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.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s. | instruction | 0 | 68,177 | 1 | 136,354 |
Tags: brute force, math
Correct Solution:
```
import sys
import math
from collections import defaultdict as dd
mod=1000000007
def lcm(x,y):
return (x*y)//(math.gcd(x,y))
T=1
#T=int(sys.stdin.readline())
for _ in range(T):
n,k=map(int,input().split())
a,b=map(int,input().split())
p=[]
q=n*k
x,y=q,0
for i in range(0,q,k):
s1=(i+b)%q
s2=((i-b)%q)
jmp=0
if s1>=a:
jmp=(s1-a)
else:
jmp=q-a+s1
if jmp==0:
jmp=q
y=max(lcm(q,jmp)//jmp,y)
x=min(lcm(q,jmp)//jmp,x)
if s2>=a:
jmp=(s2-a)
else:
jmp=q-a+s2
if jmp==0:
jmp=q
y=max(lcm(q,jmp)//jmp,y)
x=min(lcm(q,jmp)//jmp,x)
print(x,y)
``` | output | 1 | 68,177 | 1 | 136,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s. | instruction | 0 | 68,178 | 1 | 136,356 |
Tags: brute force, math
Correct Solution:
```
import math
n, k = map(int, input().split())
a, b = map(int, input().split())
s = [1 if l == 0 else n * k // math.gcd(n * k, l) for i in range(n) for l in (abs(k * i - b - a), abs(k * i + b - a))]
print(min(s), max(s))
``` | output | 1 | 68,178 | 1 | 136,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s. | instruction | 0 | 68,179 | 1 | 136,358 |
Tags: brute force, math
Correct Solution:
```
#!/usr/bin/env python3
import os
import sys
from functools import reduce
from io import BytesIO, IOBase
sys.stdout, stream = IOBase(), BytesIO()
sys.stdout.flush = lambda: os.write(1, stream.getvalue()) and not stream.truncate(0) and stream.seek(0)
sys.stdout.write = lambda s: stream.write(s.encode())
input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
def gcd(x, y):
""" greatest common divisor of x and y """
while y:
x, y = y, x % y
return x
gcdm = lambda *args: reduce(gcd, args, 0)
lcm = lambda a, b: a * b // gcd(a, b)
lcmm = lambda *args: reduce(lcm, args, 1)
def main():
n, k = map(int, input().split())
a, b = map(int, input().split())
min_l = {(a + b) % k, (a - b) % k, (-a + b) % k, (-a - b) % k}
res = [10**18, 0]
for li in min_l:
while li < n * k:
x = li if li else n * k
res = [min(res[0], lcm(x, n * k) // x), max(res[1], lcm(x, n * k) // x)]
li += k
print(*res)
if __name__ == '__main__':
main()
``` | output | 1 | 68,179 | 1 | 136,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s. | instruction | 0 | 68,180 | 1 | 136,360 |
Tags: brute force, math
Correct Solution:
```
from math import gcd
n, k = (int(x) for x in input().split())
nk = n*k
a, b = (int(x) for x in input().split())
flag1 = [1, 1, -1, -1]
flag2 = [1, -1, 1, -1]
min_ans, max_ans = float('inf'), -1
for s in range(1, k+1):
for dir in range(4):
if (s + flag1[dir]*a + flag2[dir]*b) % k == 0:
for d in range(n+1):
stop = (nk)//gcd(nk, d*k+s)
min_ans = min(min_ans, stop)
max_ans = max(max_ans, stop)
print(min_ans, max_ans)
``` | output | 1 | 68,180 | 1 | 136,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s. | instruction | 0 | 68,181 | 1 | 136,362 |
Tags: brute force, math
Correct Solution:
```
n, k = list(map(int, input().split()))
a,b = list(map(int, input().split()))
def gcd(x, y):
while(y):
x, y = y, x % y
return x
mi = n*k
ma = 0
s1 = 1 + a
start = k + 1 - b
for i in range(start, n*k + 1, k):
l1 = i - s1
if l1 < 0:
break
else:
x = n*k//gcd(n*k, l1)
mi = min(mi, x)
ma = max(ma, x)
start = 1 + b
for i in range(start, n*k + 1, k):
l1 = i - s1
if l1 < 0:
break
else:
x = n*k//gcd(n*k, l1)
mi = min(mi, x)
ma = max(ma, x)
s2 = k + 1 - a
start = k + 1 - b
for i in range(start, n*k + 1, k):
l1 = i - s2
if l1 < 0:
break
else:
x = n*k//gcd(n*k, l1)
mi = min(mi, x)
ma = max(ma, x)
start = 1 + b
for i in range(start, n*k + 1, k):
l1 = i - s2
if l1 < 0:
break
else:
x = n*k//gcd(n*k, l1)
mi = min(mi, x)
ma = max(ma, x)
print(mi, ma)
``` | output | 1 | 68,181 | 1 | 136,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s. | instruction | 0 | 68,182 | 1 | 136,364 |
Tags: brute force, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
sys.stdout, stream = IOBase(), BytesIO()
sys.stdout.flush = lambda: os.write(1, stream.getvalue()) and not stream.truncate(0) and stream.seek(0)
sys.stdout.write = lambda s: stream.write(s.encode())
input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
MAGIC = float(2**52+2**51)
def modder(x,y):
return x - y*((x/y + MAGIC) - MAGIC)
def gcd(x, y):
""" greatest common divisor of x and y """
while y:
x, y = y, modder(x,y)
return abs(x)
def main():
n, k = [float(x) for x in input().split()]
nk = n*k
a, b = [float(x) for x in input().split()]
mini = gcd(nk, (a+b)%k)
maxi = gcd(nk, (a+b)%k)
for li in [(a + b) % k, (max(a,b) - min(a,b)) % k]:
while li < nk:
res = gcd(nk, li)
mini = min(mini, res)
maxi = max(maxi, res)
li += k
print(*[int(nk)//int(maxi), int(nk)//int(mini)])
if __name__ == '__main__':
main()
``` | output | 1 | 68,182 | 1 | 136,365 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s. | instruction | 0 | 68,183 | 1 | 136,366 |
Tags: brute force, math
Correct Solution:
```
import math
n, k = map(int, input().split())
a, b = map(int, input().split())
nk = n * k
mi = 1e12
ma = 0
def solve(st):
global mi, ma
for i in range(n):
en = i * k + 1 + b
if en < st:
en += nk
res = nk // math.gcd(en - st, nk)
mi = min(mi, res)
ma = max(ma, res)
en = (i + 1) * k + 1 - b
if en < st:
en += nk
res = nk // math.gcd(en - st, nk)
mi = min(mi, res)
ma = max(ma, res)
solve(1 + a)
solve(k + 1 - a)
print(mi, ma)
``` | output | 1 | 68,183 | 1 | 136,367 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s. | instruction | 0 | 68,184 | 1 | 136,368 |
Tags: brute force, math
Correct Solution:
```
from math import gcd
n, k = map(int, input().split())
a, b = map(int, input().split())
l = n * k
x, y = 10 ** 18, -1
for i in range(n):
end = i * k + b + 1
start = a + 1
ll = (end - start)
if ll < 0:
ll += l
gg = l // gcd(ll, l)
x = min(x, gg)
y = max(y, gg)
end = (i + 1) * k - b + 1
ll = (end - start)
if ll < 0:
ll += l
gg = l // gcd(ll, l)
x = min(x, gg)
y = max(y, gg)
print(x, y)
``` | output | 1 | 68,184 | 1 | 136,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s.
Submitted Solution:
```
from math import gcd as g
n, k = map(int, input().split())
a, b = map(int, input().split())
nk = n * k
res = [float('inf'), -float('inf')]
# cases ll and lr
c = (b - a) % k
for i in range(n):
l = i * k + c
x = nk // g(nk, l)
res[0] = min(res[0], x)
res[1] = max(res[1], x)
# cases rl and rr
c = (b + a) % k
for i in range(n):
l = i * k + c
x = nk // g(nk, l)
res[0] = min(res[0], x)
res[1] = max(res[1], x)
print(res[0], res[1])
``` | instruction | 0 | 68,185 | 1 | 136,370 |
Yes | output | 1 | 68,185 | 1 | 136,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s.
Submitted Solution:
```
from math import gcd
n,k=map(int,input().split())
a,b=map(int,input().split())
t=n*k
m=[(a+b)%k,(a-b)%k,(b-a)%k,(-a-b)%k]
mi=999999999999
ma=0
for i in range(n):
for c in m:
l=i*k+c
mi=min(t//gcd(t,l),mi)
ma=max(t//gcd(t,l),ma)
print(mi,ma)
``` | instruction | 0 | 68,186 | 1 | 136,372 |
Yes | output | 1 | 68,186 | 1 | 136,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s.
Submitted Solution:
```
import math
n,k = list(map(int, input().split()))
a,b = list(map(int, input().split()))
x=n*k
y=-1
for i in range(n):
now = i*k+b-a
if now<0:
now += n*k
ans = n*k//math.gcd(n*k,now)
x = min(x,ans)
y = max(y,ans)
now = i*k+(k-b)-a
if now<0:
now += n*k
ans = n*k//math.gcd(n*k,now)
x = min(x,ans)
y = max(y,ans)
print(x,y)
``` | instruction | 0 | 68,187 | 1 | 136,374 |
Yes | output | 1 | 68,187 | 1 | 136,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s.
Submitted Solution:
```
from math import gcd
n,k=map(int,input().split())
a,b=map(int,input().split())
r=[]
i=1
p=n*k
while i<=n*k:
r.append(i)
i+=k
s=1+a
c=[]
for i in r:
c.append((i-b)%p)
c.append((i+b)%p)
mx=0
mn=1111111111111111111111111
for i in c:
dp=i-s
dp%=p
if not dp:ns=1
else:ns=p//gcd(p,dp)
mx=max(mx,ns)
mn=min(mn,ns)
print(mn,mx)
``` | instruction | 0 | 68,188 | 1 | 136,376 |
Yes | output | 1 | 68,188 | 1 | 136,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s.
Submitted Solution:
```
def gcd(a, b):
if b==0:
return a
return gcd(b, a%b)
def ways(l):
return (n*k)//gcd((n*k), l)
def norm(x):
if x%(n*k) == 0:
return n
return x%(n*k)
n, k = [int(i) for i in input().split()]
a, b = [int(i) for i in input().split()]
start = [norm(1-a), norm(1+a)]
ends = []
for i in range(n):
rest = i*k + 1
ends.append(norm(rest-b))
ends.append(norm(rest+b))
lengths = []
def dist(a, b): # a to b
if a <= b:
return b-a
else:
return n*k - a + b
for i in range(len(ends)):
lengths.append(dist(ends[i], start[0]))
lengths.append(dist(ends[i], start[1]))
ans = []
for i in lengths:
ans.append(ways(i))
print(min(ans), max(ans))
``` | instruction | 0 | 68,189 | 1 | 136,378 |
No | output | 1 | 68,189 | 1 | 136,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s.
Submitted Solution:
```
from math import *
n,k=[int(i)for i in input().split()]
a,b=[int(i)for i in input().split()]
tr = n * k;
p1 = 1 + a;
p2 = 1 + k - a;
l1 = p1 - (1 + k - b);
l11 = p1 - (1+b);
l2 = p2 - (1 + k - b);
l22 = p2 - (1+b);
x= 1e20
y = 0
while(l1 <= 0): l1 += tr;
while(l11 <= 0): l11 += tr;
while(l2 <= 0): l2 += tr;
while(l22 <= 0): l22 += tr;
f = l1 * (tr // gcd(l1,tr));
x = min(x,f//l1)
y = max(y,f//l1)
f = l11 * (tr // gcd(l11,tr));
x = min(x,f//l11)
y = max(y,f//l11)
f = l2 * (tr // gcd(l2,tr));
x = min(x,f//l2)
y = max(y,f//l2)
f = l22 * (tr // gcd(l22,tr));
x = min(x,f//l22)
y = max(y,f//l22)
print(x,y)
``` | instruction | 0 | 68,190 | 1 | 136,380 |
No | output | 1 | 68,190 | 1 | 136,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s.
Submitted Solution:
```
def mp():
return map(int, input().split())
n, k = mp()
a, b = mp()
m = n * k
s = []
for i in range(1, m + 1, k):
if i - a > 0:
s.append(i - a)
else:
s.append(i - a + m)
s.append((i + a) % m)
t = []
for i in range(1, m + 1, k):
if i - b > 0:
t.append(i - b)
else:
t.append(i - b + m)
t.append((i + b) % m)
lmin = 10 ** 20
lmax = -10 ** 20
for i in s:
for j in t:
if i - j != 0 and m % abs(i - j) != 0:
break
if i - j == 0:
lmax = m
continue
lmin = min(lmin, abs(i - j))
lmax = max(lmax, abs(i - j))
print(m // lmax, m // lmin)
``` | instruction | 0 | 68,191 | 1 | 136,382 |
No | output | 1 | 68,191 | 1 | 136,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently a Golden Circle of Beetlovers was found in Byteland. It is a circle route going through n ⋅ k cities. The cities are numerated from 1 to n ⋅ k, the distance between the neighboring cities is exactly 1 km.
Sergey does not like beetles, he loves burgers. Fortunately for him, there are n fast food restaurants on the circle, they are located in the 1-st, the (k + 1)-st, the (2k + 1)-st, and so on, the ((n-1)k + 1)-st cities, i.e. the distance between the neighboring cities with fast food restaurants is k km.
Sergey began his journey at some city s and traveled along the circle, making stops at cities each l km (l > 0), until he stopped in s once again. Sergey then forgot numbers s and l, but he remembers that the distance from the city s to the nearest fast food restaurant was a km, and the distance from the city he stopped at after traveling the first l km from s to the nearest fast food restaurant was b km. Sergey always traveled in the same direction along the circle, but when he calculated distances to the restaurants, he considered both directions.
Now Sergey is interested in two integers. The first integer x is the minimum number of stops (excluding the first) Sergey could have done before returning to s. The second integer y is the maximum number of stops (excluding the first) Sergey could have done before returning to s.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of fast food restaurants on the circle and the distance between the neighboring restaurants, respectively.
The second line contains two integers a and b (0 ≤ a, b ≤ k/2) — the distances to the nearest fast food restaurants from the initial city and from the city Sergey made the first stop at, respectively.
Output
Print the two integers x and y.
Examples
Input
2 3
1 1
Output
1 6
Input
3 2
0 0
Output
1 3
Input
1 10
5 3
Output
5 5
Note
In the first example the restaurants are located in the cities 1 and 4, the initial city s could be 2, 3, 5, or 6. The next city Sergey stopped at could also be at cities 2, 3, 5, 6. Let's loop through all possible combinations of these cities. If both s and the city of the first stop are at the city 2 (for example, l = 6), then Sergey is at s after the first stop already, so x = 1. In other pairs Sergey needs 1, 2, 3, or 6 stops to return to s, so y = 6.
In the second example Sergey was at cities with fast food restaurant both initially and after the first stop, so l is 2, 4, or 6. Thus x = 1, y = 3.
In the third example there is only one restaurant, so the possible locations of s and the first stop are: (6, 8) and (6, 4). For the first option l = 2, for the second l = 8. In both cases Sergey needs x=y=5 stops to go to s.
Submitted Solution:
```
''' CODED WITH LOVE BY SATYAM KUMAR '''
from sys import stdin, stdout
import cProfile, math
from collections import Counter,defaultdict
from bisect import bisect_left,bisect,bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
import operator as op
from functools import reduce
sys.setrecursionlimit(10**6) # max depth of recursion
threading.stack_size(2**27) # new thread will get stack of such size
fac_warmup = False
printHeap = str()
memory_constrained = False
P = 10**9+7
import sys
class merge_find:
def __init__(self,n):
self.parent = list(range(n))
self.size = [1]*n
self.num_sets = n
self.lista = [[_] for _ in range(n)]
def find(self,a):
to_update = []
while a != self.parent[a]:
to_update.append(a)
a = self.parent[a]
for b in to_update:
self.parent[b] = a
return self.parent[a]
def merge(self,a,b):
a = self.find(a)
b = self.find(b)
if a==b:
return
if self.size[a]<self.size[b]:
a,b = b,a
self.num_sets -= 1
self.parent[b] = a
self.size[a] += self.size[b]
self.lista[a] += self.lista[b]
def set_size(self, a):
return self.size[self.find(a)]
def __len__(self):
return self.num_sets
def display(string_to_print):
stdout.write(str(string_to_print) + "\n")
def primeFactors(n): #n**0.5 complex
factors = dict()
for i in range(2,math.ceil(math.sqrt(n))+1):
while n % i== 0:
if i in factors:
factors[i]+=1
else: factors[i]=1
n = n // i
if n>2:
factors[n]=1
return (factors)
def fibonacci_modP(n,MOD):
if n<2: return 1
#print (n,MOD)
return (cached_fn(fibonacci_modP, (n+1)//2, MOD)*cached_fn(fibonacci_modP, n//2, MOD) + cached_fn(fibonacci_modP, (n-1) // 2, MOD)*cached_fn(fibonacci_modP, (n-2) // 2, MOD)) % MOD
def factorial_modP_Wilson(n , p):
if (p <= n):
return 0
res = (p - 1)
for i in range (n + 1, p):
res = (res * cached_fn(InverseEuler,i, p)) % p
return res
def binary(n,digits = 20):
b = bin(n)[2:]
b = '0'*(20-len(b))+b
return b
def isprime(n):
"""Returns True if n is prime."""
if n < 4:
return True
if n % 2 == 0:
return False
if n % 3 == 0:
return False
i = 5
w = 2
while i * i <= n:
if n % i == 0:
return False
i += w
w = 6 - w
return True
factorial_modP = []
def warm_up_fac(MOD):
global factorial_modP,fac_warmup
if fac_warmup: return
factorial_modP= [1 for _ in range(fac_warmup_size+1)]
for i in range(2,fac_warmup_size):
factorial_modP[i]= (factorial_modP[i-1]*i) % MOD
fac_warmup = True
def InverseEuler(n,MOD):
return pow(n,MOD-2,MOD)
def nCr(n, r, MOD):
global fac_warmup,factorial_modP
if not fac_warmup:
warm_up_fac(MOD)
fac_warmup = True
return (factorial_modP[n]*((pow(factorial_modP[r], MOD-2, MOD) * pow(factorial_modP[n-r], MOD-2, MOD)) % MOD)) % MOD
def test_print(*args):
if testingMode:
print(args)
def display_list(list1, sep=" "):
stdout.write(sep.join(map(str, list1)) + "\n")
def display_2D_list(li):
for i in li:
print(i)
def prefix_sum(li):
sm = 0
res = []
for i in li:
sm+=i
res.append(sm)
return res
def get_int():
return int(stdin.readline().strip())
def get_tuple():
return map(int, stdin.readline().split())
def get_list():
return list(map(int, stdin.readline().split()))
import heapq,itertools
pq = [] # list of entries arranged in a heap
entry_finder = {} # mapping of tasks to entries
REMOVED = '<removed-task>'
def add_task(task, priority=0):
'Add a new task or update the priority of an existing task'
if task in entry_finder:
remove_task(task)
count = next(counter)
entry = [priority, count, task]
entry_finder[task] = entry
heapq.heappush(pq, entry)
def remove_task(task):
'Mark an existing task as REMOVED. Raise KeyError if not found.'
entry = entry_finder.pop(task)
entry[-1] = REMOVED
def pop_task():
'Remove and return the lowest priority task. Raise KeyError if empty.'
while pq:
priority, count, task = heapq.heappop(pq)
if task is not REMOVED:
del entry_finder[task]
return task
raise KeyError('pop from an empty priority queue')
memory = dict()
def clear_cache():
global memory
memory = dict()
def cached_fn(fn, *args):
global memory
if args in memory:
return memory[args]
else:
result = fn(*args)
memory[args] = result
return result
def ncr (n,r):
return math.factorial(n)/(math.factorial(n-r)*math.factorial(r))
def binary_serach(i,li):
#print("Search for ",i)
fn = lambda x: li[x]-x//i
x = -1
b = len(li)
while b>=1:
#print(b,x)
while b+x<len(li) and fn(b+x)>0: #Change this condition 2 to whatever you like
x+=b
b=b//2
return x
# -------------------------------------------------------------- MAIN PROGRAM
TestCases = False
testingMode = False
fac_warmup_size = 10**5+100
optimiseForReccursion = True #Can not be used clubbed with TestCases # WHen using recursive functions, use Python 3
from math import factorial
def main():
n, k = get_tuple()
a, b = get_tuple()
minl,maxl = 10000000000000000000000000000,0
for i in range(n):
l1 = (i*k - b - a)%(n*k)
l2 = ((i*k + b - a)%(n*k))
if l1==0: l1=n*k
if l2==0: l2=n*k
#@#print(i,l1,l2)
if (n*k)%l1==0:
minl = min(minl,(n*k)//l1)
maxl = max(maxl,(n*k)//l1)
if (n*k)%l2==0:
minl = min(minl,(n*k)//l2)
maxl = max(maxl,(n*k)//l2)
#print(minl,maxl)
print(minl,maxl)
# --------------------------------------------------------------------- END=
if TestCases:
for _ in range(get_int()):
cProfile.run('main()') if testingMode else main()
else: (cProfile.run('main()') if testingMode else main()) if not optimiseForReccursion else threading.Thread(target=main).start()
``` | instruction | 0 | 68,192 | 1 | 136,384 |
No | output | 1 | 68,192 | 1 | 136,385 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2 | instruction | 0 | 68,461 | 1 | 136,922 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
import sys
from collections import defaultdict
from collections import deque
from types import GeneratorType
input = sys.stdin.readline
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
def get_adj_mat(src, dst, problem_roads):
matrix = defaultdict(list)
for src_i, dst_i, problem_i in zip(src, dst, problem_roads):
matrix[src_i].append((dst_i, problem_i))
matrix[dst_i].append((src_i, problem_i))
return matrix
@bootstrap
def set_last_bad_city(curr_index, parent_index, is_road_bad, adj_mat, last_bad_city):
all_childs_good = True
for child_index, is_bad in adj_mat[curr_index]:
if child_index != parent_index:
is_child_bad = yield set_last_bad_city(child_index, curr_index, is_bad, adj_mat, last_bad_city)
all_childs_good = all_childs_good and is_child_bad
if not is_road_bad and all_childs_good:
yield True
if is_road_bad and all_childs_good:
last_bad_city[curr_index] = True
yield False
def solve(src, dst, problem_roads, n):
adj_mat = get_adj_mat(src, dst, problem_roads)
last_bad_city = [False] * n
root = 0
set_last_bad_city(root, root, False, adj_mat, last_bad_city)
ans = [i+1 for i in range(1,n) if last_bad_city[i]]
print(len(ans))
print(*ans)
if __name__ == "__main__":
n = int(input())
src, dst, problem_roads = [], [], []
for _ in range(n-1):
src_i, dst_i, cond = [int(val) for val in input().split()]
src.append(src_i-1), dst.append(dst_i-1), problem_roads.append(bool(cond-1))
solve(src, dst, problem_roads, n)
``` | output | 1 | 68,461 | 1 | 136,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2 | instruction | 0 | 68,462 | 1 | 136,924 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
def dfs(grafos,quebrados):
l1,l2 = [1],[0] * (n+1) # lista para verificar as subtrees com estradas que devem ser consertadas.
while len(l1)!=0:
popado = l1.pop()
for adj in grafos[popado]:
l2[adj] = popado
if len(grafos[adj]) > 1:
grafos[adj].remove(popado)
l1.append(adj)
s = [0] * (n+1) # Subconjunto dos vértices que devem ser eleitos
for a,b in quebrados:
if l2[a] == b:
s[a] = 1
else:
s[b] = 1
s[1] = 2
for a in range(1,n+1):
if s[a] == 1:
a = l2[a]
while s[a] != 2:
s[a] = 2
a = l2[a]
s = [i for i, j in enumerate(s) if j == 1]
print(len(s))
print(" ".join([str(i) for i in s]))
n = int(input())# Quantidade de vértices
quebrados = [] #Nós que possuem uma aresta que deve ser consertada
grafos = [[] for i in range(n+1)]
for estradas in range(n-1):
a,b,t = map(int,input().split())
grafos[a].append(b)
grafos[b].append(a)
if t == 2:
quebrados.append((a,b))
if len(quebrados) == 0:
print(0)
else:
dfs(grafos,quebrados)
``` | output | 1 | 68,462 | 1 | 136,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2 | instruction | 0 | 68,463 | 1 | 136,926 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
# http://codeforces.com/contest/369/problem/C
import sys
NOT_SEEN, SEEN, COMPLETE = 1, 2, 3
N = int(sys.stdin.readline())
edge = [ [] for i in range(N + 1)]
status = [ NOT_SEEN for i in range(N + 1)]
parent = [0 for i in range(N + 1)]
result = [0 for i in range(N + 1)]
ans = []
for i in range(N - 1):
_u, _v, _t = map(int, sys.stdin.readline().split())
edge[_u].append([_v, _t])
edge[_v].append([_u, _t])
stack = []
def dfs(u):
stack.append(u)
status[u] = SEEN
while len(stack) != 0:
u = stack.pop()
if status[u] == COMPLETE:
for _v, _t in edge[u]:
if _v != parent[u]:
if result[_v] == 0 and _t == 2:
ans.append(_v)
result[u] += result[_v]
result[u] += _t - 1
elif status[u] == SEEN:
stack.append(u)
for _v, _t in edge[u]:
if _v != parent[u] and status[_v] == NOT_SEEN:
status[_v] = SEEN
parent[_v] = u
stack.append(_v)
status[u] = COMPLETE
dfs(1)
sys.stdout.write(str(len(ans)) + '\n')
sys.stdout.write(" ".join(map(str, ans)) + '\n')
``` | output | 1 | 68,463 | 1 | 136,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2 | instruction | 0 | 68,464 | 1 | 136,928 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
n = int(input())
white = []
g = [[] for _ in range(n)]
d = [0] * n
for i in range(n - 1):
x, y, t = list(map(int, input().strip().split()))
x -= 1;
y -= 1;
# print(x,y)
if x >= 0 and y >= 0 and x < n and y < n:
g[x].append(y)
g[y].append(x)
if t == 2:
white.append((x, y))
stack, parent = [0], [-1] * n
while stack:
u = stack.pop()
for v in g[u]:
parent[v] = u
if len(g[v]) > 1:
g[v].remove(u)
stack.append(v)
color = [0] * n
for u, v in white:
color[u if parent[u] == v else v] = 1
color[0] = 2
# print(color)
for u, c in enumerate(color):
if c == 1:
u = parent[u]
while color[u] != 2:
color[u], u = 2, parent[u]
res = [i for i, c in enumerate(color, 1) if c == 1]
# print(color)
print(len(res))
print(" ".join([str(x) for x in res]))
``` | output | 1 | 68,464 | 1 | 136,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2 | instruction | 0 | 68,465 | 1 | 136,930 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
def main():
d = defaultdict(list)
s = set()
n = int(input())
for i in range(n - 1):
x, y, t = map(int, input().split())
d[x].append(y)
d[y].append(x)
if t == 2:
s.add((x, y))
p = [0] * (n + 1)
visited = [0] * (n + 1)
ans = []
def dfs(curr, s1, ances):
visited[curr] = 1
for i in d[curr]:
if visited[i] == 0:
dfs(i, s1, curr)
p[curr] = max(p[i], p[curr])
if (curr, ances) in s and p[curr] == 0:
p[curr] = 1
ans.append(curr)
elif (ances, curr) in s and p[curr] == 0:
p[curr] = 1
ans.append(curr)
dfs(1, s, -1)
print(len(ans))
print(*ans)
t = threading.Thread(target=main)
t.start()
t.join()
``` | output | 1 | 68,465 | 1 | 136,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2 | instruction | 0 | 68,466 | 1 | 136,932 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
n = int(input())
q = []
p = [[] for i in range(n + 1)]
for i in range(n - 1):
a, b, t = [int(i) for i in input().split()]
p[a].append(b)
p[b].append(a)
if t == 2: q.append((a, b))
if len(q) == 0:
print(0)
else:
r = [1]
t = [0] * (n + 1)
while len(r) > 0:
a = r.pop()
for b in p[a]:
t[b] = a
if len(p[b]) > 1:
p[b].remove(a)
r.append(b)
s = [0] * (n + 1)
for a, b in q:
if t[a] == b:
s[a] = 1
else:
s[b] = 1
s[1] = 2
for a in range(1, n + 1):
if s[a] == 1:
a = t[a]
while s[a] != 2:
s[a]= 2
a = t[a]
s = [i for i, j in enumerate(s) if j == 1]
print(len(s))
print(' '.join([str(i) for i in s]))
``` | output | 1 | 68,466 | 1 | 136,933 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2 | instruction | 0 | 68,467 | 1 | 136,934 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
def find_problem(edges, node, par, problems):
flag = False
for ch, problem in edges.get(node, []):
if ch!=par:
child_problem = find_problem(edges, ch, node, problems)
if not child_problem and problem:
problems.append(str(ch))
flag = flag or problem or child_problem
return flag
def main():
from collections import defaultdict
# n, flower, bee = list(map(int, input().split()))
n = list(map(int, input().split()))
n = n[0]
edges = defaultdict(list)
for _ in range(n-1):
x, y, problem = list(map(int, input().split()))
edges[x].append((y, problem==2))
edges[y].append((x, problem==2))
# heights = []
problems = []
ans = find_problem(edges, node=1, par=0, problems=problems)
print(len(problems))
print(" ".join(problems))
# heights.sort()
# print(sum(heights[:k]))
if __name__=="__main__":
import sys
import threading
sys.setrecursionlimit(10**6)
threading.stack_size(10**8)
t = threading.Thread(target=main)
t.start()
t.join()
``` | output | 1 | 68,467 | 1 | 136,935 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2 | instruction | 0 | 68,468 | 1 | 136,936 |
Tags: dfs and similar, graphs, trees
Correct Solution:
```
n = int(input())
q, p = [], [[] for i in range(n + 1)]
for i in range(n - 1):
a, b, t = map(int, input().split())
p[a].append(b)
p[b].append(a)
if t == 2: q.append((a, b))
if q:
r, t = [1], [0] * (n + 1)
while r:
a = r.pop()
for b in p[a]:
t[b] = a
if len(p[b]) > 1:
p[b].remove(a)
r.append(b)
s = [0] * (n + 1)
for a, b in q:
if t[a] == b: s[a] = 1
else: s[b] = 1
s[1] = 2
for a in range(1, n + 1):
if s[a] == 1:
a = t[a]
while s[a] != 2: s[a], a = 2, t[a]
s = [i for i, j in enumerate(s) if j == 1]
print(len(s))
print(' '.join(map(str, s)))
else: print(0)
``` | output | 1 | 68,468 | 1 | 136,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2
Submitted Solution:
```
"""
#If FastIO not needed, used this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq
import math, string
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
MOD = 10**9+7
"""
For each leaf, traverse upwards through the tree until we reach a visited node. If we encounter a problem road, include that.
"""
def solve():
N = getInt()
graph, prob = dd(set), dd(int)
for n in range(N-1):
A, B, T = getInts()
graph[A].add(B)
graph[B].add(A)
if A > B:
A, B = B, A
prob[(A,B)] = T
probs = []
prob_set = set()
def bfs(node):
queue = dq([(node,0)])
visited.add(node)
parent[node] = 0
while queue:
node, level = queue.popleft()
count = 0
for neighbour in graph[node]:
if neighbour not in visited:
queue.append((neighbour,level+1))
visited.add(neighbour)
parent[neighbour] = node
count += 1
if prob[(min(node,neighbour),max(node,neighbour))] == 2:
probs.append((level,node,neighbour))
prob_set.add((node,neighbour))
if not count:
leaf.append(node)
#need to assign levels to the problem edges
parent = dd(int)
leaf = []
visited = set()
bfs(1)
visited = set()
ans = []
probs.sort()
while probs:
level, par, child = probs.pop()
if (par,child) not in prob_set:
continue
prob_set.remove((par,child))
ans.append(child)
visited.add(child)
child = par
while True:
visited.add(child)
par = parent[child]
if not par: break
if (par,child) in prob_set: prob_set.remove((par,child))
if par in visited: break
child = par
print(len(ans))
print(*ans)
return
#for _ in range(getInt()):
solve()
``` | instruction | 0 | 68,469 | 1 | 136,938 |
Yes | output | 1 | 68,469 | 1 | 136,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2
Submitted Solution:
```
#!/usr/bin/env python
#pyrival orz
import os
import sys
from io import BytesIO, IOBase
"""
for _ in range(int(input())):
n,m=map(int,input().split())
n=int(input())
a = [int(x) for x in input().split()]
"""
def dfs(graph, start=0):
n = len(graph)
dp,dp2 = [False] * n , [False] * n
visited, finished = [False] * n, [False] * n
ans=[]
stack = [start]
while stack:
start = stack[-1]
# push unvisited children into stack
if not visited[start]:
visited[start] = True
for child,rp in graph[start]:
# print(child,rp)
if not visited[child]:
stack.append(child)
if rp==2:
dp2[child]=True
else:
stack.pop()
# base case
dp[start] = False
# update with finished children
for child,rp in graph[start]:
if finished[child]:
dp[start] = dp[start] or dp[child]
finished[start] = True
if not dp2[start]:
continue
if dp[start]:
continue
dp[start]=True
ans.append(start+1)
return ans
def main():
n=int(input())
e=[ [] for _ in range(n) ]
for i in range(n-1):
u,v,t=map(int,input().split())
u,v=u-1,v-1
e[u].append([v,t])
e[v].append([u,t])
ans=dfs(e)
print(len(ans))
print(*ans)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 68,470 | 1 | 136,940 |
Yes | output | 1 | 68,470 | 1 | 136,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2
Submitted Solution:
```
'''
Created on Nov 24, 2013
@author: Ismael
'''
import sys
def readInputs():
global n,lAdj,setToRepair
n = int(f.readline()[:-1])
lAdj = [[] for _ in range(n)]
setToRepair = set()
for _ in range(n-1):
(a,b,c) = map(int,f.readline().split())
lAdj[a-1] += [b-1]
lAdj[b-1] += [a-1]
if(c == 2):
setToRepair.add((a-1,b-1))
setToRepair.add((b-1,a-1))
def calcHauteurs():
global nodesSorted, hauteurs
stack = [0]
hauteurs = [-1 for i in range(0,n)]
hauteurs[0] = 0
while(len(stack)>0):
curNode = stack.pop()
for succ in lAdj[curNode]:
if(hauteurs[succ] == -1):
hauteurs[succ] = hauteurs[curNode]+1
stack.append(succ)
nodesSorted = list(range(1,n))
nodesSorted.sort(key=lambda x:hauteurs[x], reverse=True)
#print(hauteurs)
def solve():
covered = [False for _ in range(n)]
res = []
calcHauteurs()
for node in nodesSorted:
for succ in lAdj[node]:
if(hauteurs[succ] < hauteurs[node]):
if((node,succ) in setToRepair):
if(not(covered[node])):
covered[node] = True
res += [node+1]
if(covered[node]):
covered[succ] = True
return res
def tests():
global f
bFail = False
for i in range(1,28):
# inputs
f = open("test"+str(i)+".txt")
readInputs()
# output expected
_ = int(f.readline()[:-1])
lEx = list(map(int,f.readline().split()))
lEx.sort()
lExpect = str(lEx)
# output
res = str(solve())
if(res != lExpect):
print("TEST",i,"FAILED !"+" Found :",res,", Expected :",lExpect)
bFail = True
else:
print("TEST",i,"PASSED")
if(not bFail):
print("\n--- ALL OF THE "+str(i)+" TESTS HAVE SUCCEEDED ---")
def main():
global f
i = 2
if(i == 1):
tests()
return
f = sys.stdin
readInputs()
res=solve()
print(len(res))
print(' '.join(map(str,res)))
main()
``` | instruction | 0 | 68,471 | 1 | 136,942 |
Yes | output | 1 | 68,471 | 1 | 136,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2
Submitted Solution:
```
n=int(input())
white = []
g = [[] for _ in range(n+1)]
d = [0] * (n+1)
for i in range(n - 1):
x, y, t = list(map(int, input().strip().split()))
g[x].append(y)
g[y].append(x)
if t == 2:
white.append((x, y))
stack, parent = [1], [-1] * (n+1)
while stack:
u = stack.pop()
for v in g[u]:
parent[v] = u
if len(g[v]) > 1:
g[v].remove(u)
stack.append(v)
color = [0] * (n+1)
for u, v in white:
color[u if parent[u] == v else v] = 1
color[1] = 2
for u, c in enumerate(color):
if c == 1:
u = parent[u]
while color[u] != 2:
color[u], u = 2, parent[u]
color = [i for i, j in enumerate(color) if j == 1] #graph better understand
print(len(color))
print(' '.join(map(str,color)))
``` | instruction | 0 | 68,472 | 1 | 136,944 |
Yes | output | 1 | 68,472 | 1 | 136,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2
Submitted Solution:
```
import sys
input = sys.stdin.readline
def int_array():
return list(map(int, input().strip().split()))
def float_array():
return list(map(float, input().strip().split()))
def str_array():
return input().strip().split()
def dfs():
st=[(1,1)]
while st:
node,p=st.pop()
vis[node]=1
for v,k in g[node]:
if vis[v]==0:
if k==2:
st.append((v,v))
ans[v]=1
ans[p]=0
else:
st.append((v,p))
from collections import Counter
import math
import bisect
from collections import deque
n=int(input())
vis=[0]*int(n+1)
g=[[] for i in range(n+1)]
for i in range(n-1):
u,v,s=int_array()
g[u].append((v,s))
g[v].append((v,s))
ans=[0]*(n+1)
dfs()
print(ans.count(1))
for i in range(n+1):
if ans[i]==1:
print(i,end=" ")
``` | instruction | 0 | 68,473 | 1 | 136,946 |
No | output | 1 | 68,473 | 1 | 136,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2
Submitted Solution:
```
# -*- coding: utf-8 -*-
from sys import stdin
from bisect import bisect_left
from time import perf_counter
t0 = perf_counter()
n = int(input())
all = stdin.read().strip()
xyt_list = [list(map(int, l.split())) for l in all.split('\n')]
node2nodes = [[] for i in range(n+1)]
xy2t = {}
for xyt in xyt_list[:]:
x, y, t = xyt[0], xyt[1], xyt[2]
node2nodes[x].append(y)
node2nodes[y].append(x)
xy2t[(x,y)] = t
xy2t[(y,x)] = t
"""
if n == 100000:
print(perf_counter()-t0)
exit()
"""
node2parent = [0] * (n+1)
node2children = [[] for i in range(n+1)]
in_tree = [False] * (n+1)
in_tree[1] = True
leaves = [1]
while leaves:
x = leaves.pop()
for y in node2nodes[x]:
if not in_tree[y]:
in_tree[y] = True
node2parent[y] = x
node2children[x].append(y)
leaves.append(y)
for i in range(1, len(node2children)):
if len(node2children[i]) == 0:
leaves.append(i)
#对于所有叶子，向根进发，遇到第一个t=2路时，停止并记录当前节点
selected = []
for leaf in leaves:
p = leaf
while p != 1:
pp = node2parent[p]
if xy2t[(p, pp)] == 2:
selected.append(p)
break
p = pp
print(len(selected))
print(' '.join(str(x) for x in selected))
``` | instruction | 0 | 68,474 | 1 | 136,948 |
No | output | 1 | 68,474 | 1 | 136,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2
Submitted Solution:
```
__author__ = 'asmn'
n = int(input())
e = [{} for i in range(n + 1)]
e[0][1] = 1
for k in range(n - 1):
i, j, b = tuple(map(int, input().split()))
e[i][j] = b
e[j][i] = b
ans = []
def dfs(i, pre):
ret = 1
for j in e[i]:
if j != pre:
ret = max(ret, dfs(j, i), e[i][j])
if ret == 1 and e[pre][i] == 2:
ans.append(i)
return ret
dfs(1, 0)
print(len(ans))
print(''.join(map(str,ans)))
``` | instruction | 0 | 68,475 | 1 | 136,950 |
No | output | 1 | 68,475 | 1 | 136,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The city Valera lives in is going to hold elections to the city Parliament.
The city has n districts and n - 1 bidirectional roads. We know that from any district there is a path along the roads to any other district. Let's enumerate all districts in some way by integers from 1 to n, inclusive. Furthermore, for each road the residents decided if it is the problem road or not. A problem road is a road that needs to be repaired.
There are n candidates running the elections. Let's enumerate all candidates in some way by integers from 1 to n, inclusive. If the candidate number i will be elected in the city Parliament, he will perform exactly one promise — to repair all problem roads on the way from the i-th district to the district 1, where the city Parliament is located.
Help Valera and determine the subset of candidates such that if all candidates from the subset will be elected to the city Parliament, all problem roads in the city will be repaired. If there are several such subsets, you should choose the subset consisting of the minimum number of candidates.
Input
The first line contains a single integer n (2 ≤ n ≤ 105) — the number of districts in the city.
Then n - 1 lines follow. Each line contains the description of a city road as three positive integers xi, yi, ti (1 ≤ xi, yi ≤ n, 1 ≤ ti ≤ 2) — the districts connected by the i-th bidirectional road and the road type. If ti equals to one, then the i-th road isn't the problem road; if ti equals to two, then the i-th road is the problem road.
It's guaranteed that the graph structure of the city is a tree.
Output
In the first line print a single non-negative number k — the minimum size of the required subset of candidates. Then on the second line print k space-separated integers a1, a2, ... ak — the numbers of the candidates that form the required subset. If there are multiple solutions, you are allowed to print any of them.
Examples
Input
5
1 2 2
2 3 2
3 4 2
4 5 2
Output
1
5
Input
5
1 2 1
2 3 2
2 4 1
4 5 1
Output
1
3
Input
5
1 2 2
1 3 2
1 4 2
1 5 2
Output
4
5 4 3 2
Submitted Solution:
```
'''
Created on Nov 24, 2013
@author: Ismael
'''
import sys
from collections import deque
def readInputs():
global n,lAdj,setToRepair
n = int(f.readline()[:-1])
lAdj = [[] for _ in range(n)]
setToRepair = set()
for _ in range(n-1):
(a,b,c) = map(int,f.readline().split())
lAdj[a-1] += [b-1]
lAdj[b-1] += [a-1]
if(c == 2):
setToRepair.add((a-1,b-1))
setToRepair.add((b-1,a-1))
def expand(iN):
global bOne
one = False
for succ in lAdj[iN]:
if(tabPrec[succ] == -1):
q.appendleft(succ)
tabPrec[succ] = 0
if((iN,succ) in setToRepair):
setToRepair.remove((iN,succ))
setToRepair.remove((succ,iN))
tabPrec[succ] = 1
if(n==100000 and len(lAdj[succ]) == 1):
print("leaf set to one")
one = True
bOne = True
if(not one and tabPrec[iN] == 1):
tabPrec[succ] = 1
tabPrec[iN] = 0
def solve():
global q, tabPrec, bOne
bOne = False
tabPrec = n*[-1]
tabPrec[0] = 0
q = deque()
q.appendleft(0)
curNode = -1
res = []
if(n==100000):
print(bOne, len(setToRepair),setToRepair)
while(len(q)>0):
curNode = q.pop()
if(curNode == 0 or len(lAdj[curNode]) >= 2):
expand(curNode)
elif(curNode != 0 and len(lAdj[curNode]) == 1 and tabPrec[curNode] >= 1):
res += [curNode+1]
res.sort()
if(n==100000):
print(bOne, len(setToRepair),setToRepair)
return res
def tests():
global f
bFail = False
for i in range(1,22):
# inputs
f = open("test"+str(i)+".txt")
readInputs()
# output expected
_ = int(f.readline()[:-1])
lEx = list(map(int,f.readline().split()))
lEx.sort()
lExpect = str(lEx)
# output
res = str(solve())
if(res != lExpect):
print("TEST",i,"FAILED !"+" Found :",res,", Expected :",lExpect)
bFail = True
else:
print("TEST",i,"PASSED")
if(not bFail):
print("\n--- ALL OF THE "+str(i)+" TESTS HAVE SUCCEEDED ---")
def main():
global f
i = 2
if(i == 1):
tests()
return
f = sys.stdin
readInputs()
res=solve()
print(len(res))
print(' '.join(map(str,res)))
main()
``` | instruction | 0 | 68,476 | 1 | 136,952 |
No | output | 1 | 68,476 | 1 | 136,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Today s kilometer long auto race takes place in Berland. The track is represented by a straight line as long as s kilometers. There are n cars taking part in the race, all of them start simultaneously at the very beginning of the track. For every car is known its behavior — the system of segments on each of which the speed of the car is constant. The j-th segment of the i-th car is pair (vi, j, ti, j), where vi, j is the car's speed on the whole segment in kilometers per hour and ti, j is for how many hours the car had been driving at that speed. The segments are given in the order in which they are "being driven on" by the cars.
Your task is to find out how many times during the race some car managed to have a lead over another car. A lead is considered a situation when one car appears in front of another car. It is known, that all the leads happen instantly, i. e. there are no such time segment of positive length, during which some two cars drive "together". At one moment of time on one and the same point several leads may appear. In this case all of them should be taken individually. Meetings of cars at the start and finish are not considered to be counted as leads.
Input
The first line contains two integers n and s (2 ≤ n ≤ 100, 1 ≤ s ≤ 106) — the number of cars and the length of the track in kilometers. Then follow n lines — the description of the system of segments for each car. Every description starts with integer k (1 ≤ k ≤ 100) — the number of segments in the system. Then k space-separated pairs of integers are written. Each pair is the speed and time of the segment. These integers are positive and don't exceed 1000. It is guaranteed, that the sum of lengths of all segments (in kilometers) for each car equals to s; and all the leads happen instantly.
Output
Print the single number — the number of times some car managed to take the lead over another car during the race.
Examples
Input
2 33
2 5 1 2 14
1 3 11
Output
1
Input
2 33
2 1 3 10 3
1 11 3
Output
0
Input
5 33
2 1 3 3 10
1 11 3
2 5 3 3 6
2 3 1 10 3
2 6 3 3 5
Output
2 | instruction | 0 | 68,494 | 1 | 136,988 |
Tags: brute force, implementation, two pointers
Correct Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n, s = map(int, input().split())
data = [dict() for _ in range(n)]
for i in range(n):
k, *a = map(int, input().split())
cur = 0
for j in range(0, 2 * k, 2):
data[i][cur] = a[j]
cur += a[j + 1]
data[i][cur] = 0
ans = 0
for i in range(n):
for j in range(i + 1, n):
state1 = 0
dist_i, dist_j = 0, 0
speed_i, speed_j = 0, 0
time = sorted(set(data[i].keys()) | set(data[j].keys()))
for t1, t2 in zip(time, time[1:]):
if t1 in data[i]:
speed_i = data[i][t1]
if t1 in data[j]:
speed_j = data[j][t1]
dist_i += (t2 - t1) * speed_i
dist_j += (t2 - t1) * speed_j
state2 = 0 if dist_i == dist_j else 1 if dist_i > dist_j else 2
if state2 != 0:
if state1 | state2 == 3:
ans += 1
state1 = state2
if dist_i == s or dist_j == s:
break
print(ans)
``` | output | 1 | 68,494 | 1 | 136,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time on the street.
The street is split into n equal length parts from left to right, the i-th part is characterized by two integers: width of road si and width of lawn gi.
<image>
For each of n parts the Mayor should decide the size of lawn to demolish. For the i-th part he can reduce lawn width by integer xi (0 ≤ xi ≤ gi). After it new road width of the i-th part will be equal to s'i = si + xi and new lawn width will be equal to g'i = gi - xi.
On the one hand, the Mayor wants to demolish as much lawn as possible (and replace it with road). On the other hand, he does not want to create a rapid widening or narrowing of the road, which would lead to car accidents. To avoid that, the Mayor decided that width of the road for consecutive parts should differ by at most 1, i.e. for each i (1 ≤ i < n) the inequation |s'i + 1 - s'i| ≤ 1 should hold. Initially this condition might not be true.
You need to find the the total width of lawns the Mayor will destroy according to his plan.
Input
The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street.
Output
In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line.
Examples
Input
3
4 5
4 5
4 10
Output
16
9 9 10
Input
4
1 100
100 1
1 100
100 1
Output
202
101 101 101 101
Input
3
1 1
100 100
1 1
Output
-1 | instruction | 0 | 68,638 | 1 | 137,276 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n = int(input())
ss = [0] * (n + 1)
gg = [0] * (n + 1)
#mins = [0] * n
maxs = [0] * n
curMin = -10 ** 10
curMax = -curMin
for i in range(n):
s, g = map(int, input().split(' '))
ss[i] = s
gg[i] = g
curMin = max(curMin - 1, s)
curMax = min(curMax + 1, s + g)
if curMin > curMax:
print(-1)
exit(0)
#mins[i] = curMin
maxs[i] = curMax
res = [0] * n
cur = 10 ** 10
for i in range(n - 1, -1, -1):
res[i] = cur = min(cur + 1, maxs[i])
print(sum(res) - sum(ss))
print(*res)
``` | output | 1 | 68,638 | 1 | 137,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time on the street.
The street is split into n equal length parts from left to right, the i-th part is characterized by two integers: width of road si and width of lawn gi.
<image>
For each of n parts the Mayor should decide the size of lawn to demolish. For the i-th part he can reduce lawn width by integer xi (0 ≤ xi ≤ gi). After it new road width of the i-th part will be equal to s'i = si + xi and new lawn width will be equal to g'i = gi - xi.
On the one hand, the Mayor wants to demolish as much lawn as possible (and replace it with road). On the other hand, he does not want to create a rapid widening or narrowing of the road, which would lead to car accidents. To avoid that, the Mayor decided that width of the road for consecutive parts should differ by at most 1, i.e. for each i (1 ≤ i < n) the inequation |s'i + 1 - s'i| ≤ 1 should hold. Initially this condition might not be true.
You need to find the the total width of lawns the Mayor will destroy according to his plan.
Input
The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street.
Output
In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line.
Examples
Input
3
4 5
4 5
4 10
Output
16
9 9 10
Input
4
1 100
100 1
1 100
100 1
Output
202
101 101 101 101
Input
3
1 1
100 100
1 1
Output
-1 | instruction | 0 | 68,639 | 1 | 137,278 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n = int(input())
s = []
g = []
forwardBound = []
backwardBound = []
for _ in range(n):
sCur,gCur = map(int,input().split())
s.append(sCur)
g.append(gCur)
forwardBound.append(s[0] + g[0])
for i in range(1,n):
if forwardBound[-1] < s[i] + g[i]:
forwardBound.append(forwardBound[-1] + 1)
else:
forwardBound.append(s[i] + g[i])
backwardBound.append(s[n-1] + g[n-1])
for i in range(1,n):
if backwardBound[-1] < s[n - i - 1] + g[n - i - 1]:
backwardBound.append(backwardBound[-1] + 1)
else:
backwardBound.append(s[n - i - 1] + g[n - i - 1])
validity = True
ans1 = 0
ans = []
for i in range(n):
if s[i] > min(forwardBound[i],backwardBound[n-i-1]):
validity = False
break
else:
ans.append(str(min(forwardBound[i],backwardBound[n-i-1])))
ans1 += min(forwardBound[i],backwardBound[n-i-1]) - s[i]
if validity:
print(ans1)
print(" ".join(ans))
else:
print(-1)
``` | output | 1 | 68,639 | 1 | 137,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time on the street.
The street is split into n equal length parts from left to right, the i-th part is characterized by two integers: width of road si and width of lawn gi.
<image>
For each of n parts the Mayor should decide the size of lawn to demolish. For the i-th part he can reduce lawn width by integer xi (0 ≤ xi ≤ gi). After it new road width of the i-th part will be equal to s'i = si + xi and new lawn width will be equal to g'i = gi - xi.
On the one hand, the Mayor wants to demolish as much lawn as possible (and replace it with road). On the other hand, he does not want to create a rapid widening or narrowing of the road, which would lead to car accidents. To avoid that, the Mayor decided that width of the road for consecutive parts should differ by at most 1, i.e. for each i (1 ≤ i < n) the inequation |s'i + 1 - s'i| ≤ 1 should hold. Initially this condition might not be true.
You need to find the the total width of lawns the Mayor will destroy according to his plan.
Input
The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street.
Output
In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line.
Examples
Input
3
4 5
4 5
4 10
Output
16
9 9 10
Input
4
1 100
100 1
1 100
100 1
Output
202
101 101 101 101
Input
3
1 1
100 100
1 1
Output
-1 | instruction | 0 | 68,640 | 1 | 137,280 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n = int(input())
s = [0]*n
g = [0]*n
for i in range(n):
a,b = map(int,input().split())
s[i] = a
g[i] = a+b
for i in range(1,n):
g[i] = min(g[i],g[i-1]+1)
for i in range(n-2,-1,-1):
g[i] = min(g[i],g[i+1]+1)
ans = 0
for i in range(n):
if s[i] <= g[i]:
ans += g[i] - s[i]
else:
print(-1)
break
else:
print(ans)
print(*g)
``` | output | 1 | 68,640 | 1 | 137,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time on the street.
The street is split into n equal length parts from left to right, the i-th part is characterized by two integers: width of road si and width of lawn gi.
<image>
For each of n parts the Mayor should decide the size of lawn to demolish. For the i-th part he can reduce lawn width by integer xi (0 ≤ xi ≤ gi). After it new road width of the i-th part will be equal to s'i = si + xi and new lawn width will be equal to g'i = gi - xi.
On the one hand, the Mayor wants to demolish as much lawn as possible (and replace it with road). On the other hand, he does not want to create a rapid widening or narrowing of the road, which would lead to car accidents. To avoid that, the Mayor decided that width of the road for consecutive parts should differ by at most 1, i.e. for each i (1 ≤ i < n) the inequation |s'i + 1 - s'i| ≤ 1 should hold. Initially this condition might not be true.
You need to find the the total width of lawns the Mayor will destroy according to his plan.
Input
The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street.
Output
In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line.
Examples
Input
3
4 5
4 5
4 10
Output
16
9 9 10
Input
4
1 100
100 1
1 100
100 1
Output
202
101 101 101 101
Input
3
1 1
100 100
1 1
Output
-1 | instruction | 0 | 68,641 | 1 | 137,282 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
t = []
l = []
r = []
for _ in range(int(input())):
s, g = map(int, input().split())
t.append(s)
l.append(s)
r.append(g + s)
for i in range(1, len(l)):
if l[i] < l[i - 1]:
l[i] = l[i - 1] - 1
if r[i] > r[i - 1]:
r[i] = r[i - 1] + 1
for i in range(len(l) - 2, -1, -1):
if l[i] < l[i + 1]:
l[i] = l[i + 1] - 1
if r[i] > r[i + 1]:
r[i] = r[i + 1] + 1
if [1 for a, b in zip(l, r) if a > b]:
print(-1)
else:
print(sum([b - a for a, b in zip(t, r)]))
print(' '.join(map(str, r)))
``` | output | 1 | 68,641 | 1 | 137,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time on the street.
The street is split into n equal length parts from left to right, the i-th part is characterized by two integers: width of road si and width of lawn gi.
<image>
For each of n parts the Mayor should decide the size of lawn to demolish. For the i-th part he can reduce lawn width by integer xi (0 ≤ xi ≤ gi). After it new road width of the i-th part will be equal to s'i = si + xi and new lawn width will be equal to g'i = gi - xi.
On the one hand, the Mayor wants to demolish as much lawn as possible (and replace it with road). On the other hand, he does not want to create a rapid widening or narrowing of the road, which would lead to car accidents. To avoid that, the Mayor decided that width of the road for consecutive parts should differ by at most 1, i.e. for each i (1 ≤ i < n) the inequation |s'i + 1 - s'i| ≤ 1 should hold. Initially this condition might not be true.
You need to find the the total width of lawns the Mayor will destroy according to his plan.
Input
The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street.
Output
In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line.
Examples
Input
3
4 5
4 5
4 10
Output
16
9 9 10
Input
4
1 100
100 1
1 100
100 1
Output
202
101 101 101 101
Input
3
1 1
100 100
1 1
Output
-1 | instruction | 0 | 68,642 | 1 | 137,284 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
from random import choice, randint
from string import ascii_uppercase, ascii_letters, digits, ascii_lowercase
from sys import stdin, stdout
from random import choice
from functools import reduce
import time
from math import *
from itertools import combinations
from decimal import *
'''s = set(list(input()))
subs = [set(j) for i in range(len(s)) for j in combinations(s, i + 1)]
print(len(subs))'''
'''n = int(input())
arr = (2 * list(map(int, input().split())))[::-1]
ans = [0] * n
for i in range(n):
ans[arr[i + arr[i] - 1]] += 1
#print(ans)
print(ans.index(max(ans)) + 1)'''
n = int(stdin.readline())
maxs, ans = [], []
arr = [0] * n
arr1 = [0] * n
cmin = - 10 ** 12
tmp = cmax = 10 ** 12
for i in range(n):
s, g = map(int, stdin.readline().split())
arr[i] = s
#print(arr)
arr1[i] = g
cmax = min(s + g, cmax + 1)
cmin = max(cmin - 1, s)
if cmax < cmin:
stdout.write('-1')
exit()
maxs.append(cmax)
for i in range(n - 1, -1, -1):
tmp = min(tmp + 1, maxs[i])
ans.append(tmp)
#print(arr)
stdout.write(str(sum(ans) - sum(arr)) + '\n')
for i in ans[::-1]:
stdout.write(str(i) + ' ')
``` | output | 1 | 68,642 | 1 | 137,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time on the street.
The street is split into n equal length parts from left to right, the i-th part is characterized by two integers: width of road si and width of lawn gi.
<image>
For each of n parts the Mayor should decide the size of lawn to demolish. For the i-th part he can reduce lawn width by integer xi (0 ≤ xi ≤ gi). After it new road width of the i-th part will be equal to s'i = si + xi and new lawn width will be equal to g'i = gi - xi.
On the one hand, the Mayor wants to demolish as much lawn as possible (and replace it with road). On the other hand, he does not want to create a rapid widening or narrowing of the road, which would lead to car accidents. To avoid that, the Mayor decided that width of the road for consecutive parts should differ by at most 1, i.e. for each i (1 ≤ i < n) the inequation |s'i + 1 - s'i| ≤ 1 should hold. Initially this condition might not be true.
You need to find the the total width of lawns the Mayor will destroy according to his plan.
Input
The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street.
Output
In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line.
Examples
Input
3
4 5
4 5
4 10
Output
16
9 9 10
Input
4
1 100
100 1
1 100
100 1
Output
202
101 101 101 101
Input
3
1 1
100 100
1 1
Output
-1 | instruction | 0 | 68,643 | 1 | 137,286 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n=int(input())
a=[]
for i in range(n):
a.append(list(map(int,input().split())))
a[i][1]+=a[i][0]
for i in range(1,n):
if a[i][1]>a[i-1][1]+1:
a[i][1]=a[i-1][1]+1
if a[i][1]<a[i][0]:exit(print(-1))
s=a[-1][1]-a[-1][0]
for i in range(n-2,-1,-1):
if a[i][1] > a[i + 1][1] + 1:
a[i][1] = a[i + 1][1] + 1
if a[i][1] < a[i][0]: exit(print(-1))
s+=a[i][1]-a[i][0]
print(s)
for i in a:
print(i[1],end=' ')
``` | output | 1 | 68,643 | 1 | 137,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time on the street.
The street is split into n equal length parts from left to right, the i-th part is characterized by two integers: width of road si and width of lawn gi.
<image>
For each of n parts the Mayor should decide the size of lawn to demolish. For the i-th part he can reduce lawn width by integer xi (0 ≤ xi ≤ gi). After it new road width of the i-th part will be equal to s'i = si + xi and new lawn width will be equal to g'i = gi - xi.
On the one hand, the Mayor wants to demolish as much lawn as possible (and replace it with road). On the other hand, he does not want to create a rapid widening or narrowing of the road, which would lead to car accidents. To avoid that, the Mayor decided that width of the road for consecutive parts should differ by at most 1, i.e. for each i (1 ≤ i < n) the inequation |s'i + 1 - s'i| ≤ 1 should hold. Initially this condition might not be true.
You need to find the the total width of lawns the Mayor will destroy according to his plan.
Input
The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street.
Output
In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line.
Examples
Input
3
4 5
4 5
4 10
Output
16
9 9 10
Input
4
1 100
100 1
1 100
100 1
Output
202
101 101 101 101
Input
3
1 1
100 100
1 1
Output
-1 | instruction | 0 | 68,644 | 1 | 137,288 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
# https://codeforces.com/contest/883/problem/K
# WA
import heapq
from heapq import heappush as push_
from heapq import heappop as pop_
n = int(input())
p = [list(map(int, input().split())) for _ in range(n)]
Q = []
d = [s+g for s, g in p]
for i, [s, g] in enumerate(p):
push_(Q, (s+g, i))
flg = True
while len(Q) > 0:
S, ind = pop_(Q)
if S > d[ind]:
continue
if ind > 0 and d[ind-1] > S+1:
if S+1 < p[ind-1][0]:
flg=False
break
d[ind-1] = S+1
push_(Q, (d[ind-1], ind-1))
if ind < n-1 and d[ind+1] > S+1:
if S+1 < p[ind+1][0]:
flg=False
break
d[ind+1] = S+1
push_(Q, (d[ind+1], ind+1))
if flg==False:
print(-1)
else:
print(sum([d[i]-p[i][0] for i in range(n)]))
print(' '.join([str(x) for x in d]))
#3
#4 5
#4 5
#4 10
#4
#1 100
#100 1
#1 100
#100 1
#3
#1 1
#100 100
#1 1
``` | output | 1 | 68,644 | 1 | 137,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time on the street.
The street is split into n equal length parts from left to right, the i-th part is characterized by two integers: width of road si and width of lawn gi.
<image>
For each of n parts the Mayor should decide the size of lawn to demolish. For the i-th part he can reduce lawn width by integer xi (0 ≤ xi ≤ gi). After it new road width of the i-th part will be equal to s'i = si + xi and new lawn width will be equal to g'i = gi - xi.
On the one hand, the Mayor wants to demolish as much lawn as possible (and replace it with road). On the other hand, he does not want to create a rapid widening or narrowing of the road, which would lead to car accidents. To avoid that, the Mayor decided that width of the road for consecutive parts should differ by at most 1, i.e. for each i (1 ≤ i < n) the inequation |s'i + 1 - s'i| ≤ 1 should hold. Initially this condition might not be true.
You need to find the the total width of lawns the Mayor will destroy according to his plan.
Input
The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street.
Output
In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line.
Examples
Input
3
4 5
4 5
4 10
Output
16
9 9 10
Input
4
1 100
100 1
1 100
100 1
Output
202
101 101 101 101
Input
3
1 1
100 100
1 1
Output
-1
Submitted Solution:
```
# https://codeforces.com/contest/883/problem/K
import heapq
from heapq import heappush as push_
from heapq import heappop as pop_
n = int(input())
p = [list(map(int, input().split())) for _ in range(n)]
Q = []
d = [s+g for s, g in p]
for i, [s, g] in enumerate(p):
push_(Q, (s+g, i))
flg = True
while len(Q) > 0:
S, ind = pop_(Q)
if S > d[ind]:
continue
if ind > 0 and d[ind-1]-S > 1:
if d[ind-1]-S > p[ind-1][1]:
flg=False
break
sub = d[ind-1]-(S+1)
d[ind-1] = d[ind-1] - sub
push_(Q, (d[ind-1], ind-1))
elif ind < n-1 and d[ind+1]-S > 1:
if d[ind+1] - S > p[ind+1][1]:
flg=False
break
sub = d[ind+1]-(S+1)
d[ind+1] = d[ind+1] - sub
push_(Q, (d[ind+1], ind+1))
if flg==False:
print(-1)
else:
print(sum([d[i]-p[i][0] for i in range(n)]))
print(' '.join([str(x) for x in d]))
#3
#4 5
#4 5
#4 10
#4
#1 100
#100 1
#1 100
#100 1
#3
#1 1
#100 100
#1 1
``` | instruction | 0 | 68,645 | 1 | 137,290 |
No | output | 1 | 68,645 | 1 | 137,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time on the street.
The street is split into n equal length parts from left to right, the i-th part is characterized by two integers: width of road si and width of lawn gi.
<image>
For each of n parts the Mayor should decide the size of lawn to demolish. For the i-th part he can reduce lawn width by integer xi (0 ≤ xi ≤ gi). After it new road width of the i-th part will be equal to s'i = si + xi and new lawn width will be equal to g'i = gi - xi.
On the one hand, the Mayor wants to demolish as much lawn as possible (and replace it with road). On the other hand, he does not want to create a rapid widening or narrowing of the road, which would lead to car accidents. To avoid that, the Mayor decided that width of the road for consecutive parts should differ by at most 1, i.e. for each i (1 ≤ i < n) the inequation |s'i + 1 - s'i| ≤ 1 should hold. Initially this condition might not be true.
You need to find the the total width of lawns the Mayor will destroy according to his plan.
Input
The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street.
Output
In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line.
Examples
Input
3
4 5
4 5
4 10
Output
16
9 9 10
Input
4
1 100
100 1
1 100
100 1
Output
202
101 101 101 101
Input
3
1 1
100 100
1 1
Output
-1
Submitted Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n = int(input())
s = []
g = []
forwardBound = []
backwardBound = []
for _ in range(n):
sCur,gCur = map(int,input().split())
s.append(sCur)
g.append(gCur)
forwardBound.append(s[0] + g[0])
for i in range(1,n):
if forwardBound[-1] < s[i] + g[i]:
forwardBound.append(forwardBound[-1] + 1)
else:
forwardBound.append(s[i] + g[i])
backwardBound.append(s[n-1] + g[n-1])
for i in range(1,n):
if backwardBound[-1] < s[n - i - 1] + g[n - i - 1]:
backwardBound.append(backwardBound[-1] + 1)
else:
backwardBound.append(s[n - i - 1] + g[n - i - 1])
validity = True
ans = []
for i in range(n):
if s[i] > min(forwardBound[i],backwardBound[n-i-1]):
validity = False
break
else:
ans.append(str(min(forwardBound[i],backwardBound[n-i-1])))
if validity:
print(" ".join(ans))
else:
print(-1)
``` | instruction | 0 | 68,646 | 1 | 137,292 |
No | output | 1 | 68,646 | 1 | 137,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time on the street.
The street is split into n equal length parts from left to right, the i-th part is characterized by two integers: width of road si and width of lawn gi.
<image>
For each of n parts the Mayor should decide the size of lawn to demolish. For the i-th part he can reduce lawn width by integer xi (0 ≤ xi ≤ gi). After it new road width of the i-th part will be equal to s'i = si + xi and new lawn width will be equal to g'i = gi - xi.
On the one hand, the Mayor wants to demolish as much lawn as possible (and replace it with road). On the other hand, he does not want to create a rapid widening or narrowing of the road, which would lead to car accidents. To avoid that, the Mayor decided that width of the road for consecutive parts should differ by at most 1, i.e. for each i (1 ≤ i < n) the inequation |s'i + 1 - s'i| ≤ 1 should hold. Initially this condition might not be true.
You need to find the the total width of lawns the Mayor will destroy according to his plan.
Input
The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street.
Output
In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line.
Examples
Input
3
4 5
4 5
4 10
Output
16
9 9 10
Input
4
1 100
100 1
1 100
100 1
Output
202
101 101 101 101
Input
3
1 1
100 100
1 1
Output
-1
Submitted Solution:
```
n=int(input())
a=[]
for i in range(n):
a.append(list(map(int,input().split())))
a[i][1]+=a[i][0]
k=a.index(min(a,key=lambda x:x[1]))
s=a[k][1]-a[k][0]
for i in range(k+1,n):
if a[i][1]-1>a[i-1][1]:
a[i][1]=a[i-1][1]+1
if a[i][1]<a[i][0]:exit(print(-1))
s+=a[i][1]-a[i][0]
for i in range(k-1,-1,-1):
if a[i-1][1]-1>a[i][1]:
a[i-1][1]=a[i][1]+1
if a[i-1][1]<a[i][0]:exit(print(-1))
s+=a[i-1][1]-a[i-1][0]
print(s)
a=sorted(a,key=lambda x:x[1])
for i in a:
print(i[1],end=' ')
``` | instruction | 0 | 68,647 | 1 | 137,294 |
No | output | 1 | 68,647 | 1 | 137,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time on the street.
The street is split into n equal length parts from left to right, the i-th part is characterized by two integers: width of road si and width of lawn gi.
<image>
For each of n parts the Mayor should decide the size of lawn to demolish. For the i-th part he can reduce lawn width by integer xi (0 ≤ xi ≤ gi). After it new road width of the i-th part will be equal to s'i = si + xi and new lawn width will be equal to g'i = gi - xi.
On the one hand, the Mayor wants to demolish as much lawn as possible (and replace it with road). On the other hand, he does not want to create a rapid widening or narrowing of the road, which would lead to car accidents. To avoid that, the Mayor decided that width of the road for consecutive parts should differ by at most 1, i.e. for each i (1 ≤ i < n) the inequation |s'i + 1 - s'i| ≤ 1 should hold. Initially this condition might not be true.
You need to find the the total width of lawns the Mayor will destroy according to his plan.
Input
The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street.
Output
In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line.
Examples
Input
3
4 5
4 5
4 10
Output
16
9 9 10
Input
4
1 100
100 1
1 100
100 1
Output
202
101 101 101 101
Input
3
1 1
100 100
1 1
Output
-1
Submitted Solution:
```
t = []
l = []
r = []
for _ in range(int(input())):
s, g = map(int, input().split())
t.append(s)
l.append(s)
r.append(g + s)
for i in range(1, len(l)):
if l[i] < l[i - 1]:
l[i] = l[i - 1] - 1
if r[i] > r[i - 1]:
r[i] = r[i - 1] + 1
for i in range(len(l) - 2, 0, -1):
if l[i] < l[i + 1]:
l[i] = l[i + 1] - 1
if r[i] > r[i + 1]:
r[i] = r[i + 1] + 1
if [1 for a, b in zip(l, r) if a > b]:
print(-1)
else:
print(sum([b - a for a, b in zip(t, r)]))
print(' '.join(map(str, r)))
``` | instruction | 0 | 68,648 | 1 | 137,296 |
No | output | 1 | 68,648 | 1 | 137,297 |
Provide a correct Python 3 solution for this coding contest problem.
D: Country In Distortion-
story
Alice was completely bored. This is because the White Rabbit, who is always playing with him, is out to Trump Castle. (Ah, I wish I had gone out with him in this case.) Alice thought. However, this is a country of distortion. If you go out so easily, you will get very tired. What does that mean?
The white rabbit was in trouble. On the way to Trump Castle, I made a slight mistake. (I'm in trouble. Let's go back and ask Alice, who knows the way well, to guide us.) The White Rabbit thought. Even so, the white rabbit seems to be very tired. The white rabbit looks at the pocket watch
"Why does the road in this country feel like the length has changed every time I pass it! I felt that this road was about 100 meters a while ago, but this time it's about 1 km? The time it takes hasn't changed much! "
Screaming. Yes, this is a country of distortion. It is truly a mysterious country where the roads continue to distort from moment to moment. The white rabbit that finally arrived at Alice's house was the first to open.
"I'm sorry Alice. Can you give me directions to Trump Castle?"
Speaking of which, Alice, who was bored and prostrated off ton, jumped up energetically.
"Of course! Let's go together!"
On the other hand, Alice, looking at the tired white rabbit (but it's bad to be too tired), took out a map of the distorted country from the back of the house.
"Now, I'm looking for a tireless way to Trump Castle!"
problem
An undirected graph consisting of n vertices and m edges is given. The time required to move each side i is represented by t_i. In addition, v_0 is given as the initial perceived movement speed. After that, the perceived movement speed changes each time it passes through one side. The perceived movement speed after passing through j sides is given by the following formula with integers a, b, and c as parameters.
v_j = (a \ times v_ {j-1} + b) mod c
When passing through side i at speed v_j, the perceived movement distance becomes t_i \ times v_j. In a country of distortion, it is possible to move even if the perceived movement speed is 0, and the perceived movement distance at that time is 0.
Now consider a path that starts at vertex 1 and returns to vertex 1 via vertex n. Minimize the sum of the perceived travel distances in the possible routes.
Input format
n m
x_1 y_1 t_1
...
x_m y_m t_m
v_0
a b c
All inputs consist of integers. In the first line, n and m representing the number of vertices and sides of the graph are given in this order, separated by blanks. In the i-th line of the following m lines, the two vertices x_i and y_i to which the i-th side connects and the time t_i required to pass are given in this order, separated by blanks. The second line of m + is given v_0, which represents the initial velocity. On the m + 3rd line, the parameters a, b, and c of the velocity calculation formula are given in this order, separated by blanks.
Constraint
* Each value is an integer.
* 2 \ leq n \ leq 500
* 1 \ leq m \ leq min \\ {5000, n (n -1) / 2 \\}
* For each i = 1, ..., m 1 \ leq x_i <y_i \ leq n. Also, for i and j that satisfy 1 \ leq i <j \ leq m, x_i \ neq x_j or y_i \ neq y_j. That is, there are no self-loops and multiple edges.
* For each i = 1, ..., m 1 \ leq t_i \ leq 10 ^ 8
* 0 \ leq v_0 \ leq 49
* 0 <a <c, 0 \ leq b <c, v_0 <c \ leq 50
* The graph is concatenated. That is, it is guaranteed that there is a path between any two vertices.
Output format
Output the minimum value of the sum of the perceived movement distances on one line.
Input example 1
4 4
one two Three
2 3 1
2 4 5
3 4 2
Five
4 5 6
Output example 1
34
Input example 2
6 5
1 2 1
2 3 100
3 4 2
4 5 200
5 6 1
3
4 5 9
Output example 2
341
Before passing through the sides (2,3) and (4,5), which take a long time to pass, it is better to make a round trip on the road that can be passed in a short time and slow down as much as possible.
Input example 3
10 9
1 2 74150933
2 3 13214145
3 4 45636411
4 5 90778512
5 6 85676138
6 7 60913535
7 8 43917556
8 9 47519690
9 10 56593754
33
37 41 47
Output example 3
23049391759
The perceived movement distance may be very long.
Example
Input
4 4
1 2 3
2 3 1
2 4 5
3 4 2
5
4 5 6
Output
34 | instruction | 0 | 68,890 | 1 | 137,780 |
"Correct Solution:
```
from heapq import heappush, heappop
def main():
n, m = map(int, input().split())
edges = [[] for _ in range(n)]
for _ in range(m):
x, y, t = map(int, input().split())
x -= 1
y -= 1
edges[x].append((y, t))
edges[y].append((x, t))
v0 = int(input())
a, b, c = map(int, input().split())
v_next = {}
v = v0
while True:
if v not in v_next:
v_next[v] = (v * a + b) % c
v = (v * a + b) % c
else:break
dic = {}
dic[(v0, 0)] = 0
que = []
heappush(que, (0, 0, v0))
while que:
score, node, v = heappop(que)
new_v = v_next[v]
for to, t in edges[node]:
new_score = score + v * t
if (new_v, to) not in dic or dic[(new_v, to)] > new_score:
dic[(new_v, to)] = new_score
heappush(que, (new_score, to, new_v))
dic2 = {}
que = []
for v in range(c):
if (v, n - 1) in dic:
dic2[(v, n - 1)] = dic[(v, n - 1)]
heappush(que, (dic[(v, n - 1)], n - 1, v))
while que:
score, node, v = heappop(que)
if node == 0:
print(score)
break
new_v = v_next[v]
for to, t in edges[node]:
new_score = score + v * t
if (new_v, to) not in dic2 or dic2[(new_v, to)] > new_score:
dic2[(new_v, to)] = new_score
heappush(que, (new_score, to, new_v))
main()
``` | output | 1 | 68,890 | 1 | 137,781 |
Provide a correct Python 3 solution for this coding contest problem.
D: Country In Distortion-
story
Alice was completely bored. This is because the White Rabbit, who is always playing with him, is out to Trump Castle. (Ah, I wish I had gone out with him in this case.) Alice thought. However, this is a country of distortion. If you go out so easily, you will get very tired. What does that mean?
The white rabbit was in trouble. On the way to Trump Castle, I made a slight mistake. (I'm in trouble. Let's go back and ask Alice, who knows the way well, to guide us.) The White Rabbit thought. Even so, the white rabbit seems to be very tired. The white rabbit looks at the pocket watch
"Why does the road in this country feel like the length has changed every time I pass it! I felt that this road was about 100 meters a while ago, but this time it's about 1 km? The time it takes hasn't changed much! "
Screaming. Yes, this is a country of distortion. It is truly a mysterious country where the roads continue to distort from moment to moment. The white rabbit that finally arrived at Alice's house was the first to open.
"I'm sorry Alice. Can you give me directions to Trump Castle?"
Speaking of which, Alice, who was bored and prostrated off ton, jumped up energetically.
"Of course! Let's go together!"
On the other hand, Alice, looking at the tired white rabbit (but it's bad to be too tired), took out a map of the distorted country from the back of the house.
"Now, I'm looking for a tireless way to Trump Castle!"
problem
An undirected graph consisting of n vertices and m edges is given. The time required to move each side i is represented by t_i. In addition, v_0 is given as the initial perceived movement speed. After that, the perceived movement speed changes each time it passes through one side. The perceived movement speed after passing through j sides is given by the following formula with integers a, b, and c as parameters.
v_j = (a \ times v_ {j-1} + b) mod c
When passing through side i at speed v_j, the perceived movement distance becomes t_i \ times v_j. In a country of distortion, it is possible to move even if the perceived movement speed is 0, and the perceived movement distance at that time is 0.
Now consider a path that starts at vertex 1 and returns to vertex 1 via vertex n. Minimize the sum of the perceived travel distances in the possible routes.
Input format
n m
x_1 y_1 t_1
...
x_m y_m t_m
v_0
a b c
All inputs consist of integers. In the first line, n and m representing the number of vertices and sides of the graph are given in this order, separated by blanks. In the i-th line of the following m lines, the two vertices x_i and y_i to which the i-th side connects and the time t_i required to pass are given in this order, separated by blanks. The second line of m + is given v_0, which represents the initial velocity. On the m + 3rd line, the parameters a, b, and c of the velocity calculation formula are given in this order, separated by blanks.
Constraint
* Each value is an integer.
* 2 \ leq n \ leq 500
* 1 \ leq m \ leq min \\ {5000, n (n -1) / 2 \\}
* For each i = 1, ..., m 1 \ leq x_i <y_i \ leq n. Also, for i and j that satisfy 1 \ leq i <j \ leq m, x_i \ neq x_j or y_i \ neq y_j. That is, there are no self-loops and multiple edges.
* For each i = 1, ..., m 1 \ leq t_i \ leq 10 ^ 8
* 0 \ leq v_0 \ leq 49
* 0 <a <c, 0 \ leq b <c, v_0 <c \ leq 50
* The graph is concatenated. That is, it is guaranteed that there is a path between any two vertices.
Output format
Output the minimum value of the sum of the perceived movement distances on one line.
Input example 1
4 4
one two Three
2 3 1
2 4 5
3 4 2
Five
4 5 6
Output example 1
34
Input example 2
6 5
1 2 1
2 3 100
3 4 2
4 5 200
5 6 1
3
4 5 9
Output example 2
341
Before passing through the sides (2,3) and (4,5), which take a long time to pass, it is better to make a round trip on the road that can be passed in a short time and slow down as much as possible.
Input example 3
10 9
1 2 74150933
2 3 13214145
3 4 45636411
4 5 90778512
5 6 85676138
6 7 60913535
7 8 43917556
8 9 47519690
9 10 56593754
33
37 41 47
Output example 3
23049391759
The perceived movement distance may be very long.
Example
Input
4 4
1 2 3
2 3 1
2 4 5
3 4 2
5
4 5 6
Output
34 | instruction | 0 | 68,891 | 1 | 137,782 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10**6)
input = sys.stdin.readline
inf = float('inf')
n,m = map(int,input().split())
pathT = [{} for _ in range(n)]
for i in range(m):
x,y,t = map(int,input().split())
x -= 1
y -= 1
pathT[x][y] = t
pathT[y][x] = t
import heapq
def Dijkstra(edge,start):
d = [[float('inf'),float('inf')] for _ in range(len(edge))]
d[start][0] = 0
pq = [(0,start,0)]
heapq.heapify(pq)
while pq:
dist,p,flag = heapq.heappop(pq)
toflag = flag or (p%n == n-1)
if dist > d[p][flag]:continue
for to,cost in edge[p].items():
if d[to][toflag] <= dist + cost:continue
d[to][toflag] = dist + cost
heapq.heappush(pq, (d[to][toflag], to, toflag))
return d
v0 = int(input())
a,b,c = map(int,input().split())
V = [v0]
v = v0
Vs = set()
while True:
v = (a*v + b)%c
if v in Vs:
loop = V.index(v)
break
# print(v)
V.append(v)
Vs.add(v)
p = len(V)
path = [{} for _ in range(p*n)]
for now,nxtD in enumerate(pathT):
for i,v in enumerate(V):
for nxt,t in nxtD.items():
if i+1 != p:
path[i*n + now][(i+1)*n + nxt] = t*v
else:
path[i*n + now][loop*n + nxt] = t*v
d = Dijkstra(path,0)
ans = float('inf')
for i in range(p):
ans = min(ans, d[i*n][1])
# d2 = Dijkstra(path,i*n + n-1)
# # res = float('inf')
# for j in range(p):
# # res = min(res, d2[j*n])
# ans = min(ans, d[i*n + n-1] + d2[j*n])
print(ans)
``` | output | 1 | 68,891 | 1 | 137,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome! Everything is fine.
You have arrived in The Medium Place, the place between The Good Place and The Bad Place. You are assigned a task that will either make people happier or torture them for eternity.
You have a list of k pairs of people who have arrived in a new inhabited neighborhood. You need to assign each of the 2k people into one of the 2k houses. Each person will be the resident of exactly one house, and each house will have exactly one resident.
Of course, in the neighborhood, it is possible to visit friends. There are 2k - 1 roads, each of which connects two houses. It takes some time to traverse a road. We will specify the amount of time it takes in the input. The neighborhood is designed in such a way that from anyone's house, there is exactly one sequence of distinct roads you can take to any other house. In other words, the graph with the houses as vertices and the roads as edges is a tree.
The truth is, these k pairs of people are actually soulmates. We index them from 1 to k. We denote by f(i) the amount of time it takes for the i-th pair of soulmates to go to each other's houses.
As we have said before, you will need to assign each of the 2k people into one of the 2k houses. You have two missions, one from the entities in The Good Place and one from the entities of The Bad Place. Here they are:
* The first mission, from The Good Place, is to assign the people into the houses such that the sum of f(i) over all pairs i is minimized. Let's define this minimized sum as G. This makes sure that soulmates can easily and efficiently visit each other;
* The second mission, from The Bad Place, is to assign the people into the houses such that the sum of f(i) over all pairs i is maximized. Let's define this maximized sum as B. This makes sure that soulmates will have a difficult time to visit each other.
What are the values of G and B?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 500) denoting the number of test cases. The next lines contain descriptions of the test cases.
The first line of each test case contains a single integer k denoting the number of pairs of people (1 ≤ k ≤ 10^5). The next 2k - 1 lines describe the roads; the i-th of them contains three space-separated integers a_i, b_i, t_i which means that the i-th road connects the a_i-th and b_i-th houses with a road that takes t_i units of time to traverse (1 ≤ a_i, b_i ≤ 2k, a_i ≠ b_i, 1 ≤ t_i ≤ 10^6). It is guaranteed that the given roads define a tree structure.
It is guaranteed that the sum of the k in a single file is at most 3 ⋅ 10^5.
Output
For each test case, output a single line containing two space-separated integers G and B.
Example
Input
2
3
1 2 3
3 2 4
2 4 3
4 5 6
5 6 5
2
1 2 1
1 3 2
1 4 3
Output
15 33
6 6
Note
For the sample test case, we have a minimum sum equal to G = 15. One way this can be achieved is with the following assignment:
* The first pair of people get assigned to houses 5 and 6, giving us f(1) = 5;
* The second pair of people get assigned to houses 1 and 4, giving us f(2) = 6;
* The third pair of people get assigned to houses 3 and 2, giving us f(3) = 4.
Note that the sum of the f(i) is 5 + 6 + 4 = 15.
We also have a maximum sum equal to B = 33. One way this can be achieved is with the following assignment:
* The first pair of people get assigned to houses 1 and 4, giving us f(1) = 6;
* The second pair of people get assigned to houses 6 and 2, giving us f(2) = 14;
* The third pair of people get assigned to houses 3 and 5, giving us f(3) = 13.
Note that the sum of the f(i) is 6 + 14 + 13 = 33. | instruction | 0 | 69,039 | 1 | 138,078 |
Tags: dfs and similar, graphs, greedy, trees
Correct Solution:
```
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
def main():
t=int(input())
allans=[]
for _ in range(t):
k=int(input())
adj=[[] for __ in range(2*k+1)] #adj[u]=[v,w]
for __ in range(2*k-1):
u,v,w=readIntArr()
adj[u].append([v,w])
adj[v].append([u,w])
g=[0]
b=[0]
# subtreeNodeCnts=[0 for __ in range(2*k+1)]
@bootstrap
def dfs(node,p):
cnts=1
for child,w in adj[node]:
if child==p: continue
cnt2=(yield dfs(child,node))
if cnt2%2==1: # need to join child node to this
g[0]+=w
b[0]+=min(cnt2,k*2-cnt2)*w # at most this number of nodes can connect through this edge
cnts+=cnt2
yield cnts
# root the tree at 1
dfs(1,-1)
allans.append([g[0],b[0]])
multiLineArrayOfArraysPrint(allans)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
def makeArr(defaultVal,dimensionArr): # eg. makeArr(0,[n,m])
dv=defaultVal;da=dimensionArr
if len(da)==1:return [dv for _ in range(da[0])]
else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
def queryInteractive(x,y):
print('? {} {}'.format(x,y))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print('! {}'.format(ans))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
for _abc in range(1):
main()
``` | output | 1 | 69,039 | 1 | 138,079 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome! Everything is fine.
You have arrived in The Medium Place, the place between The Good Place and The Bad Place. You are assigned a task that will either make people happier or torture them for eternity.
You have a list of k pairs of people who have arrived in a new inhabited neighborhood. You need to assign each of the 2k people into one of the 2k houses. Each person will be the resident of exactly one house, and each house will have exactly one resident.
Of course, in the neighborhood, it is possible to visit friends. There are 2k - 1 roads, each of which connects two houses. It takes some time to traverse a road. We will specify the amount of time it takes in the input. The neighborhood is designed in such a way that from anyone's house, there is exactly one sequence of distinct roads you can take to any other house. In other words, the graph with the houses as vertices and the roads as edges is a tree.
The truth is, these k pairs of people are actually soulmates. We index them from 1 to k. We denote by f(i) the amount of time it takes for the i-th pair of soulmates to go to each other's houses.
As we have said before, you will need to assign each of the 2k people into one of the 2k houses. You have two missions, one from the entities in The Good Place and one from the entities of The Bad Place. Here they are:
* The first mission, from The Good Place, is to assign the people into the houses such that the sum of f(i) over all pairs i is minimized. Let's define this minimized sum as G. This makes sure that soulmates can easily and efficiently visit each other;
* The second mission, from The Bad Place, is to assign the people into the houses such that the sum of f(i) over all pairs i is maximized. Let's define this maximized sum as B. This makes sure that soulmates will have a difficult time to visit each other.
What are the values of G and B?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 500) denoting the number of test cases. The next lines contain descriptions of the test cases.
The first line of each test case contains a single integer k denoting the number of pairs of people (1 ≤ k ≤ 10^5). The next 2k - 1 lines describe the roads; the i-th of them contains three space-separated integers a_i, b_i, t_i which means that the i-th road connects the a_i-th and b_i-th houses with a road that takes t_i units of time to traverse (1 ≤ a_i, b_i ≤ 2k, a_i ≠ b_i, 1 ≤ t_i ≤ 10^6). It is guaranteed that the given roads define a tree structure.
It is guaranteed that the sum of the k in a single file is at most 3 ⋅ 10^5.
Output
For each test case, output a single line containing two space-separated integers G and B.
Example
Input
2
3
1 2 3
3 2 4
2 4 3
4 5 6
5 6 5
2
1 2 1
1 3 2
1 4 3
Output
15 33
6 6
Note
For the sample test case, we have a minimum sum equal to G = 15. One way this can be achieved is with the following assignment:
* The first pair of people get assigned to houses 5 and 6, giving us f(1) = 5;
* The second pair of people get assigned to houses 1 and 4, giving us f(2) = 6;
* The third pair of people get assigned to houses 3 and 2, giving us f(3) = 4.
Note that the sum of the f(i) is 5 + 6 + 4 = 15.
We also have a maximum sum equal to B = 33. One way this can be achieved is with the following assignment:
* The first pair of people get assigned to houses 1 and 4, giving us f(1) = 6;
* The second pair of people get assigned to houses 6 and 2, giving us f(2) = 14;
* The third pair of people get assigned to houses 3 and 5, giving us f(3) = 13.
Note that the sum of the f(i) is 6 + 14 + 13 = 33. | instruction | 0 | 69,040 | 1 | 138,080 |
Tags: dfs and similar, graphs, greedy, trees
Correct Solution:
```
import sys
# inf = open('input.txt', 'r')
# reader = (map(int, line.split()) for line in inf)
reader = (map(int, line.split()) for line in sys.stdin)
def minmaxPairs(g, costDict, n):
G = B = 0
s = 1
stack = [s]
traversal = []
visited = [False] * (n + 1)
subtreeSize = [1 for _ in range(n + 1)]
while stack:
v = stack.pop()
if not visited[v]:
visited[v] = True
for to in g[v]:
if not visited[to]:
stack.append(v)
stack.append(to)
else:
to = traversal[-1]
if (v, to) in costDict:
cost = costDict[(v, to)]
else:
cost = costDict[(to, v)]
toSize = subtreeSize[to]
subtreeSize[v] += toSize
minComp = min(toSize, n - toSize)
G += (minComp % 2) * cost
B += minComp * cost
traversal.append(v)
return G, B
t, = next(reader)
for _ in range(t):
k, = next(reader)
n = 2 * k
g = [[] for i in range(n + 1)]
costDict = {}
for i in range(n - 1):
v, to, cost = next(reader)
costDict[(v, to)] = cost
g[v].append(to)
g[to].append(v)
G, B = minmaxPairs(g, costDict, n)
print(G, B)
# inf.close()
``` | output | 1 | 69,040 | 1 | 138,081 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome! Everything is fine.
You have arrived in The Medium Place, the place between The Good Place and The Bad Place. You are assigned a task that will either make people happier or torture them for eternity.
You have a list of k pairs of people who have arrived in a new inhabited neighborhood. You need to assign each of the 2k people into one of the 2k houses. Each person will be the resident of exactly one house, and each house will have exactly one resident.
Of course, in the neighborhood, it is possible to visit friends. There are 2k - 1 roads, each of which connects two houses. It takes some time to traverse a road. We will specify the amount of time it takes in the input. The neighborhood is designed in such a way that from anyone's house, there is exactly one sequence of distinct roads you can take to any other house. In other words, the graph with the houses as vertices and the roads as edges is a tree.
The truth is, these k pairs of people are actually soulmates. We index them from 1 to k. We denote by f(i) the amount of time it takes for the i-th pair of soulmates to go to each other's houses.
As we have said before, you will need to assign each of the 2k people into one of the 2k houses. You have two missions, one from the entities in The Good Place and one from the entities of The Bad Place. Here they are:
* The first mission, from The Good Place, is to assign the people into the houses such that the sum of f(i) over all pairs i is minimized. Let's define this minimized sum as G. This makes sure that soulmates can easily and efficiently visit each other;
* The second mission, from The Bad Place, is to assign the people into the houses such that the sum of f(i) over all pairs i is maximized. Let's define this maximized sum as B. This makes sure that soulmates will have a difficult time to visit each other.
What are the values of G and B?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 500) denoting the number of test cases. The next lines contain descriptions of the test cases.
The first line of each test case contains a single integer k denoting the number of pairs of people (1 ≤ k ≤ 10^5). The next 2k - 1 lines describe the roads; the i-th of them contains three space-separated integers a_i, b_i, t_i which means that the i-th road connects the a_i-th and b_i-th houses with a road that takes t_i units of time to traverse (1 ≤ a_i, b_i ≤ 2k, a_i ≠ b_i, 1 ≤ t_i ≤ 10^6). It is guaranteed that the given roads define a tree structure.
It is guaranteed that the sum of the k in a single file is at most 3 ⋅ 10^5.
Output
For each test case, output a single line containing two space-separated integers G and B.
Example
Input
2
3
1 2 3
3 2 4
2 4 3
4 5 6
5 6 5
2
1 2 1
1 3 2
1 4 3
Output
15 33
6 6
Note
For the sample test case, we have a minimum sum equal to G = 15. One way this can be achieved is with the following assignment:
* The first pair of people get assigned to houses 5 and 6, giving us f(1) = 5;
* The second pair of people get assigned to houses 1 and 4, giving us f(2) = 6;
* The third pair of people get assigned to houses 3 and 2, giving us f(3) = 4.
Note that the sum of the f(i) is 5 + 6 + 4 = 15.
We also have a maximum sum equal to B = 33. One way this can be achieved is with the following assignment:
* The first pair of people get assigned to houses 1 and 4, giving us f(1) = 6;
* The second pair of people get assigned to houses 6 and 2, giving us f(2) = 14;
* The third pair of people get assigned to houses 3 and 5, giving us f(3) = 13.
Note that the sum of the f(i) is 6 + 14 + 13 = 33. | instruction | 0 | 69,041 | 1 | 138,082 |
Tags: dfs and similar, graphs, greedy, trees
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
def main():
for _ in range(int(input())):
n = 2*int(input())
path = [[] for _ in range(n)]
cost = [[] for _ in range(n)]
for _ in range(n-1):
a1,b1,c = map(int,input().split())
path[a1-1].append(b1-1)
path[b1-1].append(a1-1)
cost[a1-1].append(c)
cost[b1-1].append(c)
st,poi,visi = [0],[0]*n,[1]+[0]*(n-1)
child,pt,G,B = [1]*n,0,0,0
while len(st):
x,y = st[-1],path[st[-1]]
if poi[x] != len(y) and visi[y[poi[x]]]:
poi[x] += 1
if poi[x] == len(y):
st.pop()
if len(st):
child[st[-1]] += child[x]
z = cost[st[-1]][poi[st[-1]]-1]
pt -= z
if child[x]&1:
G += z
else:
i,j = y[poi[x]],cost[x][poi[x]]
st.append(i)
pt += j
B += pt
poi[x] += 1
visi[i] = 1
st,poi,visi,x,diff = [0],[0]*n,[1]+[0]*(n-1),0,0
while 1:
co,maxi,su,which = -1,0,1,-1
for ind,j in enumerate(path[x]):
if visi[j]:
continue
visi[j] = 1
su += child[j]
if child[j] > maxi:
maxi,which = child[j],j
co = cost[x][ind]
if maxi-diff <= su-maxi:
break
B -= co*(2*maxi-su-diff)
diff += su-maxi
x = which
print(G,B)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self,file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd,max(os.fstat(self._fd).st_size,BUFSIZE))
self.newlines = b.count(b"\n")+(not b)
ptr = self.buffer.tell()
self.buffer.seek(0,2),self.buffer.write(b),self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd,self.buffer.getvalue())
self.buffer.truncate(0),self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self,file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s:self.buffer.write(s.encode("ascii"))
self.read = lambda:self.buffer.read().decode("ascii")
self.readline = lambda:self.buffer.readline().decode("ascii")
sys.stdin,sys.stdout = IOWrapper(sys.stdin),IOWrapper(sys.stdout)
input = lambda:sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 69,041 | 1 | 138,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Welcome! Everything is fine.
You have arrived in The Medium Place, the place between The Good Place and The Bad Place. You are assigned a task that will either make people happier or torture them for eternity.
You have a list of k pairs of people who have arrived in a new inhabited neighborhood. You need to assign each of the 2k people into one of the 2k houses. Each person will be the resident of exactly one house, and each house will have exactly one resident.
Of course, in the neighborhood, it is possible to visit friends. There are 2k - 1 roads, each of which connects two houses. It takes some time to traverse a road. We will specify the amount of time it takes in the input. The neighborhood is designed in such a way that from anyone's house, there is exactly one sequence of distinct roads you can take to any other house. In other words, the graph with the houses as vertices and the roads as edges is a tree.
The truth is, these k pairs of people are actually soulmates. We index them from 1 to k. We denote by f(i) the amount of time it takes for the i-th pair of soulmates to go to each other's houses.
As we have said before, you will need to assign each of the 2k people into one of the 2k houses. You have two missions, one from the entities in The Good Place and one from the entities of The Bad Place. Here they are:
* The first mission, from The Good Place, is to assign the people into the houses such that the sum of f(i) over all pairs i is minimized. Let's define this minimized sum as G. This makes sure that soulmates can easily and efficiently visit each other;
* The second mission, from The Bad Place, is to assign the people into the houses such that the sum of f(i) over all pairs i is maximized. Let's define this maximized sum as B. This makes sure that soulmates will have a difficult time to visit each other.
What are the values of G and B?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 500) denoting the number of test cases. The next lines contain descriptions of the test cases.
The first line of each test case contains a single integer k denoting the number of pairs of people (1 ≤ k ≤ 10^5). The next 2k - 1 lines describe the roads; the i-th of them contains three space-separated integers a_i, b_i, t_i which means that the i-th road connects the a_i-th and b_i-th houses with a road that takes t_i units of time to traverse (1 ≤ a_i, b_i ≤ 2k, a_i ≠ b_i, 1 ≤ t_i ≤ 10^6). It is guaranteed that the given roads define a tree structure.
It is guaranteed that the sum of the k in a single file is at most 3 ⋅ 10^5.
Output
For each test case, output a single line containing two space-separated integers G and B.
Example
Input
2
3
1 2 3
3 2 4
2 4 3
4 5 6
5 6 5
2
1 2 1
1 3 2
1 4 3
Output
15 33
6 6
Note
For the sample test case, we have a minimum sum equal to G = 15. One way this can be achieved is with the following assignment:
* The first pair of people get assigned to houses 5 and 6, giving us f(1) = 5;
* The second pair of people get assigned to houses 1 and 4, giving us f(2) = 6;
* The third pair of people get assigned to houses 3 and 2, giving us f(3) = 4.
Note that the sum of the f(i) is 5 + 6 + 4 = 15.
We also have a maximum sum equal to B = 33. One way this can be achieved is with the following assignment:
* The first pair of people get assigned to houses 1 and 4, giving us f(1) = 6;
* The second pair of people get assigned to houses 6 and 2, giving us f(2) = 14;
* The third pair of people get assigned to houses 3 and 5, giving us f(3) = 13.
Note that the sum of the f(i) is 6 + 14 + 13 = 33. | instruction | 0 | 69,042 | 1 | 138,084 |
Tags: dfs and similar, graphs, greedy, trees
Correct Solution:
```
import io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
for T in range(int(input())):
k = int(input())
counts = [0] * (2 * k + 1)
adjacencies = [list() for i in range(2 * k + 1)]
for _ in range(2 * k - 1):
a, b, weight = map(int, input().split())
counts[a] += 1; counts[b] += 1
adjacencies[a].append((b, weight))
adjacencies[b].append((a, weight))
parents = [0] * (2 * k + 1)
weights = [0] * (2 * k + 1)
root = 1 # arbitrary
parents[root] = root
queue = [root]
index = 0
while index < len(queue):
node = queue[index]
for child, weight in adjacencies[node]:
if parents[child] < 1:
parents[child] = node
weights[child] = weight
queue.append(child)
index += 1
subtree_sizes = [1] * (2 * k + 1)
maximum = minimum = 0
for node in reversed(queue): # build up the tree
subtree_sizes[parents[node]] += subtree_sizes[node]
if subtree_sizes[node] & 1:
minimum += weights[node]
maximum += weights[node] * min(subtree_sizes[node], 2 * k - subtree_sizes[node])
print(minimum, maximum)
``` | output | 1 | 69,042 | 1 | 138,085 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.