(b)>s):
print("No")
else:
if(abs(a)+abs(b)-s)%2==0:
print("Yes")
else:
print("No")
``` | output | 1 | 32,062 | 1 | 64,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.
Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?
Input
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
Output
If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Examples
Input
5 5 11
Output
No
Input
10 15 25
Output
Yes
Input
0 5 1
Output
No
Input
0 0 2
Output
Yes
Note
In fourth sample case one possible route is: <image>. | instruction | 0 | 32,063 | 1 | 64,126 |
Tags: math
Correct Solution:
```
a,b,s=map(int,input().split())
s-=abs(a)+abs(b)
if s%2==1 or s<0:
print("No")
else:
print("Yes")
``` | output | 1 | 32,063 | 1 | 64,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.
Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?
Input
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
Output
If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Examples
Input
5 5 11
Output
No
Input
10 15 25
Output
Yes
Input
0 5 1
Output
No
Input
0 0 2
Output
Yes
Note
In fourth sample case one possible route is: <image>.
Submitted Solution:
```
import math
arr = input().split()
a = int(arr[0])
b = int(arr[1])
s = int(arr[2])
if a < 0:
a *= -1
if b < 0:
b *= -1
if s >= a + b and (s % 2 - a % 2 - b % 2) % 2 == 0:
print('Yes')
else:
print('No')
``` | instruction | 0 | 32,064 | 1 | 64,128 |
Yes | output | 1 | 32,064 | 1 | 64,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.
Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?
Input
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
Output
If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Examples
Input
5 5 11
Output
No
Input
10 15 25
Output
Yes
Input
0 5 1
Output
No
Input
0 0 2
Output
Yes
Note
In fourth sample case one possible route is: <image>.
Submitted Solution:
```
n,m,s=map(int,input().split())
if (s-(abs(n)+abs(m)))%2==0 and (s-(abs(n)+abs(m)))>=0:
print("Yes")
else:
print("No")
``` | instruction | 0 | 32,065 | 1 | 64,130 |
Yes | output | 1 | 32,065 | 1 | 64,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.
Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?
Input
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
Output
If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Examples
Input
5 5 11
Output
No
Input
10 15 25
Output
Yes
Input
0 5 1
Output
No
Input
0 0 2
Output
Yes
Note
In fourth sample case one possible route is: <image>.
Submitted Solution:
```
a, b, c = map(int, input().split())
a, b = abs(a), abs(b)
print("Yes" if (c >= a + b) and ((c - (a + b)) % 2 == 0) else "No")
``` | instruction | 0 | 32,066 | 1 | 64,132 |
Yes | output | 1 | 32,066 | 1 | 64,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.
Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?
Input
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
Output
If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Examples
Input
5 5 11
Output
No
Input
10 15 25
Output
Yes
Input
0 5 1
Output
No
Input
0 0 2
Output
Yes
Note
In fourth sample case one possible route is: <image>.
Submitted Solution:
```
a, b, s = list(map(int, input().split()))
print('YES') if (s - (abs(a) + abs(b)) ) % 2 == 0 and s >= (abs(a) + abs(b)) else print('NO')
``` | instruction | 0 | 32,067 | 1 | 64,134 |
Yes | output | 1 | 32,067 | 1 | 64,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.
Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?
Input
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
Output
If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Examples
Input
5 5 11
Output
No
Input
10 15 25
Output
Yes
Input
0 5 1
Output
No
Input
0 0 2
Output
Yes
Note
In fourth sample case one possible route is: <image>.
Submitted Solution:
```
a,b,s=map(int,input().split())
a,b=map(abs,[a,b])
if (s-(a+b))%2==0:
print("Yes")
else:
print("No")
``` | instruction | 0 | 32,068 | 1 | 64,136 |
No | output | 1 | 32,068 | 1 | 64,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.
Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?
Input
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
Output
If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Examples
Input
5 5 11
Output
No
Input
10 15 25
Output
Yes
Input
0 5 1
Output
No
Input
0 0 2
Output
Yes
Note
In fourth sample case one possible route is: <image>.
Submitted Solution:
```
# import sys
# sys.stdin = open("#input.txt", "r")
a,b,s = list(map(int, input().split()))
print("YNeos"[(s-a-b)<0 or (s-a-b)&1::2])
``` | instruction | 0 | 32,069 | 1 | 64,138 |
No | output | 1 | 32,069 | 1 | 64,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.
Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?
Input
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
Output
If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Examples
Input
5 5 11
Output
No
Input
10 15 25
Output
Yes
Input
0 5 1
Output
No
Input
0 0 2
Output
Yes
Note
In fourth sample case one possible route is: <image>.
Submitted Solution:
```
a, b, s = input().split()
a = int(a); b = int(b); s = int(s);
if s < a+b:
print("No")
elif (s-a-b)%2 == 0:
print("Yes")
else:
print("No")
``` | instruction | 0 | 32,070 | 1 | 64,140 |
No | output | 1 | 32,070 | 1 | 64,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Someday, Drazil wanted to go on date with Varda. Drazil and Varda live on Cartesian plane. Drazil's home is located in point (0, 0) and Varda's home is located in point (a, b). In each step, he can move in a unit distance in horizontal or vertical direction. In other words, from position (x, y) he can go to positions (x + 1, y), (x - 1, y), (x, y + 1) or (x, y - 1).
Unfortunately, Drazil doesn't have sense of direction. So he randomly chooses the direction he will go to in each step. He may accidentally return back to his house during his travel. Drazil may even not notice that he has arrived to (a, b) and continue travelling.
Luckily, Drazil arrived to the position (a, b) successfully. Drazil said to Varda: "It took me exactly s steps to travel from my house to yours". But Varda is confused about his words, she is not sure that it is possible to get from (0, 0) to (a, b) in exactly s steps. Can you find out if it is possible for Varda?
Input
You are given three integers a, b, and s ( - 109 ≤ a, b ≤ 109, 1 ≤ s ≤ 2·109) in a single line.
Output
If you think Drazil made a mistake and it is impossible to take exactly s steps and get from his home to Varda's home, print "No" (without quotes).
Otherwise, print "Yes".
Examples
Input
5 5 11
Output
No
Input
10 15 25
Output
Yes
Input
0 5 1
Output
No
Input
0 0 2
Output
Yes
Note
In fourth sample case one possible route is: <image>.
Submitted Solution:
```
a, b, number_of_steps = [int(x) for x in input().split()]
if (a + b) < number_of_steps:
if ((number_of_steps-(a+b)) % 2) == 0:
print("Yes")
else:
print("No")
elif (a + b) == number_of_steps:
print("Yes")
else:
print("No")
``` | instruction | 0 | 32,071 | 1 | 64,142 |
No | output | 1 | 32,071 | 1 | 64,143 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling. | instruction | 0 | 32,174 | 1 | 64,348 |
Tags: greedy, implementation, math
Correct Solution:
```
def main():
A, B, F, K = map(int, input().split())
gas = B
refuel = 0
i = 1
while i <= K:
if i % 2:
gas -= F
if gas >= 0:
if K - i > 0 and gas < (A - F) * 2 or \
K - i == 0 and gas < A - F:
gas = B
refuel += 1
gas -= A - F
else:
gas -= A - F
if gas >= 0:
if K - i > 0 and gas < F * 2 or \
K - i == 0 and gas < F:
gas = B
refuel += 1
gas -= F
if gas < 0:
print(-1)
return
i += 1
print(refuel)
main()
``` | output | 1 | 32,174 | 1 | 64,349 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling. | instruction | 0 | 32,175 | 1 | 64,350 |
Tags: greedy, implementation, math
Correct Solution:
```
def main():
a, b, f, k = input().split(" ")
endpoint = int(a)
capacity = int(b)
station = int(f)
journeys = int(k)
print(betterBus(endpoint, capacity, station, journeys))
def bus(endpoint, capacity, station, journeys):
refuels = 0
current = capacity
to = True
for j in range(journeys):
r = range(endpoint)
if not to:
r = range(endpoint, 0, -1)
to = True
else:
to = False
for e in r:
if e == station:
if j == journeys - 1:
if (current - e) < 0:
current = capacity
refuels += 1
elif (current - ((endpoint - station) * 2)) < 0:
current = capacity
refuels += 1
else:
if current == 0:
return -1
current -= 1
return refuels
def betterBus(endpoint, capacity, station, journeys):
refuels = 0
current = capacity
to = True
for j in range(journeys):
if to:
current -= station
if current < 0:
return -1
if j == journeys - 1:
if (current - (endpoint-station)) < 0:
current = capacity
if (current - (endpoint-station)) < 0:
return -1
refuels += 1
elif (current - ((endpoint - station) * 2)) < 0:
current = capacity
refuels += 1
current -= endpoint - station
to = False
else:
current -= endpoint - station
if current < 0:
return -1
if j == journeys - 1:
if (current - station) < 0:
current = capacity
if (current - station) < 0:
return -1
refuels += 1
elif (current - (station * 2)) < 0:
current = capacity
refuels += 1
current -= station
to = True
return refuels
main()
``` | output | 1 | 32,175 | 1 | 64,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling. | instruction | 0 | 32,176 | 1 | 64,352 |
Tags: greedy, implementation, math
Correct Solution:
```
import os
import sys
import re
if 'PYCHARM' in os.environ:
sys.stdin = open('in', 'r')
a, b, f, k = map(int, input().split())
x = b
ans = 0
def go(l):
if l > b:
print(-1)
sys.exit(0)
global ans, x
if x < l:
ans += 1
x = b
x -= l
go(f)
fw = True
for i in range(k - 1):
go(2 * (a - f if fw else f))
fw = not fw
go(a - f if fw else f)
print(ans)
``` | output | 1 | 32,176 | 1 | 64,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling. | instruction | 0 | 32,177 | 1 | 64,354 |
Tags: greedy, implementation, math
Correct Solution:
```
a,b,f,k=map(int,input().split())
tmp=b
flag=0
c=0
for i in range(k):
#print ("i",i,k)
if i%2==0:
#print ("even")
if f>b:
#print (-1)
flag=1
break
else:
b=b-f
if i!=(k-1):
if 2*(a-f)>b:
c+=1
b=tmp
b=b-(a-f)
#print (b)
else:
if (a-f)>b:
b=tmp
if (a-f)>b:
flag=1
else:
c+=1
else:
#print (b)
#print ("odd")
if b<a-f:
#print (-1)
flag=1
break
else:
b=b-(a-f)
#print (b)
if i!=(k-1):
#print ("Hurray")
if 2*f>b:
c+=1
b=tmp
b=b-f
#print (b)
else:
if f>b:
b=tmp
if f>b:
flag=1
else:
c+=1
#print (c)
if flag==0:
print (c)
else:
print (-1)
``` | output | 1 | 32,177 | 1 | 64,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling. | instruction | 0 | 32,178 | 1 | 64,356 |
Tags: greedy, implementation, math
Correct Solution:
```
def solve(a, b, f, k):
origb = b
disa = 2*f
disb = 2*(a-f)
trava = False
counter = 0
b -= disa/2
if b < 0:
return -1
while k > 0:
if trava:
trava = False
if k == 1:
disa = disa / 2
if disa > b:
counter += 1
b = origb -disa
if b < 0:
return -1
k -= 1
else:
b -= disa
if b < 0:
return -1
k -= 1
else:
trava = True
if k == 1:
disb= disb/2
if disb > b:
counter += 1
b = origb - disb
if b < 0:
return -1
k -= 1
else:
b -= disb
if b < 0:
return -1
k -= 1
return counter
inputs = input("")
afind = inputs.find(" ")
a = int(inputs[0:afind])
inputs = inputs[afind+1:]
afind = inputs.find(" ")
b = int(inputs[0:afind])
inputs = inputs[afind+1:]
afind = inputs.find(" ")
f = int(inputs[0:afind])
inputs = inputs[afind+1:]
k = int(inputs[0:])
print(solve(a, b, f, k))
``` | output | 1 | 32,178 | 1 | 64,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling. | instruction | 0 | 32,179 | 1 | 64,358 |
Tags: greedy, implementation, math
Correct Solution:
```
a, b, f, k = map(int, input().split())
ans = 0
t = b
l = a - f
x = 0
if b < max(f, l):
ans = -1
else:
while k > 0:
if x == 0:
t -= f
if t < 0:
ans = -1
break
t -= l
if k == 1:
l = 0
if b - l < 0:
ans = -1
break
if t - l < 0:
t = b - l
ans += 1
k -= 1
x = a
else:
t -= l
if t < 0:
ans = -1
break
t -= f
if k == 1:
f = 0
if b - f < 0:
ans = -1
break
if t - f < 0:
t = b - f
ans += 1
k -= 1
x = 0
print(ans)
``` | output | 1 | 32,179 | 1 | 64,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling. | instruction | 0 | 32,180 | 1 | 64,360 |
Tags: greedy, implementation, math
Correct Solution:
```
def impossible():
print(-1)
exit()
dest, cap, station, journey = map(int, input().strip().split())
left, right = 2 * station, 2*(dest-station)
if journey == 1 and (cap < station or cap < dest-station):
impossible()
if journey == 2 and (cap < station or cap < right):
impossible()
if journey >= 3 and (cap < left or cap < right):
impossible()
time= 0
consume = station
for i in range(journey-1):
go = left if (i&1) else right
if consume + go > cap:
time += 1
consume = go
else:
consume += go
if ((journey & 1) == 0 and consume + station > cap) or ((journey & 1) and (consume + dest - station > cap)):
print(time+1)
else:
print(time)
``` | output | 1 | 32,180 | 1 | 64,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling. | instruction | 0 | 32,181 | 1 | 64,362 |
Tags: greedy, implementation, math
Correct Solution:
```
DEBUG = False
def cover_distance():
"""
Distance [a*k] likes [f]+[2*(a-f)]+[2*f]+[2*(a-f)]+[2*f]+...\
+OR([2*(a-f)]+[f],[2*f]+[a-f])
Last add depends on even/odd of k.
cuts <= b
Cover distance [a*k] by cuts with greedy alg.
"""
a, b, f, k = map(int, input().split())
fuel = b
add_fuel_count = 0
fuel_on_next_stop = fuel - f
if fuel_on_next_stop < 0:
return -1
distance_to_destination = 0
distance_to_gas_station = 0
for i in range(1, k + 1):
if DEBUG:
print("i= {} : ".format(i))
print("\t fuel_on_station = {} ".format(fuel_on_next_stop))
if i == k:
# init distance_to_destination:
if i % 2:
distance_to_destination = a - f
else:
distance_to_destination = f
# calculate fuel
fuel = fuel_on_next_stop
if fuel < distance_to_destination:
fuel = b
add_fuel_count += 1
fuel_on_next_stop = fuel - distance_to_destination
else:
# init distance_to_gas_station:
if i % 2:
distance_to_gas_station = 2 * (a - f)
else:
distance_to_gas_station = 2 * f
# calculate fuel
fuel = fuel_on_next_stop
if fuel < distance_to_gas_station:
fuel = b
add_fuel_count += 1
fuel_on_next_stop = fuel - distance_to_gas_station
if DEBUG:
# init distance
if distance_to_destination:
distance = distance_to_destination
else:
distance = distance_to_gas_station
print("\t distance = {} ".format(distance))
print("\t fuel_on_next_stop = {} ".format(fuel_on_next_stop))
if fuel == b:
print("\t YES")
if fuel_on_next_stop < 0:
return -1
return add_fuel_count
if __name__ == "__main__":
print(cover_distance())
``` | output | 1 | 32,181 | 1 | 64,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
Submitted Solution:
```
a,b,f,k=map(int,input().split())
t=0
lt=b
ans=0
while t<k:
if t&1:
lt-=a-f
if lt<0:exit(print(-1))
if lt<f*2:
ans+=1
if t == k - 1 and lt >=f: ans -= 1
lt=b
lt-=f
if lt < 0: exit(print(-1))
else:
lt-=f
if lt < 0: exit(print(-1))
if lt<(a-f)*2:
ans += 1
if t == k - 1 and lt >= a - f:ans-=1
lt=b
lt-=a-f
if lt < 0: exit(print(-1))
t+=1
print(ans)
``` | instruction | 0 | 32,182 | 1 | 64,364 |
Yes | output | 1 | 32,182 | 1 | 64,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
Submitted Solution:
```
a, b, f, k = map(int, input().split())
x,y=f,a-f
if k==1:
if b<max(x,y):
print(-1)
quit()
elif k==2:
if b<max(x,2*y):
print(-1)
quit()
else:
if b<max(2*x,2*y):
print(-1)
quit()
gas=b
refill=0
test=True
while k>0:
if test:
if gas>=x+2*y:
gas-=a
elif gas>=x:
if k==1:
if gas>=a:
break
gas=b-y
refill+=1
else:
print(-1)
quit()
test= not test
else:
if gas>=2*x+y:
gas-=a
elif gas>=y:
if k==1:
if gas>=a:
break
gas=b-x
refill+=1
else:
print(-1)
quit()
test= not test
k-=1
print(refill)
``` | instruction | 0 | 32,183 | 1 | 64,366 |
Yes | output | 1 | 32,183 | 1 | 64,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
Submitted Solution:
```
a, b, f, k = [int(z) for z in input().split()]
to_fuel = f
from_fuel = a - f
alrdy_fueled = 0
petrol = b
for i in range(k):
petrol -= to_fuel
if (petrol < 0):
print(-1)
exit(0)
if petrol < 2 * from_fuel and i != k - 1:
petrol = b
alrdy_fueled += 1
elif i == k - 1:
if petrol < from_fuel:
petrol = b
alrdy_fueled += 1
petrol -= from_fuel
if (petrol < 0):
print(-1)
exit(0)
to_fuel, from_fuel = from_fuel, to_fuel
print(alrdy_fueled)
``` | instruction | 0 | 32,184 | 1 | 64,368 |
Yes | output | 1 | 32,184 | 1 | 64,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
Submitted Solution:
```
a, b, f, k = map(int, input().split())
now = b
da = [0, f, a]
an = 0
for r in range(k):
if r == k - 1:
if r % 2 == 0:
if now < f:
print(-1)
exit(0)
now -= f
if now < a - f:
now = b
an += 1
if now < a - f:
print(-1)
exit(0)
now -= a - f
else:
if now < a - f:
print(-1)
exit(0)
now -= a - f
if now < f:
now = b
an += 1
if now < f:
print(-1)
exit(0)
now -= f
else:
if r % 2 == 0:
if now < f:
print(-1)
exit(0)
now -= f
if now < 2 * (a - f):
now = b
an += 1
if now < (a - f):
print(-1)
exit(0)
now -= a - f
else:
if now < a - f:
print(-1)
exit(0)
now -= a - f
if now < 2 * f:
now = b
an += 1
if now < f:
print(-1)
exit(0)
now -= f
print(an)
``` | instruction | 0 | 32,185 | 1 | 64,370 |
Yes | output | 1 | 32,185 | 1 | 64,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
Submitted Solution:
```
a, b, f, k = map(int, input().split(' '))
l, r = f, a-f
if b < l or b < 2*r or b < 2*l:
print(-1)
else:
remain = b
cnt = 0
i = 0
while i < k:
remain -= l
if i < k-1 and remain < 2*r:
remain = b
cnt += 1
elif i == k-1 and remain < r:
remain = b
cnt += 1
remain -= 2 * r
i += 1
if i == k:
break
if i < k-1 and remain < 2*l:
remain = b
cnt += 1
elif i == k-1 and remain < l:
remain = b
cnt += 1
i += 1
print(cnt)
``` | instruction | 0 | 32,186 | 1 | 64,372 |
No | output | 1 | 32,186 | 1 | 64,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
Submitted Solution:
```
#python3
# utf-8
a, b, f, k = (int(x) for x in input().split())
z = 0
curr_petrol = b
curr_races_made = 0
ans = 0
z___f = f - z
f___a = a - f
if 2 * z___f > b or 2 * f___a > b:
print(-1)
exit()
#direction, races_made, petrol
curr_save = None
route = [z___f, f___a, f___a, z___f]
while curr_races_made < 2 * k:
curr_pos = curr_races_made % 4
curr_petrol -= route[curr_pos]
curr_races_made += 1
if curr_petrol < 0:
curr_petrol = b
ans += 1
curr_races_made = curr_save
continue
if curr_pos == 0 or curr_pos == 2:
curr_save = curr_races_made
print(ans)
``` | instruction | 0 | 32,187 | 1 | 64,374 |
No | output | 1 | 32,187 | 1 | 64,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
Submitted Solution:
```
a, b, f, k = [int(i) for i in input().split()]
##tank = b
##journeys = 0
##refuels = 0
##current = 0
##while(journeys != k):
## print("current = %d, tank = %d, refuels = %d, journeys = %d" % (current, tank, refuels, journeys))
## if (tank // a >= k):
## print(refuels)
## exit(0)
## if current == 0:
## if tank >= a + (a-f):
## tank -= a
## elif tank >= f and b >= a-f:
## refuels += 1
## tank = b-(a-f)
## else:
## break
##
## current = a
##
## elif current == a:
## if tank >= a + f:
## tank -= a
## elif tank >= a-f and b >= f:
## refuels += 1
## tank = b-f
## else:
## break
##
## current = 0
##
## journeys += 1
if b < f:
print(-1)
exit()
journeys = 0
previous = 0
refuels1 = 0
tank = b-f
cant1 = False
while(journeys != k):
if previous == 0:
if tank >= a-f + a*(k-journeys-1):
break
if tank >= 2*(a-f):
tank -= 2*(a-f)
elif b >= 2*(a-f):
refuels1 += 1
tank = b - 2*(a-f)
else:
cant1 = True
break
journeys += 1
previous = a
if previous == a:
if tank >= f + a*(k-journeys-1):
break
if tank >= 2*f:
tank -= 2*f
elif b >= 2*f:
refuels1 += 1
tank = b-2*f
else:
cant1 = True
break
journeys += 1
previous = 0
journeys = 0
previous = 0
refuels2 = 1
tank = b
cant2 = False
while(journeys != k):
if previous == 0:
if tank >= a-f + a*(k-journeys-1):
break
if tank >= 2*(a-f):
tank -= 2*(a-f)
elif b >= 2*(a-f):
refuels2 += 1
tank = b - 2*(a-f)
else:
cant2 = True
break
journeys += 1
previous = a
if previous == a:
if tank >= f + a*(k-journeys-1):
break
if tank >= 2*f:
tank -= 2*f
elif b >= 2*f:
refuels2 += 1
tank = b-2*f
else:
cant2 = True
break
journeys += 1
previous = 0
if cant1 and not cant2:
print(refuels2)
elif not cant1 and cant2:
print(refuels1)
elif cant1 and cant2:
print(-1)
else:
print(min(refuels1, refuels2))
``` | instruction | 0 | 32,188 | 1 | 64,376 |
No | output | 1 | 32,188 | 1 | 64,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A bus moves along the coordinate line Ox from the point x = 0 to the point x = a. After starting from the point x = 0, it reaches the point x = a, immediately turns back and then moves to the point x = 0. After returning to the point x = 0 it immediately goes back to the point x = a and so on. Thus, the bus moves from x = 0 to x = a and back. Moving from the point x = 0 to x = a or from the point x = a to x = 0 is called a bus journey. In total, the bus must make k journeys.
The petrol tank of the bus can hold b liters of gasoline. To pass a single unit of distance the bus needs to spend exactly one liter of gasoline. The bus starts its first journey with a full petrol tank.
There is a gas station in point x = f. This point is between points x = 0 and x = a. There are no other gas stations on the bus route. While passing by a gas station in either direction the bus can stop and completely refuel its tank. Thus, after stopping to refuel the tank will contain b liters of gasoline.
What is the minimum number of times the bus needs to refuel at the point x = f to make k journeys? The first journey starts in the point x = 0.
Input
The first line contains four integers a, b, f, k (0 < f < a ≤ 106, 1 ≤ b ≤ 109, 1 ≤ k ≤ 104) — the endpoint of the first bus journey, the capacity of the fuel tank of the bus, the point where the gas station is located, and the required number of journeys.
Output
Print the minimum number of times the bus needs to refuel to make k journeys. If it is impossible for the bus to make k journeys, print -1.
Examples
Input
6 9 2 4
Output
4
Input
6 10 2 4
Output
2
Input
6 5 4 3
Output
-1
Note
In the first example the bus needs to refuel during each journey.
In the second example the bus can pass 10 units of distance without refueling. So the bus makes the whole first journey, passes 4 units of the distance of the second journey and arrives at the point with the gas station. Then it can refuel its tank, finish the second journey and pass 2 units of distance from the third journey. In this case, it will again arrive at the point with the gas station. Further, he can refill the tank up to 10 liters to finish the third journey and ride all the way of the fourth journey. At the end of the journey the tank will be empty.
In the third example the bus can not make all 3 journeys because if it refuels during the second journey, the tanks will contain only 5 liters of gasoline, but the bus needs to pass 8 units of distance until next refueling.
Submitted Solution:
```
inp = input().split(' ')
a = int(inp[0])
b = int(inp[1])
f = int(inp[2])
k = int(inp[3])
b_temp = b
i = 0
doz = 0
while i < k:
if b<f or b<a-f:
print(-1)
break
elif b>=f:
b = b - f
# print('1-', b,doz,i)
if b<=a-f:
i = i + 1
if k - i != 0:
doz = doz + 1
b = b_temp
b = b - (a-i)
# print('2-', b,doz,i)
if b<=a-f:
print(-1)
break
else:
b = b - (a-f)
i = i + 1
# print('2-', b,doz,i)
elif b>a-f and b<2*(a-f):
doz = doz + 1
b = b_temp
i = i + 1
b = b - (a-f)
# print('2-', b,doz,i)
elif b>=2*(a-f):
b = b - (a-f)
i = i + 1
# print('2-', b,doz,i)
if i == k:
break
if b<a-f:
if i < k:
print(-1)
break
elif b>=a-f:
b = b - (a-f)
# print('3-', b,doz,i)
if b<=f:
if b == f:
i = i + 1
b = b - f
else:
i = i + 1
doz = doz + 1
b = b_temp
b = b - f
# print('4-', b,doz,i)
elif b>f and b<2*f:
doz = doz + 1
b = b_temp
i = i + 1
b = b - f
# print('4-', b,doz,i)
elif b>=2*f:
b = b-f
i = i + 1
# print('4-', b,doz,i)
if i == k:
print(doz)
``` | instruction | 0 | 32,189 | 1 | 64,378 |
No | output | 1 | 32,189 | 1 | 64,379 |
Provide a correct Python 3 solution for this coding contest problem.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688 | instruction | 0 | 32,430 | 1 | 64,860 |
"Correct Solution:
```
def s():
for e in iter(input,'0 0'):
w,h=map(int,e.split())
M=[[[1,0]*2 for _ in[0]*h]for _ in[0]*w]
for i in range(1,w):
for j in range(1,h):
a,b=M[i-1][j][:2]
c,d=M[i][j-1][2:]
M[i][j]=[d,a+b,b,c+d]
print((sum(M[w-2][h-1][:2])+sum(M[w-1][h-2][2:]))%10**5)
if'__main__'==__name__:s()
``` | output | 1 | 32,430 | 1 | 64,861 |
Provide a correct Python 3 solution for this coding contest problem.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688 | instruction | 0 | 32,431 | 1 | 64,862 |
"Correct Solution:
```
for e in iter(input,'0 0'):
w,h=map(int,e.split())
M=[[[1,0]*2 for _ in[0]*h]for _ in[0]*w]
for i in range(1,w):
for j in range(1,h):
a,b=M[i-1][j][:2]
c,d=M[i][j-1][2:]
M[i][j]=[d,a+b,b,c+d]
print((sum(M[w-2][h-1][:2])+sum(M[w-1][h-2][2:]))%10**5)
``` | output | 1 | 32,431 | 1 | 64,863 |
Provide a correct Python 3 solution for this coding contest problem.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688 | instruction | 0 | 32,432 | 1 | 64,864 |
"Correct Solution:
```
while True:
w, h = map(int, input().split())
if not w:
break
district = [[0] * w for _ in range(h)]
district[0] = [(1, 0, 0, 0)] * w
for i in range(1, h):
row = district[i]
row[0] = (0, 1, 0, 0)
for j in range(1, w):
left, under = row[j - 1], district[i - 1][j]
row[j] = tuple(map(lambda x: x % 100000, (left[0] + left[2], under[1] + under[3], left[1], under[0])))
print(sum(district[-1][-1]) % 100000)
``` | output | 1 | 32,432 | 1 | 64,865 |
Provide a correct Python 3 solution for this coding contest problem.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688 | instruction | 0 | 32,433 | 1 | 64,866 |
"Correct Solution:
```
def s():
for e in iter(input,'0 0'):
w,h=map(int,e.split())
M=[[[1,0]*2 for _ in[0]*h]for _ in[0]*w]
for i in range(1,w):
for j in range(1,h):
a,b=M[i-1][j][:2]
c,d=M[i][j-1][2:]
M[i][j]=[d,a+b,b,c+d]
print((sum(M[w-2][h-1][:2])+sum(M[w-1][h-2][2:]))%10**5)
s()
``` | output | 1 | 32,433 | 1 | 64,867 |
Provide a correct Python 3 solution for this coding contest problem.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688 | instruction | 0 | 32,434 | 1 | 64,868 |
"Correct Solution:
```
for e in iter(input,'0 0'):
w,h=map(int,e.split())
M=[[[1,0]*2 for _ in[0]*h]for _ in[0]*w]
for i in range(1,w):
for j in range(1,h):
a,b,c,d=[*M[i-1][j][:2],*M[i][j-1][2:]]
M[i][j]=[d,a+b,b,c+d]
print((sum(M[w-2][h-1][:2])+sum(M[w-1][h-2][2:]))%10**5)
``` | output | 1 | 32,434 | 1 | 64,869 |
Provide a correct Python 3 solution for this coding contest problem.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688 | instruction | 0 | 32,435 | 1 | 64,870 |
"Correct Solution:
```
# AOJ 0547: Commute routes
# Python3 2018.7.1 bal4u
import sys
from sys import stdin
input = stdin.readline
S, W = 0, 1
while True:
w, h = map(int, input().split())
if w == 0: break
p = [[[[0 for a in range(2)] for b in range(2)] for c in range(h+1)] for d in range(w+1)]
for x in range(1, w+1): p[x][1][W][S] = p[x][1][W][W] = 1
for y in range(1, h+1): p[1][y][S][W] = p[1][y][S][S] = 1
p[w][1][W][W] = p[1][h][S][S] = 0
for y in range(2, h+1):
for x in range(2, w+1):
p[x][y][S][S] += p[x][y-1][S][S] + p[x][y-1][W][S]
p[x][y][W][W] += p[x-1][y][W][W] + p[x-1][y][S][W]
p[x][y][S][W] += p[x][y-1][S][S]
p[x][y][W][S] += p[x-1][y][W][W]
p[x][y][S][S] %= 100000
p[x][y][W][W] %= 100000
p[x][y][S][W] %= 100000
p[x][y][W][S] %= 100000
print((p[w][h][S][S]+p[w][h][W][W]) % 100000)
``` | output | 1 | 32,435 | 1 | 64,871 |
Provide a correct Python 3 solution for this coding contest problem.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688 | instruction | 0 | 32,436 | 1 | 64,872 |
"Correct Solution:
```
def main():
while True:
import sys
input=sys.stdin.readline
w,h=map(int,input().split())
if w==0:
break
dp=[[[0,0,0,0] for _ in [0]*(w+1)] for _ in [0]*(h+1)] #右直進、上直進、右、上
dp[1][0]=[0,1,0,0]
dp[0][1]=[1,0,0,0]
mod=10**5
for k in range(1,w+h-2):
for j in range(max(0,k-h+1),min(w,k+1)):
i=k-j
dp[i+1][j][1]=(dp[i+1][j][1]+dp[i][j][1]+dp[i][j][3])%mod
dp[i+1][j][3]=(dp[i+1][j][3]+dp[i][j][0])%mod
dp[i][j+1][0]=(dp[i][j+1][0]+dp[i][j][0]+dp[i][j][2])%mod
dp[i][j+1][2]=(dp[i][j+1][2]+dp[i][j][1])%mod
print(sum(dp[h-1][w-1])%mod)
if __name__=='__main__':
main()
``` | output | 1 | 32,436 | 1 | 64,873 |
Provide a correct Python 3 solution for this coding contest problem.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688 | instruction | 0 | 32,437 | 1 | 64,874 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Commute routes
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0547
"""
import sys
def solve(width, height):
# 北向き 直前に回転あり・回転なし、東向 直前に回転あり・回転なし
dp = [[None] * width for _ in range(height)]
dp[0][0] = [0, 1, 0, 1]
for h in range(height):
for w in range(width):
if h == 0:
dp[0][w] = [0, 0, 0, 1]
elif w == 0:
dp[h][0] = [0, 1, 0, 0]
else:
dp[h][w] = [dp[h-1][w][3], dp[h-1][w][0]+dp[h-1][w][1], dp[h][w-1][1], dp[h][w-1][2]+dp[h][w-1][3]]
return sum(dp[-1][-1]) % 100000
def main(args):
while True:
w, h = map(int, input().split())
if w == 0 and h == 0:
break
result = solve(w, h)
print(result)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 32,437 | 1 | 64,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688
Submitted Solution:
```
while True:
w,h = map(int,input().split())
if w==0: break
mapp = [[[0]*2 for _ in range(w+2)] for _ in range(h+2)]
mapp[0][0][0] = mapp[0][0][1] = 1
mapp[0][2][0] = mapp[2][0][1] = -1
for i in range(0,h):
for j in range(0,w):
if i==h-1 and j==w-1 : break
mapp[i][j+1][0] += mapp[i][j][0]
if mapp[i][j+1][0] >= 100000: mapp[i][j+1][0] -= 100000
mapp[i][j+2][0] += mapp[i][j][1]
if mapp[i][j+2][0] >= 100000: mapp[i][j+2][0] -= 100000
mapp[i+1][j][1] += mapp[i][j][1]
if mapp[i+1][j][1] >= 100000: mapp[i+1][j][0] -= 100000
mapp[i+2][j][1] += mapp[i][j][0]
if mapp[i+2][j][1] >= 100000: mapp[i+2][j][0] -= 100000
print(sum(mapp[h-1][w-1]+mapp[h][w-1]+mapp[h-1][w])%100000)
``` | instruction | 0 | 32,438 | 1 | 64,876 |
Yes | output | 1 | 32,438 | 1 | 64,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688
Submitted Solution:
```
while True:
h,w = map(int, input().split())
if not h:
break
#dp[R][U][""]
h -= 1
w -= 1
mp = {(0,2,"UU") : 1, (2,0,"RR") : 1, (1,1,"UR") : 1, (1,1,"RU") : 1}
for i in range(h + 1):
for j in range(w + 1):
if i + j <= 2:
continue
mp[(i, j, 'UU')] = 0
if (i, j - 1, 'UU') in mp:
mp[(i, j, 'UU')] += mp[(i, j - 1, 'UU')]
if (i, j - 1, 'RU') in mp:
mp[(i, j, 'UU')] += mp[(i, j - 1, 'RU')]
mp[(i, j, 'RR')] = 0
if (i - 1, j, 'RR') in mp:
mp[(i, j, 'RR')] += mp[(i - 1, j, 'RR')]
if (i - 1, j, 'UR') in mp:
mp[(i, j, 'RR')] += mp[(i - 1, j, 'UR')]
mp[(i, j, 'UR')] = 0
if (i - 1, j, 'UU') in mp:
mp[(i, j, 'UR')] += mp[(i - 1, j, 'UU')]
mp[(i, j, 'RU')] = 0
if (i, j - 1, 'RR') in mp:
mp[(i, j, 'RU')] += mp[(i, j - 1, 'RR')]
print((mp[(h, w, 'UU')] + mp[(h, w, 'UR')] + mp[(h, w, 'RU')] + mp[(h, w, 'RR')]) % 100000)
``` | instruction | 0 | 32,439 | 1 | 64,878 |
Yes | output | 1 | 32,439 | 1 | 64,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688
Submitted Solution:
```
for e in iter(input,'0 0'):
w,h=map(int,e.split())
M=[[[1,0]*2 for _ in[0]*h]for _ in[0]*w]
for i in range(1,w):
for j in range(1,h):
a,b=M[i-1][j][:2]
c,d=M[i][j-1][2:]
M[i][j]=[d,a+b,b,c+d]
for x in M:print(x)
print((sum(M[w-2][h-1][:2])+sum(M[w-1][h-2][2:]))%10**5)
``` | instruction | 0 | 32,440 | 1 | 64,880 |
No | output | 1 | 32,440 | 1 | 64,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688
Submitted Solution:
```
for e in iter(input,'0 0'):
w,h=map(int,e.split())
M=[[[1,0]*2 for _ in[0]*h]for _ in[0]*w]
for i in range(1,w):
for j in range(1,h):
a=M[i-1][j][:2]
b=M[i][j-1][2:]
M[i][j]=[b[1],a,a[1],b]
print((sum(M[w-2][h-1][:2])+sum(M[w-1][h-2][2:]))%10**5)
``` | instruction | 0 | 32,441 | 1 | 64,882 |
No | output | 1 | 32,441 | 1 | 64,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
A city in Canada where JOI lives is divided into a grid pattern by w roads that extend straight in the north-south direction and h roads that extend straight in the east-west direction.
The w roads in the north-south direction are numbered 1, 2, ..., w in order from the west. In addition, h roads in the east-west direction are numbered 1, 2, ..., h in order from the south. The intersection of the i-th north-south road from the west and the j-th east-west road from the south is represented by (i, j).
JOI lives near the intersection (1, 1) and drives to a company near the intersection (w, h). Cars can only move along the road. JOI travels only to the east or north to shorten his commute time. The city also has the following traffic rules to reduce traffic accidents:
* A car that turns at an intersection cannot turn at the intersection immediately after that.
That is, it is not permissible to go one block after turning at an intersection and turn again. At this time, how many possible commuting routes for Mr. JOI?
Given w and h, create a program that outputs the remainder of JOI's number of commuting routes divided by 100000.
input
The input consists of multiple datasets. Each dataset consists of one line, and two integers w, h (2 ≤ w ≤ 100, 2 ≤ h ≤ 100) are written, separated by a blank. w represents the number of roads in the north-south direction, and h represents the number of roads in the east-west direction.
When both w and h are 0, it indicates the end of input. The number of data sets does not exceed 5.
output
For each data set, the remainder of JOI's number of commuting routes divided by 100000 is output on one line.
Example
Input
3 4
15 15
0 0
Output
5
43688
Submitted Solution:
```
for e in iter(input,'0 0'):
w,h=map(int,e.split())
M=[[[1,0]*2 for _ in[0]*h]for _ in[0]*w]
for i in range(1,w):
for j in range(1,h):
a,b=M[i-1][j][:2]
c,d=M[i][j-1][2:]
M[i][j]=[d,a+b,b,c+d]
for x in M:print(x)
print(sum(M[w-2][h-1][:2])+sum(M[w-1][h-2][2:]))
``` | instruction | 0 | 32,442 | 1 | 64,884 |
No | output | 1 | 32,442 | 1 | 64,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
Input
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, ..., an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output
If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".
Examples
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
NO
Note
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | instruction | 0 | 32,922 | 1 | 65,844 |
Tags: dfs and similar, graphs, implementation
Correct Solution:
```
n,t = map(int,input().split())
a = list(map(int,input().split()))
v = []
i = 0
while i <= n-1:
v.append(i+1)
if i == n-1:
break
elif i < n-1:
i += a[i]
print(['NO','YES'][t in v])
``` | output | 1 | 32,922 | 1 | 65,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
Input
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, ..., an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output
If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".
Examples
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
NO
Note
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | instruction | 0 | 32,923 | 1 | 65,846 |
Tags: dfs and similar, graphs, implementation
Correct Solution:
```
from sys import stdin, stdout
n, t = stdin.readline().split()
n = int(n)
t = int(t) - 1
arr = stdin.readline().split()
arr.append('0')
ans = False
idx = 0
while not ans:
if idx == t:
ans = True
if idx == n - 1:
break
idx += int(arr[idx])
if ans:
stdout.write('YES\n')
else:
stdout.write('NO\n')
``` | output | 1 | 32,923 | 1 | 65,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
Input
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, ..., an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output
If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".
Examples
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
NO
Note
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | instruction | 0 | 32,924 | 1 | 65,848 |
Tags: dfs and similar, graphs, implementation
Correct Solution:
```
b = int(input().split() [1])
f = [0] + [int(x) for x in input().split()]
i = 1
while i < b:
i = i + f[i]
if i == b:
print("YES")
quit()
print("NO")
``` | output | 1 | 32,924 | 1 | 65,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
Input
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, ..., an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output
If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".
Examples
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
NO
Note
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | instruction | 0 | 32,925 | 1 | 65,850 |
Tags: dfs and similar, graphs, implementation
Correct Solution:
```
def newYear(n,t,list):
i = 1
while i < t:
i += list[i-1]
if i == t:
print("YES")
return
print("NO")
return
params = input().split()
n = int(params[0])
t = int(params[1])
params = input().split()
for i in range(len(params)):
params[i] = int(params[i])
newYear(n,t,params)
``` | output | 1 | 32,925 | 1 | 65,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
Input
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, ..., an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output
If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".
Examples
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
NO
Note
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | instruction | 0 | 32,926 | 1 | 65,852 |
Tags: dfs and similar, graphs, implementation
Correct Solution:
```
# n,t=map(int,input().split())
n,t=map(int,input().split())
arr=list(map(int,input().split()))
visit=[1]
i=0
while True:
try:
visit.append(i+arr[i]+1)
i=i+arr[i]
except:
break
if t in visit:
print("YES")
else:
print("NO")
``` | output | 1 | 32,926 | 1 | 65,853 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
Input
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, ..., an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output
If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".
Examples
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
NO
Note
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | instruction | 0 | 32,927 | 1 | 65,854 |
Tags: dfs and similar, graphs, implementation
Correct Solution:
```
n,t=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
def f(z):
return z+a[z-1]
i=1
while i<=n:
i=f(i)
if i==t:
print('YES')
break
elif i>t:
print('NO')
break
``` | output | 1 | 32,927 | 1 | 65,855 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
Input
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, ..., an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output
If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".
Examples
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
NO
Note
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | instruction | 0 | 32,928 | 1 | 65,856 |
Tags: dfs and similar, graphs, implementation
Correct Solution:
```
n, t = map(int, input().split())
a = list(map(int, input().split()))
visited = [False] * n
visited[0] = False
u = 0
while u < n-1:
visited[u] = True
u = u + a[u]
visited[u] = True
if visited[t-1]:
print("YES")
else:
print("NO")
``` | output | 1 | 32,928 | 1 | 65,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
Input
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, ..., an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output
If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".
Examples
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
NO
Note
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | instruction | 0 | 32,929 | 1 | 65,858 |
Tags: dfs and similar, graphs, implementation
Correct Solution:
```
n,t=[int(i) for i in input().split()]
lst=[int(i) for i in input().split()]
c=1
nk=0
while True:
if c<t:
c=lst[c-1]+c
elif c==t:
print("YES")
break
else:
nk=1
break
if nk==1:
print('NO')
``` | output | 1 | 32,929 | 1 | 65,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
Input
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, ..., an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output
If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".
Examples
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
NO
Note
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
Submitted Solution:
```
k=lambda:map(int,input().split())
t=list(k())[1];*l,=k();s=0
while s<t-1:s+=l[s]
print('YNEOS'[s>t-1::2])
``` | instruction | 0 | 32,930 | 1 | 65,860 |
Yes | output | 1 | 32,930 | 1 | 65,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a1, a2, ..., an - 1. For every integer i where 1 ≤ i ≤ n - 1 the condition 1 ≤ ai ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + ai), and one can travel from cell i to cell (i + ai) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + ai) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ ai ≤ n - i one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.
Input
The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 104) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.
The second line contains n - 1 space-separated integers a1, a2, ..., an - 1 (1 ≤ ai ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output
If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".
Examples
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
NO
Note
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
Submitted Solution:
```
b=[int(x) for x in input().split()]
n=b[0]
t=b[1]
a=[int(x) for x in input().split()]
tren=0
uvjet=0
t-=1
while True:
if tren==t:
print("YES")
uvjet=1
break
if tren>t:
break
else:
tren=tren+a[tren]
if uvjet==0:
print("NO")
``` | instruction | 0 | 32,931 | 1 | 65,862 |
Yes | output | 1 | 32,931 | 1 | 65,863 |
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