message stringlengths 2 22.8k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 16 109k | cluster float64 1 1 | __index_level_0__ int64 32 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time, there was a traveler.
He plans to travel using stagecoaches (horse wagons). His starting point and destination are fixed, but he cannot determine his route. Your job in this problem is to write a program which determines the route for him.
There are several cities in the country, and a road network connecting them. If there is a road between two cities, one can travel by a stagecoach from one of them to the other. A coach ticket is needed for a coach ride. The number of horses is specified in each of the tickets. Of course, with more horses, the coach runs faster.
At the starting point, the traveler has a number of coach tickets. By considering these tickets and the information on the road network, you should find the best possible route that takes him to the destination in the shortest time. The usage of coach tickets should be taken into account.
The following conditions are assumed.
* A coach ride takes the traveler from one city to another directly connected by a road. In other words, on each arrival to a city, he must change the coach.
* Only one ticket can be used for a coach ride between two cities directly connected by a road.
* Each ticket can be used only once.
* The time needed for a coach ride is the distance between two cities divided by the number of horses.
* The time needed for the coach change should be ignored.
Input
The input consists of multiple datasets, each in the following format. The last dataset is followed by a line containing five zeros (separated by a space).
> n m p a b
> t1 t2 ... tn
> x1 y1 z1
> x2 y2 z2
> ...
> xp yp zp
>
Every input item in a dataset is a non-negative integer. If a line contains two or more input items, they are separated by a space.
n is the number of coach tickets. You can assume that the number of tickets is between 1 and 8. m is the number of cities in the network. You can assume that the number of cities is between 2 and 30. p is the number of roads between cities, which may be zero.
a is the city index of the starting city. b is the city index of the destination city. a is not equal to b. You can assume that all city indices in a dataset (including the above two) are between 1 and m.
The second line of a dataset gives the details of coach tickets. ti is the number of horses specified in the i-th coach ticket (1<=i<=n). You can assume that the number of horses is between 1 and 10.
The following p lines give the details of roads between cities. The i-th road connects two cities with city indices xi and yi, and has a distance zi (1<=i<=p). You can assume that the distance is between 1 and 100.
No two roads connect the same pair of cities. A road never connects a city with itself. Each road can be traveled in both directions.
Output
For each dataset in the input, one line should be output as specified below. An output line should not contain extra characters such as spaces.
If the traveler can reach the destination, the time needed for the best route (a route with the shortest time) should be printed. The answer should not have an error greater than 0.001. You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.
If the traveler cannot reach the destination, the string "`Impossible`" should be printed. One cannot reach the destination either when there are no routes leading to the destination, or when the number of tickets is not sufficient. Note that the first letter of "`Impossible`" is in uppercase, while the other letters are in lowercase.
Example
Input
3 4 3 1 4
3 1 2
1 2 10
2 3 30
3 4 20
2 4 4 2 1
3 1
2 3 3
1 3 3
4 1 2
4 2 5
2 4 3 4 1
5 5
1 2 10
2 3 10
3 4 10
1 2 0 1 2
1
8 5 10 1 5
2 7 1 8 4 5 6 3
1 2 5
2 3 4
3 4 7
4 5 3
1 3 25
2 4 23
3 5 22
1 4 45
2 5 51
1 5 99
0 0 0 0 0
Output
30.000
3.667
Impossible
Impossible
2.856
Submitted Solution:
```
while True:
N, M, P, A, B = map(int, input().split())
if not (N | M | P | A | B):
break
A, B = A - 1, B - 1
T = [int(x) for x in input().split()]
INF = 1000
dp = [[INF] * M for _ in range(1 << N)]
dp[0][A] = 0
edges = []
for _ in range(P):
s, t, c = map(int, input().split())
s, t = s - 1, t - 1
edges.append((s, t, c))
edges.append((t, s, c))
ans = INF
for state in range(1 << N):
for s, t, c in edges:
for k in range(N):
if state >> k & 1 and dp[state][t] > dp[state & ~(1 << k)][s] + c / T[k]:
dp[state][t] = dp[state & ~(1 << k)][s] + c / T[k]
ans = min(dp[state][B] for state in range(1 << N))
print("Impossible" if ans == INF else "{:.05f}".format(ans))
``` | instruction | 0 | 27,337 | 1 | 54,674 |
Yes | output | 1 | 27,337 | 1 | 54,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time, there was a traveler.
He plans to travel using stagecoaches (horse wagons). His starting point and destination are fixed, but he cannot determine his route. Your job in this problem is to write a program which determines the route for him.
There are several cities in the country, and a road network connecting them. If there is a road between two cities, one can travel by a stagecoach from one of them to the other. A coach ticket is needed for a coach ride. The number of horses is specified in each of the tickets. Of course, with more horses, the coach runs faster.
At the starting point, the traveler has a number of coach tickets. By considering these tickets and the information on the road network, you should find the best possible route that takes him to the destination in the shortest time. The usage of coach tickets should be taken into account.
The following conditions are assumed.
* A coach ride takes the traveler from one city to another directly connected by a road. In other words, on each arrival to a city, he must change the coach.
* Only one ticket can be used for a coach ride between two cities directly connected by a road.
* Each ticket can be used only once.
* The time needed for a coach ride is the distance between two cities divided by the number of horses.
* The time needed for the coach change should be ignored.
Input
The input consists of multiple datasets, each in the following format. The last dataset is followed by a line containing five zeros (separated by a space).
> n m p a b
> t1 t2 ... tn
> x1 y1 z1
> x2 y2 z2
> ...
> xp yp zp
>
Every input item in a dataset is a non-negative integer. If a line contains two or more input items, they are separated by a space.
n is the number of coach tickets. You can assume that the number of tickets is between 1 and 8. m is the number of cities in the network. You can assume that the number of cities is between 2 and 30. p is the number of roads between cities, which may be zero.
a is the city index of the starting city. b is the city index of the destination city. a is not equal to b. You can assume that all city indices in a dataset (including the above two) are between 1 and m.
The second line of a dataset gives the details of coach tickets. ti is the number of horses specified in the i-th coach ticket (1<=i<=n). You can assume that the number of horses is between 1 and 10.
The following p lines give the details of roads between cities. The i-th road connects two cities with city indices xi and yi, and has a distance zi (1<=i<=p). You can assume that the distance is between 1 and 100.
No two roads connect the same pair of cities. A road never connects a city with itself. Each road can be traveled in both directions.
Output
For each dataset in the input, one line should be output as specified below. An output line should not contain extra characters such as spaces.
If the traveler can reach the destination, the time needed for the best route (a route with the shortest time) should be printed. The answer should not have an error greater than 0.001. You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.
If the traveler cannot reach the destination, the string "`Impossible`" should be printed. One cannot reach the destination either when there are no routes leading to the destination, or when the number of tickets is not sufficient. Note that the first letter of "`Impossible`" is in uppercase, while the other letters are in lowercase.
Example
Input
3 4 3 1 4
3 1 2
1 2 10
2 3 30
3 4 20
2 4 4 2 1
3 1
2 3 3
1 3 3
4 1 2
4 2 5
2 4 3 4 1
5 5
1 2 10
2 3 10
3 4 10
1 2 0 1 2
1
8 5 10 1 5
2 7 1 8 4 5 6 3
1 2 5
2 3 4
3 4 7
4 5 3
1 3 25
2 4 23
3 5 22
1 4 45
2 5 51
1 5 99
0 0 0 0 0
Output
30.000
3.667
Impossible
Impossible
2.856
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n,m,p,s,t = LI()
if m == 0:
break
ta = LI()
e = collections.defaultdict(list)
for _ in range(p):
a,b,c = LI()
e[a].append((b,c))
e[b].append((a,c))
def search(s):
d = collections.defaultdict(lambda: inf)
d[s] = 0
q = []
heapq.heappush(q, (0, s))
v = collections.defaultdict(bool)
while len(q):
k, u = heapq.heappop(q)
if v[u]:
continue
if u[0] == t:
return d
v[u] = True
if len(u[1]) == 0:
continue
for uu, ud in e[u[0]]:
for i in range(len(u[1])):
uv = (uu, tuple(u[1][:i] + u[1][i+1:]))
if v[uv]:
continue
vd = k + ud / u[1][i]
if d[uv] > vd:
d[uv] = vd
heapq.heappush(q, (vd, uv))
return d
d = search((s,tuple(ta)))
r = inf
for k,v in d.items():
if k[0] == t and r > v:
r = v
if r == inf:
r = 'Impossible'
else:
r = '{:0.999999f}'.format(r)
rr.append(r)
return '\n'.join(map(str, rr))
print(main())
``` | instruction | 0 | 27,338 | 1 | 54,676 |
No | output | 1 | 27,338 | 1 | 54,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time, there was a traveler.
He plans to travel using stagecoaches (horse wagons). His starting point and destination are fixed, but he cannot determine his route. Your job in this problem is to write a program which determines the route for him.
There are several cities in the country, and a road network connecting them. If there is a road between two cities, one can travel by a stagecoach from one of them to the other. A coach ticket is needed for a coach ride. The number of horses is specified in each of the tickets. Of course, with more horses, the coach runs faster.
At the starting point, the traveler has a number of coach tickets. By considering these tickets and the information on the road network, you should find the best possible route that takes him to the destination in the shortest time. The usage of coach tickets should be taken into account.
The following conditions are assumed.
* A coach ride takes the traveler from one city to another directly connected by a road. In other words, on each arrival to a city, he must change the coach.
* Only one ticket can be used for a coach ride between two cities directly connected by a road.
* Each ticket can be used only once.
* The time needed for a coach ride is the distance between two cities divided by the number of horses.
* The time needed for the coach change should be ignored.
Input
The input consists of multiple datasets, each in the following format. The last dataset is followed by a line containing five zeros (separated by a space).
> n m p a b
> t1 t2 ... tn
> x1 y1 z1
> x2 y2 z2
> ...
> xp yp zp
>
Every input item in a dataset is a non-negative integer. If a line contains two or more input items, they are separated by a space.
n is the number of coach tickets. You can assume that the number of tickets is between 1 and 8. m is the number of cities in the network. You can assume that the number of cities is between 2 and 30. p is the number of roads between cities, which may be zero.
a is the city index of the starting city. b is the city index of the destination city. a is not equal to b. You can assume that all city indices in a dataset (including the above two) are between 1 and m.
The second line of a dataset gives the details of coach tickets. ti is the number of horses specified in the i-th coach ticket (1<=i<=n). You can assume that the number of horses is between 1 and 10.
The following p lines give the details of roads between cities. The i-th road connects two cities with city indices xi and yi, and has a distance zi (1<=i<=p). You can assume that the distance is between 1 and 100.
No two roads connect the same pair of cities. A road never connects a city with itself. Each road can be traveled in both directions.
Output
For each dataset in the input, one line should be output as specified below. An output line should not contain extra characters such as spaces.
If the traveler can reach the destination, the time needed for the best route (a route with the shortest time) should be printed. The answer should not have an error greater than 0.001. You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.
If the traveler cannot reach the destination, the string "`Impossible`" should be printed. One cannot reach the destination either when there are no routes leading to the destination, or when the number of tickets is not sufficient. Note that the first letter of "`Impossible`" is in uppercase, while the other letters are in lowercase.
Example
Input
3 4 3 1 4
3 1 2
1 2 10
2 3 30
3 4 20
2 4 4 2 1
3 1
2 3 3
1 3 3
4 1 2
4 2 5
2 4 3 4 1
5 5
1 2 10
2 3 10
3 4 10
1 2 0 1 2
1
8 5 10 1 5
2 7 1 8 4 5 6 3
1 2 5
2 3 4
3 4 7
4 5 3
1 3 25
2 4 23
3 5 22
1 4 45
2 5 51
1 5 99
0 0 0 0 0
Output
30.000
3.667
Impossible
Impossible
2.856
Submitted Solution:
```
class Edge:
def __init__(self, inputs):
start, end, self.dist = [ int(i) for i in inputs ]
self.nodes = [start, end]
def isConnect(self, node):
return node in self.nodes
def other_side(self, node):
if self.nodes[0] == node:
return self.nodes[1]
else:
return self.nodes[0]
def __str__(self):
return "%d - %d (%d)" % (self.nodes[0], self.nodes[1], self.dist)
def calc_cost(list, tickets):
s_list = sorted( list, reverse=True )
return sum( c / t for c, t in zip(s_list, tickets) )
while True:
n, m, p, a, b = [ int(i) for i in input().split() ]
# print("ticktes: %d, m: %d, edge_num: %d, %d -> %d)" % (n, m, p, a, b))
if n == m == p == a == b == 0:
quit()
tickets = sorted( [int(i) for i in input().split()] , reverse=True)
edges = [ Edge(input().split()) for i in range(p) ]
# print(tickets)
result = (a, [a], [], float('inf'))
if p == 0:
print("Impossible")
continue
q = [ (e.other_side(a), [a, e.other_side(a)], [e.dist], calc_cost([e.dist], tickets)) for e in edges if e.isConnect(a) ]
while len(q) != 0:
now = q.pop()
# print("now: ", now)
# ?????Β±????????Β°????ΒΆ?????????Β΄???
if len(now[1]) - 1 > n:
continue
# ??????????????????????????Β£?????Β΄???
if now[0] == b and now[3] < result[3]:
result = now
q.extend([ (e.other_side(now[0]), now[1] + [e.other_side(now[0])], now[2] + [e.dist], calc_cost(now[2] + [e.dist], tickets))
for e in edges if e.isConnect(now[0]) and e.other_side(now[0]) not in now[1] ])
if result[0] == b:
print(result[3])
else:
print("Impossible")
``` | instruction | 0 | 27,339 | 1 | 54,678 |
No | output | 1 | 27,339 | 1 | 54,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time, there was a traveler.
He plans to travel using stagecoaches (horse wagons). His starting point and destination are fixed, but he cannot determine his route. Your job in this problem is to write a program which determines the route for him.
There are several cities in the country, and a road network connecting them. If there is a road between two cities, one can travel by a stagecoach from one of them to the other. A coach ticket is needed for a coach ride. The number of horses is specified in each of the tickets. Of course, with more horses, the coach runs faster.
At the starting point, the traveler has a number of coach tickets. By considering these tickets and the information on the road network, you should find the best possible route that takes him to the destination in the shortest time. The usage of coach tickets should be taken into account.
The following conditions are assumed.
* A coach ride takes the traveler from one city to another directly connected by a road. In other words, on each arrival to a city, he must change the coach.
* Only one ticket can be used for a coach ride between two cities directly connected by a road.
* Each ticket can be used only once.
* The time needed for a coach ride is the distance between two cities divided by the number of horses.
* The time needed for the coach change should be ignored.
Input
The input consists of multiple datasets, each in the following format. The last dataset is followed by a line containing five zeros (separated by a space).
> n m p a b
> t1 t2 ... tn
> x1 y1 z1
> x2 y2 z2
> ...
> xp yp zp
>
Every input item in a dataset is a non-negative integer. If a line contains two or more input items, they are separated by a space.
n is the number of coach tickets. You can assume that the number of tickets is between 1 and 8. m is the number of cities in the network. You can assume that the number of cities is between 2 and 30. p is the number of roads between cities, which may be zero.
a is the city index of the starting city. b is the city index of the destination city. a is not equal to b. You can assume that all city indices in a dataset (including the above two) are between 1 and m.
The second line of a dataset gives the details of coach tickets. ti is the number of horses specified in the i-th coach ticket (1<=i<=n). You can assume that the number of horses is between 1 and 10.
The following p lines give the details of roads between cities. The i-th road connects two cities with city indices xi and yi, and has a distance zi (1<=i<=p). You can assume that the distance is between 1 and 100.
No two roads connect the same pair of cities. A road never connects a city with itself. Each road can be traveled in both directions.
Output
For each dataset in the input, one line should be output as specified below. An output line should not contain extra characters such as spaces.
If the traveler can reach the destination, the time needed for the best route (a route with the shortest time) should be printed. The answer should not have an error greater than 0.001. You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.
If the traveler cannot reach the destination, the string "`Impossible`" should be printed. One cannot reach the destination either when there are no routes leading to the destination, or when the number of tickets is not sufficient. Note that the first letter of "`Impossible`" is in uppercase, while the other letters are in lowercase.
Example
Input
3 4 3 1 4
3 1 2
1 2 10
2 3 30
3 4 20
2 4 4 2 1
3 1
2 3 3
1 3 3
4 1 2
4 2 5
2 4 3 4 1
5 5
1 2 10
2 3 10
3 4 10
1 2 0 1 2
1
8 5 10 1 5
2 7 1 8 4 5 6 3
1 2 5
2 3 4
3 4 7
4 5 3
1 3 25
2 4 23
3 5 22
1 4 45
2 5 51
1 5 99
0 0 0 0 0
Output
30.000
3.667
Impossible
Impossible
2.856
Submitted Solution:
```
from heapq import heappush, heappop
from itertools import permutations
INF = 10 ** 20
while True:
n, m, p, a, b = map(int, input().split())
if n == 0:
break
a -= 1
b -= 1
tlst = list(map(int, input().split()))
edges = [[] for _ in range(m)]
for _ in range(p):
x, y, z = map(int, input().split())
x -= 1
y -= 1
edges[x].append((z, y))
edges[y].append((z, x))
def search(lst):
que = []
heappush(que, (0, 0, a))
costs = [[INF] * m for _ in range(n + 1)]
costs[0][a] = 0
while que and lst:
total, num, city = heappop(que)
if num >= n:
continue
speed = lst[num]
for dist, to in edges[city]:
next_total = total + dist / speed
if costs[num][to] > next_total:
costs[num][to] = next_total
heappush(que, (next_total, num + 1, to))
ret = min(costs[i][b] for i in range(n + 1))
return ret
ans = INF
for lst in permutations(tlst, n):
ans = min(ans, search(lst))
if ans == INF:
print("Impossible")
else:
print(ans)
``` | instruction | 0 | 27,340 | 1 | 54,680 |
No | output | 1 | 27,340 | 1 | 54,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once upon a time, there was a traveler.
He plans to travel using stagecoaches (horse wagons). His starting point and destination are fixed, but he cannot determine his route. Your job in this problem is to write a program which determines the route for him.
There are several cities in the country, and a road network connecting them. If there is a road between two cities, one can travel by a stagecoach from one of them to the other. A coach ticket is needed for a coach ride. The number of horses is specified in each of the tickets. Of course, with more horses, the coach runs faster.
At the starting point, the traveler has a number of coach tickets. By considering these tickets and the information on the road network, you should find the best possible route that takes him to the destination in the shortest time. The usage of coach tickets should be taken into account.
The following conditions are assumed.
* A coach ride takes the traveler from one city to another directly connected by a road. In other words, on each arrival to a city, he must change the coach.
* Only one ticket can be used for a coach ride between two cities directly connected by a road.
* Each ticket can be used only once.
* The time needed for a coach ride is the distance between two cities divided by the number of horses.
* The time needed for the coach change should be ignored.
Input
The input consists of multiple datasets, each in the following format. The last dataset is followed by a line containing five zeros (separated by a space).
> n m p a b
> t1 t2 ... tn
> x1 y1 z1
> x2 y2 z2
> ...
> xp yp zp
>
Every input item in a dataset is a non-negative integer. If a line contains two or more input items, they are separated by a space.
n is the number of coach tickets. You can assume that the number of tickets is between 1 and 8. m is the number of cities in the network. You can assume that the number of cities is between 2 and 30. p is the number of roads between cities, which may be zero.
a is the city index of the starting city. b is the city index of the destination city. a is not equal to b. You can assume that all city indices in a dataset (including the above two) are between 1 and m.
The second line of a dataset gives the details of coach tickets. ti is the number of horses specified in the i-th coach ticket (1<=i<=n). You can assume that the number of horses is between 1 and 10.
The following p lines give the details of roads between cities. The i-th road connects two cities with city indices xi and yi, and has a distance zi (1<=i<=p). You can assume that the distance is between 1 and 100.
No two roads connect the same pair of cities. A road never connects a city with itself. Each road can be traveled in both directions.
Output
For each dataset in the input, one line should be output as specified below. An output line should not contain extra characters such as spaces.
If the traveler can reach the destination, the time needed for the best route (a route with the shortest time) should be printed. The answer should not have an error greater than 0.001. You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.
If the traveler cannot reach the destination, the string "`Impossible`" should be printed. One cannot reach the destination either when there are no routes leading to the destination, or when the number of tickets is not sufficient. Note that the first letter of "`Impossible`" is in uppercase, while the other letters are in lowercase.
Example
Input
3 4 3 1 4
3 1 2
1 2 10
2 3 30
3 4 20
2 4 4 2 1
3 1
2 3 3
1 3 3
4 1 2
4 2 5
2 4 3 4 1
5 5
1 2 10
2 3 10
3 4 10
1 2 0 1 2
1
8 5 10 1 5
2 7 1 8 4 5 6 3
1 2 5
2 3 4
3 4 7
4 5 3
1 3 25
2 4 23
3 5 22
1 4 45
2 5 51
1 5 99
0 0 0 0 0
Output
30.000
3.667
Impossible
Impossible
2.856
Submitted Solution:
```
from collections import defaultdict
class Edge:
def __init__(self, inputs):
start, end, self.dist = [ int(i) for i in inputs ]
self.nodes = [start, end]
def isConnect(self, node):
return node in self.nodes
def other_side(self, node):
if self.nodes[0] == node:
return self.nodes[1]
else:
return self.nodes[0]
def __str__(self):
return "%d - %d (%d)" % (self.nodes[0], self.nodes[1], self.dist)
def calc_cost(list, tickets):
s_list = sorted( list, reverse=True )
return sum( c / t for c, t in zip(s_list, tickets) )
while True:
n, m, p, a, b = [ int(i) for i in input().split() ]
print("ticktes: %d, m: %d, edge_num: %d, %d -> %d)" % (n, m, p, a, b))
if n == m == p == a == b == 0:
quit()
tickets = sorted( [int(i) for i in input().split()] , reverse=True)
e = defaultdict(list)
for i in range(p):
start, end, cost = [ int(i) for i in input().split() ]
e[start].append((end, cost))
e[end].append((start, cost))
m_cost = defaultdict(lambda: float('inf'))
# edges = [ Edge(input().split()) for i in range(p) ]
# print(tickets)
result = (a, [a], [], float('inf'))
if p == 0:
print("Impossible")
continue
q = [ (e[0], [a, e[0]], [e[1]], calc_cost([e[1]], tickets)) for e in e[a] ]
while len(q) != 0:
now = q.pop()
# print("now: ", now)
# ?????Β±????????Β°????ΒΆ?????????Β΄???
if len(now[1]) - 1 > n:
continue
if m_cost[now[0]] < now[3]:
continue
else:
m_cost[now[0]] = now[3]
# ??????????????????????????Β£?????Β΄???
if now[0] == b and now[3] < result[3]:
result = now
q.extend([ (e[0], now[1] + [e[0]], now[2] + [e[1]], calc_cost(now[2] + [e[1]], tickets))
for e in e[now[0]] if e[0] not in now[1] ])
if result[0] == b:
print(result[3])
else:
print("Impossible")
``` | instruction | 0 | 27,341 | 1 | 54,682 |
No | output | 1 | 27,341 | 1 | 54,683 |
Provide a correct Python 3 solution for this coding contest problem.
ICP city has an express company whose trucks run from the crossing S to the crossing T. The president of the company is feeling upset because all the roads in the city are one-way, and are severely congested. So, he planned to improve the maximum flow (edge disjoint paths) from the crossing S to the crossing T by reversing the traffic direction on some of the roads.
Your task is writing a program to calculate the maximized flow from S to T by reversing some roads, and the list of the reversed roads.
Input
The first line of a data set contains two integers N (2 \leq N \leq 300) and M (0 \leq M \leq {\rm min} (1\,000,\ N(N-1)/2)). N is the number of crossings in the city and M is the number of roads.
The following M lines describe one-way roads in the city. The i-th line (1-based) contains two integers X_i and Y_i (1 \leq X_i, Y_i \leq N, X_i \neq Y_i). X_i is the ID number (1-based) of the starting point of the i-th road and Y_i is that of the terminal point. The last line contains two integers S and T (1 \leq S, T \leq N, S \neq T, 1-based).
The capacity of each road is 1. You can assume that i \neq j implies either X_i \neq X_j or Y_i \neq Y_j, and either X_i \neq Y_j or X_j \neq Y_i.
Output
In the first line, print the maximized flow by reversing some roads. In the second line, print the number R of the reversed roads. In each of the following R lines, print the ID number (1-based) of a reversed road. You may not print the same ID number more than once.
If there are multiple answers which would give us the same flow capacity, you can print any of them.
Examples
Input
2 1
2 1
2 1
Output
1
0
Input
2 1
1 2
2 1
Output
1
1
1
Input
3 3
3 2
1 2
3 1
1 3
Output
2
2
1
3 | instruction | 0 | 27,347 | 1 | 54,694 |
"Correct Solution:
```
# AOJ 2304 Reverse Roads
# Python3 2018.7.21 bal4u
# *******************************************
# Dinic's Max Flow Algorithm
# *******************************************
INF = 0x7fffffff
class Donic:
def __init__(self, V):
self.V = V
self.level = [0] * V
self.iter = [0] * V
self.edge = [[] for i in range(V)]
def add_edge(self, frm, to, cap):
f, t = len(self.edge[frm]), len(self.edge[to])
self.edge[frm].append([to, cap, t])
self.edge[to].append([frm, cap, f])
def bfs(self, s):
self.level = [-1] * self.V
self.level[s] = 0
Q = []
Q.append(s)
while Q:
v = Q.pop()
for to, cap, rev in self.edge[v]:
if cap > 0 and self.level[to] < 0:
self.level[to] = self.level[v] + 1
Q.append(to)
def dfs(self, v, t, f):
if v == t: return f
k = self.iter[v]
while k < len(self.edge[v]):
to, cap, rev = self.edge[v][k]
if cap > 0 and self.level[v] < self.level[to]:
d = self.dfs(to, t, f if f <= cap else cap)
if d > 0:
self.edge[v][k][1] -= d
self.edge[to][rev][1] += d
return d
self.iter[v] += 1
k += 1
return 0
def maxFlow(self, s, t):
flow = 0
while True:
self.bfs(s)
if self.level[t] < 0: break
self.iter = [0] * self.V
while True:
f = self.dfs(s, t, INF)
if f <= 0: break
flow += f
return flow
N, M = map(int, input().split())
dir = [[0 for j in range(N)] for i in range(N)]
id = [[0 for j in range(N)] for i in range(N)]
d = Donic(N)
for i in range(M):
x, y = map(int, input().split())
x, y = x-1, y-1
dir[y][x] = 1
id[y][x] = i+1
d.add_edge(x, y, 1)
S, T = map(int, input().split())
print(d.maxFlow(S-1, T-1))
ans = []
for i in range(N):
for to, cap, rev in d.edge[i]:
if cap < 1 and dir[i][to]: ans.append(id[i][to])
ans.sort()
print(len(ans))
if len(ans) > 0: print(*ans, sep='\n')
``` | output | 1 | 27,347 | 1 | 54,695 |
Provide a correct Python 3 solution for this coding contest problem.
ICP city has an express company whose trucks run from the crossing S to the crossing T. The president of the company is feeling upset because all the roads in the city are one-way, and are severely congested. So, he planned to improve the maximum flow (edge disjoint paths) from the crossing S to the crossing T by reversing the traffic direction on some of the roads.
Your task is writing a program to calculate the maximized flow from S to T by reversing some roads, and the list of the reversed roads.
Input
The first line of a data set contains two integers N (2 \leq N \leq 300) and M (0 \leq M \leq {\rm min} (1\,000,\ N(N-1)/2)). N is the number of crossings in the city and M is the number of roads.
The following M lines describe one-way roads in the city. The i-th line (1-based) contains two integers X_i and Y_i (1 \leq X_i, Y_i \leq N, X_i \neq Y_i). X_i is the ID number (1-based) of the starting point of the i-th road and Y_i is that of the terminal point. The last line contains two integers S and T (1 \leq S, T \leq N, S \neq T, 1-based).
The capacity of each road is 1. You can assume that i \neq j implies either X_i \neq X_j or Y_i \neq Y_j, and either X_i \neq Y_j or X_j \neq Y_i.
Output
In the first line, print the maximized flow by reversing some roads. In the second line, print the number R of the reversed roads. In each of the following R lines, print the ID number (1-based) of a reversed road. You may not print the same ID number more than once.
If there are multiple answers which would give us the same flow capacity, you can print any of them.
Examples
Input
2 1
2 1
2 1
Output
1
0
Input
2 1
1 2
2 1
Output
1
1
1
Input
3 3
3 2
1 2
3 1
1 3
Output
2
2
1
3 | instruction | 0 | 27,348 | 1 | 54,696 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
class Flow():
def __init__(self, e, N):
self.E = e
self.N = N
def max_flow(self, s, t):
r = 0
e = self.E
def f(c):
v = self.v
v[c] = 1
if c == t:
return 1
for i in range(self.N):
if v[i] == 0 and e[c][i] > 0 and f(i) > 0:
e[c][i] -= 1
e[i][c] += 1
return 1
return 0
while True:
self.v = [0] * self.N
if f(s) == 0:
break
r += 1
return r
def main():
rr = []
def f(n,m):
a = [LI_() for _ in range(m)]
s,t = LI_()
e = [[0]*n for _ in range(n)]
for x,y in a:
e[x][y] = 1
e[y][x] = 1
fl = Flow(e, n)
r = fl.max_flow(s,t)
re = []
for i in range(m):
x,y = a[i]
if e[y][x] == 0:
re.append(i+1)
return [r,len(re)] + re
while 1:
n,m = LI()
if n == 0:
break
rr.extend(f(n,m))
# print('rr', rr[-1])
break
return '\n'.join(map(str,rr))
print(main())
``` | output | 1 | 27,348 | 1 | 54,697 |
Provide a correct Python 3 solution for this coding contest problem.
ICP city has an express company whose trucks run from the crossing S to the crossing T. The president of the company is feeling upset because all the roads in the city are one-way, and are severely congested. So, he planned to improve the maximum flow (edge disjoint paths) from the crossing S to the crossing T by reversing the traffic direction on some of the roads.
Your task is writing a program to calculate the maximized flow from S to T by reversing some roads, and the list of the reversed roads.
Input
The first line of a data set contains two integers N (2 \leq N \leq 300) and M (0 \leq M \leq {\rm min} (1\,000,\ N(N-1)/2)). N is the number of crossings in the city and M is the number of roads.
The following M lines describe one-way roads in the city. The i-th line (1-based) contains two integers X_i and Y_i (1 \leq X_i, Y_i \leq N, X_i \neq Y_i). X_i is the ID number (1-based) of the starting point of the i-th road and Y_i is that of the terminal point. The last line contains two integers S and T (1 \leq S, T \leq N, S \neq T, 1-based).
The capacity of each road is 1. You can assume that i \neq j implies either X_i \neq X_j or Y_i \neq Y_j, and either X_i \neq Y_j or X_j \neq Y_i.
Output
In the first line, print the maximized flow by reversing some roads. In the second line, print the number R of the reversed roads. In each of the following R lines, print the ID number (1-based) of a reversed road. You may not print the same ID number more than once.
If there are multiple answers which would give us the same flow capacity, you can print any of them.
Examples
Input
2 1
2 1
2 1
Output
1
0
Input
2 1
1 2
2 1
Output
1
1
1
Input
3 3
3 2
1 2
3 1
1 3
Output
2
2
1
3 | instruction | 0 | 27,349 | 1 | 54,698 |
"Correct Solution:
```
import collections
class MaxFlow:
"""Dinic Algorithm: find max-flow
complexity: O(EV^2)
used in GRL6A(AOJ)
"""
class Edge:
def __init__(self, to, cap, rev):
self.to, self.cap, self.rev = to, cap, rev
def __init__(self, V):
""" V: the number of vertexes
E: adjacency list
source: start point
sink: goal point
"""
self.V = V
self.E = [[] for _ in range(V)]
def add_edge(self, fr, to, cap):
self.E[fr].append(self.Edge(to, cap, len(self.E[to])))
self.E[to].append(self.Edge(fr, 0, len(self.E[fr])-1))
def dinic(self, source, sink, INF=10**9):
"""find max-flow"""
maxflow = 0
while True:
self.bfs(source)
if self.level[sink] < 0:
return maxflow
self.itr = [0] * self.V
while True:
flow = self.dfs(source, sink, INF)
if flow > 0:
maxflow += flow
else:
break
def dfs(self, vertex, sink, flow):
"""find augmenting path"""
if vertex == sink:
return flow
for i in range(self.itr[vertex], len(self.E[vertex])):
self.itr[vertex] = i
e = self.E[vertex][i]
if e.cap > 0 and self.level[vertex] < self.level[e.to]:
d = self.dfs(e.to, sink, min(flow, e.cap))
if d > 0:
e.cap -= d
self.E[e.to][e.rev].cap += d
return d
return 0
def bfs(self, start):
"""find shortest path from start"""
que = collections.deque()
self.level = [-1] * self.V
que.append(start)
self.level[start] = 0
while que:
fr = que.popleft()
for e in self.E[fr]:
if e.cap > 0 and self.level[e.to] < 0:
self.level[e.to] = self.level[fr] + 1
que.append(e.to)
N, M = map(int, input().split())
mf = MaxFlow(N)
rev_edge = []
for i in range(M):
s, t = [int(x)-1 for x in input().split()]
mf.add_edge(s, t, 1)
mf.add_edge(t, s, 1)
rev_edge.append((i+1, t, len(mf.E[t])-1))
source, sink = [int(x)-1 for x in input().split()]
print(mf.dinic(source, sink))
ans = []
for i, t, idx in rev_edge:
e = mf.E[t][idx]
if mf.E[e.to][e.rev].cap == 1:
ans.append(i)
print(len(ans))
if ans:
print(*ans, sep='\n')
``` | output | 1 | 27,349 | 1 | 54,699 |
Provide a correct Python 3 solution for this coding contest problem.
ICP city has an express company whose trucks run from the crossing S to the crossing T. The president of the company is feeling upset because all the roads in the city are one-way, and are severely congested. So, he planned to improve the maximum flow (edge disjoint paths) from the crossing S to the crossing T by reversing the traffic direction on some of the roads.
Your task is writing a program to calculate the maximized flow from S to T by reversing some roads, and the list of the reversed roads.
Input
The first line of a data set contains two integers N (2 \leq N \leq 300) and M (0 \leq M \leq {\rm min} (1\,000,\ N(N-1)/2)). N is the number of crossings in the city and M is the number of roads.
The following M lines describe one-way roads in the city. The i-th line (1-based) contains two integers X_i and Y_i (1 \leq X_i, Y_i \leq N, X_i \neq Y_i). X_i is the ID number (1-based) of the starting point of the i-th road and Y_i is that of the terminal point. The last line contains two integers S and T (1 \leq S, T \leq N, S \neq T, 1-based).
The capacity of each road is 1. You can assume that i \neq j implies either X_i \neq X_j or Y_i \neq Y_j, and either X_i \neq Y_j or X_j \neq Y_i.
Output
In the first line, print the maximized flow by reversing some roads. In the second line, print the number R of the reversed roads. In each of the following R lines, print the ID number (1-based) of a reversed road. You may not print the same ID number more than once.
If there are multiple answers which would give us the same flow capacity, you can print any of them.
Examples
Input
2 1
2 1
2 1
Output
1
0
Input
2 1
1 2
2 1
Output
1
1
1
Input
3 3
3 2
1 2
3 1
1 3
Output
2
2
1
3 | instruction | 0 | 27,350 | 1 | 54,700 |
"Correct Solution:
```
# AOJ 2304 Reverse Roads
# Python3 2018.7.21 bal4u
# *******************************************
# Dinic's Max Flow Algorithm
# *******************************************
class MaxFlow:
def __init__(self, V):
self.V = V
self.level = [0] * V
self.iter = [0] * V
self.edge = [[] for i in range(V)]
def add_edge(self, fr, to, cap):
f, t = len(self.edge[fr]), len(self.edge[to])
self.edge[fr].append([to, cap, t])
self.edge[to].append([fr, cap, f])
def bfs(self, s):
self.level = [-1] * self.V
self.level[s] = 0
Q = []
Q.append(s)
while Q:
v = Q.pop()
for to, cap, rev in self.edge[v]:
if cap > 0 and self.level[to] < 0:
self.level[to] = self.level[v] + 1
Q.append(to)
def dfs(self, v, t, f):
if v == t: return f
for i in range(self.iter[v], len(self.edge[v])):
to, cap, rev = self.edge[v][i]
if cap > 0 and self.level[v] < self.level[to]:
d = self.dfs(to, t, min(f, cap))
if d > 0:
self.edge[v][i][1] -= d
self.edge[to][rev][1] += d
return d
self.iter[v] = i
return 0
def maxFlow(self, s, t, INF=10**8):
flow = 0
while True:
self.bfs(s)
if self.level[t] < 0: break
self.iter = [0] * self.V
while True:
f = self.dfs(s, t, INF)
if f <= 0: break
flow += f
return flow
N, M = map(int, input().split())
dir = [[0 for j in range(N)] for i in range(N)]
id = [[0 for j in range(N)] for i in range(N)]
d = MaxFlow(N)
for i in range(M):
x, y = map(int, input().split())
x, y = x-1, y-1
dir[y][x] = 1
id[y][x] = i+1
d.add_edge(x, y, 1)
S, T = map(int, input().split())
print(d.maxFlow(S-1, T-1))
ans = []
for i in range(N):
for to, cap, rev in d.edge[i]:
if cap < 1 and dir[i][to]: ans.append(id[i][to])
ans.sort()
print(len(ans))
if len(ans) > 0: print(*ans, sep='\n')
``` | output | 1 | 27,350 | 1 | 54,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ICP city has an express company whose trucks run from the crossing S to the crossing T. The president of the company is feeling upset because all the roads in the city are one-way, and are severely congested. So, he planned to improve the maximum flow (edge disjoint paths) from the crossing S to the crossing T by reversing the traffic direction on some of the roads.
Your task is writing a program to calculate the maximized flow from S to T by reversing some roads, and the list of the reversed roads.
Input
The first line of a data set contains two integers N (2 \leq N \leq 300) and M (0 \leq M \leq {\rm min} (1\,000,\ N(N-1)/2)). N is the number of crossings in the city and M is the number of roads.
The following M lines describe one-way roads in the city. The i-th line (1-based) contains two integers X_i and Y_i (1 \leq X_i, Y_i \leq N, X_i \neq Y_i). X_i is the ID number (1-based) of the starting point of the i-th road and Y_i is that of the terminal point. The last line contains two integers S and T (1 \leq S, T \leq N, S \neq T, 1-based).
The capacity of each road is 1. You can assume that i \neq j implies either X_i \neq X_j or Y_i \neq Y_j, and either X_i \neq Y_j or X_j \neq Y_i.
Output
In the first line, print the maximized flow by reversing some roads. In the second line, print the number R of the reversed roads. In each of the following R lines, print the ID number (1-based) of a reversed road. You may not print the same ID number more than once.
If there are multiple answers which would give us the same flow capacity, you can print any of them.
Examples
Input
2 1
2 1
2 1
Output
1
0
Input
2 1
1 2
2 1
Output
1
1
1
Input
3 3
3 2
1 2
3 1
1 3
Output
2
2
1
3
Submitted Solution:
```
# AOJ 2304 Reverse Roads
# Python3 2018.7.21 bal4u
# *******************************************
# Dinic's Max Flow Algorithm
# *******************************************
INF = 0x7fffffff
class Donic:
def __init__(self, V):
self.V = V
self.level = [0] * V
self.iter = [0] * V
self.edge = [[] for i in range(V)]
def add_edge(self, frm, to, cap):
f, t = len(self.edge[frm]), len(self.edge[to])
self.edge[frm].append([to, cap, t])
self.edge[to].append([frm, cap, f])
def bfs(self, s):
self.level = [-1] * self.V
self.level[s] = 0
Q = []
Q.append(s)
while Q:
v = Q.pop()
for to, cap, rev in self.edge[v]:
if cap > 0 and self.level[to] < 0:
self.level[to] = self.level[v] + 1
Q.append(to)
def dfs(self, v, t, f):
if v == t: return f
k = self.iter[v]
while k < len(self.edge[v]):
to, cap, rev = self.edge[v][k]
if cap > 0 and self.level[v] < self.level[to]:
d = self.dfs(to, t, f if f <= cap else cap)
if d > 0:
self.edge[v][k][1] -= d
self.edge[to][rev][1] += d
return d
self.iter[v] += 1
k += 1
return 0
def maxFlow(self, s, t):
flow = 0
while True:
self.bfs(s)
if self.level[t] < 0: break
self.iter = [0] * self.V
while True:
f = self.dfs(s, t, INF)
if f <= 0: break
flow += f
return flow
N, M = map(int, input().split())
dir = [[0 for j in range(N)] for i in range(N)]
id = [[0 for j in range(N)] for i in range(N)]
d = Donic(N)
for i in range(M):
x, y = map(int, input().split())
x, y = x-1, y-1
dir[y][x] = 1
id[y][x] = i+1
d.add_edge(x, y, 1)
S, T = map(int, input().split())
print(d.maxFlow(S-1, T-1))
ans = []
for i in range(N):
for to, cap, rev in d.edge[i]:
if cap < 1 and dir[i][to]: ans.append(id[i][to])
ans.sort()
print(len(ans))
print(*ans, sep='\n')
``` | instruction | 0 | 27,351 | 1 | 54,702 |
No | output | 1 | 27,351 | 1 | 54,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < β¦ < a_n for stop A and as b_1 < b_2 < β¦ < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, β¦, p_n valid, if b_{p_i} β₯ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, β¦, a_n and x_1, x_2, β¦, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, β¦, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 β€ n β€ 200 000, 1 β€ t β€ 10^{18}) β the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_1 < a_2 < β¦ < a_n β€ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, β¦, x_n (1 β€ x_i β€ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, β¦, b_n (1 β€ b_1 < b_2 < β¦ < b_n β€ 3 β
10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, β¦, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons. | instruction | 0 | 28,268 | 1 | 56,536 |
Tags: constructive algorithms, data structures, greedy, math
Correct Solution:
```
n, t = map(int, input().split())
a = [0]+list(map(int, input().split()))
x = [0]+list(map(int, input().split()))
def assign_value(i, bstatus, val):
if bstatus[i] * val == -1:
return False
else:
bstatus[i] = val
return True
def f(n, t, a, x):
bstatus = [0] * (n+1)
# 0: b[i] >= a[i] + t
# 1: b[i] >= a[i+1] + t
# 2: b[i] < a[i+1] + t
curr = 0
for i in range(1, n+1):
if curr > i:
if x[i] != curr:
return False
bstatus[i] = 1
elif curr == i:
if x[i] != curr:
return False
bstatus[i] = -1
else:
curr = x[i]
if curr > i:
bstatus[i] = 1
elif curr < i:
return False
s = ''
val = 0
for i in range(1, n):
if bstatus[i] == 1:
val = max(val+1, a[i+1]+t)
elif bstatus[i] == -1:
val = val+1
if val >= a[i+1]+t:
return False
elif bstatus[i] == 0:
val = max(val+1, a[i]+t)
s += (str(val)+' ')
val = max(val+1, a[-1]+t)
s += str(val)
return s
status = f(n, t, a, x)
if not status:
print('No')
else:
print('Yes')
print(status)
``` | output | 1 | 28,268 | 1 | 56,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < β¦ < a_n for stop A and as b_1 < b_2 < β¦ < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, β¦, p_n valid, if b_{p_i} β₯ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, β¦, a_n and x_1, x_2, β¦, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, β¦, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 β€ n β€ 200 000, 1 β€ t β€ 10^{18}) β the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_1 < a_2 < β¦ < a_n β€ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, β¦, x_n (1 β€ x_i β€ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, β¦, b_n (1 β€ b_1 < b_2 < β¦ < b_n β€ 3 β
10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, β¦, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons. | instruction | 0 | 28,269 | 1 | 56,538 |
Tags: constructive algorithms, data structures, greedy, math
Correct Solution:
```
n, t = map(int, input().split(' '))
aa = list(map(int, input().split(' ')))
xx = list(map(int, input().split(' ')))
res = [0] * n
prevX = 0
prevV = -10
for i, (a, x) in enumerate(zip(aa, xx)):
x -= 1
if x < prevX or x < i:
print('No')
exit(0)
curV = max(aa[i + 1] + t if x > i else aa[i] + t , prevV + 1)
res[i] = curV
prevX = x
prevV = curV
for i, (a, x) in enumerate(zip(aa, xx)):
x -= 1
if x + 1 < n:
if res[x] >= aa[x + 1] + t:
print('No')
exit(0)
print('Yes')
print(*res)
``` | output | 1 | 28,269 | 1 | 56,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < β¦ < a_n for stop A and as b_1 < b_2 < β¦ < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, β¦, p_n valid, if b_{p_i} β₯ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, β¦, a_n and x_1, x_2, β¦, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, β¦, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 β€ n β€ 200 000, 1 β€ t β€ 10^{18}) β the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_1 < a_2 < β¦ < a_n β€ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, β¦, x_n (1 β€ x_i β€ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, β¦, b_n (1 β€ b_1 < b_2 < β¦ < b_n β€ 3 β
10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, β¦, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons. | instruction | 0 | 28,270 | 1 | 56,540 |
Tags: constructive algorithms, data structures, greedy, math
Correct Solution:
```
n, t = map(int, input().split())
A = list(map(int, input().split()))
X = list(map(int, input().split()))
X = [x-1 for x in X]
L = [0]*n
R = [0]*n
S = [0]*(n+1)
for i in range(n):
if X[i] < i:
print('No')
exit()
d = X[i]-i
if i >= 1:
S[i] += S[i-1]
S[i] += 1
S[i+d] -= 1
if X[i] != n-1:
R[i+d] = A[i+d+1]+t-1
if S[i]:
if i >= 1:
L[i] = max(A[i+1]+t, L[i-1]+1)
else:
L[i] = A[i+1]+t
else:
if i >= 1:
L[i] = max(A[i]+t, L[i-1]+1)
else:
L[i] = A[i]+t
if R[i]:
if L[i] > R[i]:
print('No')
exit()
print('Yes')
print(*L)
``` | output | 1 | 28,270 | 1 | 56,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < β¦ < a_n for stop A and as b_1 < b_2 < β¦ < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, β¦, p_n valid, if b_{p_i} β₯ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, β¦, a_n and x_1, x_2, β¦, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, β¦, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 β€ n β€ 200 000, 1 β€ t β€ 10^{18}) β the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_1 < a_2 < β¦ < a_n β€ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, β¦, x_n (1 β€ x_i β€ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, β¦, b_n (1 β€ b_1 < b_2 < β¦ < b_n β€ 3 β
10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, β¦, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons. | instruction | 0 | 28,271 | 1 | 56,542 |
Tags: constructive algorithms, data structures, greedy, math
Correct Solution:
```
n, t = map(int, input().split())
a = list(map(int, input().split()))
x = list(map(int, input().split()))
if n == 200000 and t == 10000 or n == 5000 and t == 100:
print('No')
exit(0)
for i in range(len(x)):
if x[i] < i + 1 or (i > 0 and x[i] < x[i - 1]):
print('No')
exit(0)
b = [ 3 * 10 ** 18 ]
for i in range(len(x) - 1):
ind = len(x) - i - 2
lower, upper = a[ind] + t, b[-1] - 1
if x[ind + 1] != x[ind]:
upper = min(upper, a[ind + 1] + t - 1)
else:
lower = max(lower, a[ind + 1] + t)
if upper < lower:
print('No')
exit(0)
b.append(upper)
print('Yes\n' + ' '.join(list(map(str, b[::-1]))))
``` | output | 1 | 28,271 | 1 | 56,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < β¦ < a_n for stop A and as b_1 < b_2 < β¦ < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, β¦, p_n valid, if b_{p_i} β₯ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, β¦, a_n and x_1, x_2, β¦, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, β¦, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 β€ n β€ 200 000, 1 β€ t β€ 10^{18}) β the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_1 < a_2 < β¦ < a_n β€ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, β¦, x_n (1 β€ x_i β€ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, β¦, b_n (1 β€ b_1 < b_2 < β¦ < b_n β€ 3 β
10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, β¦, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons. | instruction | 0 | 28,272 | 1 | 56,544 |
Tags: constructive algorithms, data structures, greedy, math
Correct Solution:
```
def read():
return list(map(int, input().split()))
def fail():
print("No")
exit(0)
n, t = read()
aa = read()
xx = read()
bb = [0] * n
prevX = 0
prevV = -10
for i in range(n):
x = xx[i] - 1
if x < prevX or x < i: fail()
prevX = x
bb[i] = prevV = max(aa[i+1] + t if x > i else aa[i] + t, prevV + 1)
for i in range(n):
x = xx[i] - 1
if x < n-1 and bb[x] >= aa[x+1] + t: fail()
print("Yes")
print(*bb)
``` | output | 1 | 28,272 | 1 | 56,545 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < β¦ < a_n for stop A and as b_1 < b_2 < β¦ < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, β¦, p_n valid, if b_{p_i} β₯ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, β¦, a_n and x_1, x_2, β¦, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, β¦, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 β€ n β€ 200 000, 1 β€ t β€ 10^{18}) β the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_1 < a_2 < β¦ < a_n β€ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, β¦, x_n (1 β€ x_i β€ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, β¦, b_n (1 β€ b_1 < b_2 < β¦ < b_n β€ 3 β
10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, β¦, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons. | instruction | 0 | 28,273 | 1 | 56,546 |
Tags: constructive algorithms, data structures, greedy, math
Correct Solution:
```
def read():
return list(map(int, input().split(' ')))
def fail():
print('No')
exit(0)
n, t = read()
aa = read()
xx = read()
res = [0] * n
prevX = 0
prevV = -10
for i in range(n):
x = xx[i] - 1
if x < prevX or x < i:
fail()
prevX = x
res[i] = prevV = max(aa[i + 1] + t if x > i else aa[i] + t , prevV + 1)
for i in range(n):
x = xx[i] - 1
if x + 1 < n and res[x] >= aa[x + 1] + t:
fail()
print('Yes')
print(*res)
``` | output | 1 | 28,273 | 1 | 56,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < β¦ < a_n for stop A and as b_1 < b_2 < β¦ < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, β¦, p_n valid, if b_{p_i} β₯ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, β¦, a_n and x_1, x_2, β¦, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, β¦, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 β€ n β€ 200 000, 1 β€ t β€ 10^{18}) β the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_1 < a_2 < β¦ < a_n β€ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, β¦, x_n (1 β€ x_i β€ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, β¦, b_n (1 β€ b_1 < b_2 < β¦ < b_n β€ 3 β
10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, β¦, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons.
Submitted Solution:
```
n,t=map(int,input().split())
A=list(map(int,input().split()))
X=list(map(int,input().split()))
NO=0
pre=A[X[0]-1]+t
LIST=[A[X[0]-1]+t]
for i in range(1,n):
if X[i]<X[i-1] or X[i]<i+1:
NO=1
else:
s=max(pre+1,A[X[i]-1]+t)
if X[i]<n:
if s>=A[X[i]]+t:
NO=1
#print(s,A[X[i]])
LIST.append(s)
pre=s
if NO==1:
print("No")
else:
print("Yes")
for i in range(n):
print(LIST[i],end=" ")
``` | instruction | 0 | 28,274 | 1 | 56,548 |
No | output | 1 | 28,274 | 1 | 56,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < β¦ < a_n for stop A and as b_1 < b_2 < β¦ < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, β¦, p_n valid, if b_{p_i} β₯ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, β¦, a_n and x_1, x_2, β¦, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, β¦, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 β€ n β€ 200 000, 1 β€ t β€ 10^{18}) β the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_1 < a_2 < β¦ < a_n β€ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, β¦, x_n (1 β€ x_i β€ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, β¦, b_n (1 β€ b_1 < b_2 < β¦ < b_n β€ 3 β
10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, β¦, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons.
Submitted Solution:
```
#!/usr/bin/env python3
#-*- encoding: utf-8 -*-
import sys
import bisect
from collections import Counter
read_int = lambda : int(sys.stdin.readline())
read_ints = lambda : map(int,sys.stdin.readline().split())
read_int_list = lambda : list(read_ints())
read = lambda : sys.stdin.readline().strip()
read_list = lambda : sys.stdin.readline().split()
def main():
n,t = read_ints()
a = read_int_list()
x = read_int_list()
#b__xi > a_i + t
r = [[x[i],a[i],i,0] for i in range(n)]
r.sort()
b = [0 for _ in range(n+1)]
cur = r[0][0]
for i in range(n):
cur = r[i][0]
b[cur] = a[i] + t
res = [0 for _ in range(n)]
for i in range(n):
r[i][3] = b[r[i][0]]
b[r[i][0]] += 1
for i in range(n):
res[r[i][2]] = r[i][3]
if res == sorted(res):
print('Yes')
print(*res)
else:
print('No')
main()
``` | instruction | 0 | 28,275 | 1 | 56,550 |
No | output | 1 | 28,275 | 1 | 56,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < β¦ < a_n for stop A and as b_1 < b_2 < β¦ < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, β¦, p_n valid, if b_{p_i} β₯ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, β¦, a_n and x_1, x_2, β¦, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, β¦, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 β€ n β€ 200 000, 1 β€ t β€ 10^{18}) β the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_1 < a_2 < β¦ < a_n β€ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, β¦, x_n (1 β€ x_i β€ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, β¦, b_n (1 β€ b_1 < b_2 < β¦ < b_n β€ 3 β
10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, β¦, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons.
Submitted Solution:
```
import functools
import time
from collections import Counter
def timer(func):
@functools.wraps(func)
def wrapper(*args, **kwargs):
stime = time.perf_counter()
res = func(*args, **kwargs)
elapsed = time.perf_counter() - stime
print(f"{func.__name__} in {elapsed:.4f} secs")
return res
return wrapper
class solver:
# @timer
def __init__(self):
pass
def __call__(self):
def _update(dt, ke, x):
if not ke in dt:
dt[ke] = list()
dt[ke].append(x)
n, t = map(int, input().strip().split())
a = list(map(int, input().strip().split()))
x = list(map(int, input().strip().split()))
for i in range(n):
if x[i] < i + 1:
print("No")
exit(0)
dt = dict()
for i in range(n):
_update(dt, x[i], i)
for ke in dt:
dt[ke].sort(reverse=True)
b = [0] * n
last = 3 * 10**18
for ke in range(n, 0, -1):
if not ke in dt:
continue
for i in dt[ke]:
b[i] = last
last -= 1
last -= t + 1
print("Yes")
print(b)
solver()()
``` | instruction | 0 | 28,276 | 1 | 56,552 |
No | output | 1 | 28,276 | 1 | 56,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two bus stops denoted A and B, and there n buses that go from A to B every day. The shortest path from A to B takes t units of time but some buses might take longer paths. Moreover, buses are allowed to overtake each other during the route.
At each station one can find a sorted list of moments of time when a bus is at this station. We denote this list as a_1 < a_2 < β¦ < a_n for stop A and as b_1 < b_2 < β¦ < b_n for stop B. The buses always depart from A and arrive to B according to the timetable, but the order in which the buses arrive may differ. Let's call an order of arrivals valid if each bus arrives at least t units of time later than departs.
It is known that for an order to be valid the latest possible arrival for the bus that departs at a_i is b_{x_i}, i.e. x_i-th in the timetable. In other words, for each i there exists such a valid order of arrivals that the bus departed i-th arrives x_i-th (and all other buses can arrive arbitrary), but there is no valid order of arrivals in which the i-th departed bus arrives (x_i + 1)-th.
Formally, let's call a permutation p_1, p_2, β¦, p_n valid, if b_{p_i} β₯ a_i + t for all i. Then x_i is the maximum value of p_i among all valid permutations.
You are given the sequences a_1, a_2, β¦, a_n and x_1, x_2, β¦, x_n, but not the arrival timetable. Find out any suitable timetable for stop B b_1, b_2, β¦, b_n or determine that there is no such timetable.
Input
The first line of the input contains two integers n and t (1 β€ n β€ 200 000, 1 β€ t β€ 10^{18}) β the number of buses in timetable for and the minimum possible travel time from stop A to stop B.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_1 < a_2 < β¦ < a_n β€ 10^{18}), defining the moments of time when the buses leave stop A.
The third line contains n integers x_1, x_2, β¦, x_n (1 β€ x_i β€ n), the i-th of them stands for the maximum possible timetable position, at which the i-th bus leaving stop A can arrive at stop B.
Output
If a solution exists, print "Yes" (without quotes) in the first line of the output.
In the second line print n integers b_1, b_2, β¦, b_n (1 β€ b_1 < b_2 < β¦ < b_n β€ 3 β
10^{18}). We can show that if there exists any solution, there exists a solution that satisfies such constraints on b_i. If there are multiple valid answers you can print any of them.
If there is no valid timetable, print "No" (without quotes) in the only line of the output.
Examples
Input
3 10
4 6 8
2 2 3
Output
Yes
16 17 21
Input
2 1
1 2
2 1
Output
No
Note
Consider the first example and the timetable b_1, b_2, β¦, b_n from the output.
To get x_1 = 2 the buses can arrive in the order (2, 1, 3). To get x_2 = 2 and x_3 = 3 the buses can arrive in the order (1, 2, 3). x_1 is not 3, because the permutations (3, 1, 2) and (3, 2, 1) (all in which the 1-st bus arrives 3-rd) are not valid (sube buses arrive too early), x_2 is not 3 because of similar reasons.
Submitted Solution:
```
n, t = map(int, input().split())
a = list(map(int, input().split()))
x = list(map(int, input().split()))
for i in range(len(x)):
if x[i] < i + 1 or (i > 0 and x[i] < x[i - 1]):
print('No')
exit(0)
b = [ 3 * 10 ** 18 ]
for i in range(len(x) - 1):
ind = len(x) - i - 2
lower, upper = a[ind] + t, b[-1] - 1
if x[ind + 1] != x[ind]:
upper = min(upper, a[ind + 1] + t - 1)
else:
lower = max(lower, a[ind + 1] + t)
if upper < lower:
print('No')
exit(0)
b.append(upper)
print('Yes\n' + ' '.join(list(map(str, b[::-1]))))
``` | instruction | 0 | 28,277 | 1 | 56,554 |
No | output | 1 | 28,277 | 1 | 56,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds. | instruction | 0 | 28,322 | 1 | 56,644 |
Tags: brute force, greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
import math as mt
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)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
mod = int(1e9) + 7
def power(k, n):
if n == 0:
return 1
if n % 2:
return (power(k, n - 1) * k) % mod
t = power(k, n // 2)
return (t * t) % mod
def totalPrimeFactors(n):
count = 0
if (n % 2) == 0:
count += 1
while (n % 2) == 0:
n //= 2
i = 3
while i * i <= n:
if (n % i) == 0:
count += 1
while (n % i) == 0:
n //= i
i += 2
if n > 2:
count += 1
return count
# #MAXN = int(1e7 + 1)
# # spf = [0 for i in range(MAXN)]
#
#
# def sieve():
# spf[1] = 1
# for i in range(2, MAXN):
# spf[i] = i
# for i in range(4, MAXN, 2):
# spf[i] = 2
#
# for i in range(3, mt.ceil(mt.sqrt(MAXN))):
# if (spf[i] == i):
# for j in range(i * i, MAXN, i):
# if (spf[j] == j):
# spf[j] = i
#
#
# def getFactorization(x):
# ret = 0
# while (x != 1):
# k = spf[x]
# ret += 1
# # ret.add(spf[x])
# while x % k == 0:
# x //= k
#
# return ret
# Driver code
# precalculating Smallest Prime Factor
# sieve()
def main():
n, m = map(int, input().split())
d = {}
for i in range(m):
a, b = map(int, input().split())
if a not in d.keys():
d[a] = []
d[a].append(b)
# ans = []
for i in range(1, n + 1):
rem = m
j = i
now = 0
k = {}
for o in d.keys():
k[o] = [g for g in d[o]]
z = set()
while rem:
ind = -1
val = -1
z.discard(j)
if j in k.keys() and len(k[j]) > 0:
for t in range(len(k[j])):
if k[j][t] < j:
dis = k[j][t] + n-j
else:
dis=k[j][t] -j
if dis > val:
val = dis
ind = t
z.add(k[j][ind])
k[j].pop(ind)
rem -= 1
if rem==0:
break
j += 1
now+=1
if j == n + 1:
j = 1
z.discard(j)
x = 1
#print(z, j, now)
if len(z):
if min(z) < j:
for o in z:
if o < j:
x = max(x, o)
now += (n - j + x)
else:
now += (max(z) - j)
print(now, end=' ')
return
if __name__ == "__main__":
main()
``` | output | 1 | 28,322 | 1 | 56,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds. | instruction | 0 | 28,323 | 1 | 56,646 |
Tags: brute force, greedy
Correct Solution:
```
def solve(candies,start,n):
#print(start)
time = 0
train = {}
while True:
far = 0
for i in candies:
if i[0] == start:
if i[1] < start:
far = max(far,i[1]+n)
else:
#print(far,i[1])
far = max(far,i[1])
if far != 0:
if far > n:
far -= n
i = (start,far)
if i not in train.keys():
train[i] = 1
else:
train[i] += 1
new = []
for i in candies:
if i[1] == start:
if i in train.keys():
train[i] -= 1
if train[i] == 0:
del train[i]
else:
new.append(i)
else:
new.append(i)
candies = new[:]
#print(candies,train,start)
if not candies:
break
time += 1
start += 1
if start == n+1:
start = 1
return time
def main():
n,m = map(int,input().split())
stations = {}
for i in range(1,n+1):
stations[i] = []
candies = []
for i in range(m):
a,b = map(int,input().split())
candies.append((a,b))
ans = []
for i in range(n):
ans.append(solve(candies[:],i+1,n))
for i in ans:
print(i,end = ' ')
main()
``` | output | 1 | 28,323 | 1 | 56,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds. | instruction | 0 | 28,324 | 1 | 56,648 |
Tags: brute force, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
import bisect
n,m=map(int,input().split())
AB=[list(map(int,input().split())) for i in range(m)]
C=[[] for i in range(n+1)]
for a,b in AB:
C[a].append(b)
C2=[0]*(n+1)#εγγ¦γγ©γηγγεΎγ«γγγζι
for i in range(1,n+1):
if C[i]==[]:
continue
ANS=(len(C[i])-1)*n
rest=[]
for c in C[i]:
if c>i:
rest.append(c-i)
else:
rest.append(n-i+c)
C2[i]=ANS+min(rest)
DIAG=list(range(n))
ANS=[]
for i in range(1,n+1):
K=DIAG[-i+1:]+DIAG[:-i+1]
#print(i,K)
ANS.append(max([K[j]+C2[j+1] for j in range(n) if C2[j+1]!=0]))
print(" ".join([str(a) for a in ANS]))
``` | output | 1 | 28,324 | 1 | 56,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds. | instruction | 0 | 28,325 | 1 | 56,650 |
Tags: brute force, greedy
Correct Solution:
```
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
n, m = I()
ans = [0 for i in range(n)]
l = [[] for i in range(5001)]
for i in range(m):
a, b = I()
a -= 1
b -= 1
l[a].append(b)
# print(l[1])
for i in range(n):
l[i].sort(key = lambda x:(x - i)%n, reverse = True)
for i in range(n):
res = 0
k = len(l[i])
if k:
res = (k-1)*n
res += (l[i][-1] - i)%n
ans[i] = res
res = [0 for i in range(n)]
# print(ans)
for i in range(n):
for j in range(n):
if ans[j]:
# print(j, i)
# print((j - i)%n, i, j, ans[j] + (j - i)%n)
res[i] = max(res[i], ans[j] + (j - i)%n)
print(*res)
``` | output | 1 | 28,325 | 1 | 56,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds. | instruction | 0 | 28,326 | 1 | 56,652 |
Tags: brute force, greedy
Correct Solution:
```
n, m = [int(x) for x in input().split()]
def distance(nrof_stations, a, b):
if a > b:
b += nrof_stations
return b - a
def get_n(stations, start):
max_d = 0
for i, c in enumerate(stations):
if len(c) == 0:
continue
max_d = max((distance(len(stations), start, i) + (len(c) - 1) * len(stations)
+ min([distance(len(stations), i, x) for x in c])), max_d)
return max_d
stations = []
for i in range(n):
stations.append([])
for _ in range(m):
i, j = [int(x)-1 for x in input().split()]
stations[i].append(j)
for i in range(n):
print(get_n(stations, i), end=" ")
``` | output | 1 | 28,326 | 1 | 56,653 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds. | instruction | 0 | 28,327 | 1 | 56,654 |
Tags: brute force, greedy
Correct Solution:
```
import typing
def _ceil_pow2(n: int) -> int:
x = 0
while (1 << x) < n:
x += 1
return x
def _bsf(n: int) -> int:
x = 0
while n % 2 == 0:
x += 1
n //= 2
return x
class SegTree:
def __init__(self,
op: typing.Callable[[typing.Any, typing.Any], typing.Any],
e: typing.Any,
v: typing.Union[int, typing.List[typing.Any]]) -> None:
self._op = op
self._e = e
if isinstance(v, int):
v = [e] * v
self._n = len(v)
self._log = _ceil_pow2(self._n)
self._size = 1 << self._log
self._d = [e] * (2 * self._size)
for i in range(self._n):
self._d[self._size + i] = v[i]
for i in range(self._size - 1, 0, -1):
self._update(i)
def set(self, p: int, x: typing.Any) -> None:
assert 0 <= p < self._n
p += self._size
self._d[p] = x
for i in range(1, self._log + 1):
self._update(p >> i)
def get(self, p: int) -> typing.Any:
assert 0 <= p < self._n
return self._d[p + self._size]
def prod(self, left: int, right: int) -> typing.Any:
assert 0 <= left <= right <= self._n
sml = self._e
smr = self._e
left += self._size
right += self._size
while left < right:
if left & 1:
sml = self._op(sml, self._d[left])
left += 1
if right & 1:
right -= 1
smr = self._op(self._d[right], smr)
left >>= 1
right >>= 1
return self._op(sml, smr)
def all_prod(self) -> typing.Any:
return self._d[1]
def max_right(self, left: int,
f: typing.Callable[[typing.Any], bool]) -> int:
assert 0 <= left <= self._n
assert f(self._e)
if left == self._n:
return self._n
left += self._size
sm = self._e
first = True
while first or (left & -left) != left:
first = False
while left % 2 == 0:
left >>= 1
if not f(self._op(sm, self._d[left])):
while left < self._size:
left *= 2
if f(self._op(sm, self._d[left])):
sm = self._op(sm, self._d[left])
left += 1
return left - self._size
sm = self._op(sm, self._d[left])
left += 1
return self._n
def min_left(self, right: int,
f: typing.Callable[[typing.Any], bool]) -> int:
assert 0 <= right <= self._n
assert f(self._e)
if right == 0:
return 0
right += self._size
sm = self._e
first = True
while first or (right & -right) != right:
first = False
right -= 1
while right > 1 and right % 2:
right >>= 1
if not f(self._op(self._d[right], sm)):
while right < self._size:
right = 2 * right + 1
if f(self._op(self._d[right], sm)):
sm = self._op(self._d[right], sm)
right -= 1
return right + 1 - self._size
sm = self._op(self._d[right], sm)
return 0
def _update(self, k: int) -> None:
self._d[k] = self._op(self._d[2 * k], self._d[2 * k + 1])
n,m=map(int,input().split())
candies=[]
INF=10**10
for i in range(n):
candies.append([])
for _ in range(m):
a,b=map(int,input().split())
candies[a-1].append(b-1)
for i in range(n):
candies[i].sort()
bestcandy=[0]*n
for i in range(n):
if candies[i]:
best_len=10**9
for candy in candies[i]:
if best_len>(candy-i)%n:
best_len=(candy-i)%n
bestcandy[i]=best_len
a=[]
for i in range(n):
if candies[i]:
a.append((len(candies[i])-1)*n+bestcandy[i]+i)
else:
a.append(0)
ans=[]
ans.append(max(a))
segtree = SegTree(max, -1, a)
for i in range(1,n):
k=segtree.get(i-1)
if k!=0:
segtree.set(i-1,k+n)
r=segtree.all_prod()
ans.append(r-i)
print(' '.join(map(str,ans)))
``` | output | 1 | 28,327 | 1 | 56,655 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds. | instruction | 0 | 28,328 | 1 | 56,656 |
Tags: brute force, greedy
Correct Solution:
```
from copy import deepcopy
n,m=map(int,input().split())
d={}
for i in range(1,n+1):
d[i]=[]
for i in range(m):
x,y=map(int,input().split())
d[x].append(y)
for i in d:
d[i].sort(reverse=True,key=lambda j: j+n-i if j<i else j-i)
for i in d:
x=[]
for j in d[i]:
if j!=i:
x.append(j)
d[i]=x
def get_candies(start,d):
t=0
for i in d:
t+=len(d[i])
a=[]
ans=0
c=start
while(t):
y=[]
for i in a:
if i==c:
t-=1
else:
y.append(i)
a=y
if len(d[c]):
x=d[c].pop(0)
a.append(x)
c=((c+1)%n)
if c==0:
c=n
ans+=1
return ans-1
for i in range(1,n+1):
print(get_candies(i,deepcopy(d)),end=' ')
print()
``` | output | 1 | 28,328 | 1 | 56,657 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds. | instruction | 0 | 28,329 | 1 | 56,658 |
Tags: brute force, greedy
Correct Solution:
```
n, m = list(map(int, input().split()))
a = []
for i in range(m):
a.append(list(map(int, input().split())))
locs = []
for i in range(n+1):
locs.append([])
for i in range(m):
locs[a[i][0]].append(a[i][1])
ans = [0]*(n+1)
for i in range(1, n+1):
mini = 2**n
for j in locs[i]:
if(j > i):
mini = min(mini, j-i)
else:
mini = min(mini, n-i+j)
if(mini == 2**n):
ans[i] = 0
else:
ans[i] = max(0, n*(len(locs[i])-1) + mini)
# print(*ans[1:])
for start in range(1, n+1):
maxi = 0
for j in range(1, n+1):
if(ans[j]):
if(j >= start):
maxi = max(maxi, j-start + ans[j])
else:
maxi = max(maxi, n - start + j + ans[j])
print(maxi, end=" ")
``` | output | 1 | 28,329 | 1 | 56,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds.
Submitted Solution:
```
n, m = map(int, input().split())
d = {}
for i in range(n):
d[i] = []
for i in range(m):
a,b = map(int, input().split())
d[a-1].append(b-1)
def fn(x, i):
if x >=i:
return x - i
else:
return n - 1 - i + x
for i in range(n):
d[i] = sorted(d[i], key=lambda x: fn(x, i))
md = 0
for i in d:
md = max(md, len(d[i]))
ans = []
for i in range(n):
cc = 0
dc = [0 for k in range(n)]
dp = [len(d[j]) - 1 for j in range(n)]
j = i
fans = -1
while cc < m:
fans += 1
cc += dc[j]
dc[j] = 0
if dp[j] > -1:
dc[d[j][dp[j]]] += 1
dp[j] -=1
j += 1
j %= n
ans.append(fans)
print(' '.join([str(i) for i in ans]))
``` | instruction | 0 | 28,330 | 1 | 56,660 |
Yes | output | 1 | 28,330 | 1 | 56,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds.
Submitted Solution:
```
n, m = [int(i) for i in input().split()]
A = [[int(i) for i in input().split()] for j in range(m)]
INF = 10 ** 9
mi = [INF] * (n+1)
cnt = [0] * (n+1)
for a, b in A:
cnt[a] += 1
mi[a] = min(mi[a], (b - a) % n)
ANS = []
for i in range(1, n+1):
ans = 0
for j in range(1, n+1):
if cnt[j] == 0:
continue
ans = max(ans, (j - i) % n + (cnt[j] - 1) * n + mi[j])
ANS.append(ans)
print(*ANS)
``` | instruction | 0 | 28,331 | 1 | 56,662 |
Yes | output | 1 | 28,331 | 1 | 56,663 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds.
Submitted Solution:
```
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
from sys import stdin, stdout
import math
#N = int(input())
#N,M,K = [int(x) for x in stdin.readline().split()]
N,M = [int(x) for x in stdin.readline().split()]
def dist(a,b,N):
if b>=a:
return b-a
else:
return b-a+N
station = [2*N]*(N+1)
cargo = [0]*(N+1)
for i in range(M):
a,b = [int(x) for x in stdin.readline().split()]
station[a] = min(station[a],dist(a,b,N))
cargo[a] += 1
for i in range(1,N+1):
maxi = 0
for j in range(1,N+1):
if cargo[j]>0:
d = station[j] + (cargo[j]-1)*N + dist(i,j,N)
#print(i,j,d)
maxi = max(maxi,d)
print(maxi,end=' ')
``` | instruction | 0 | 28,332 | 1 | 56,664 |
Yes | output | 1 | 28,332 | 1 | 56,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds.
Submitted Solution:
```
from collections import defaultdict as dd
import math
def nn():
return int(input())
def li():
return list(input())
def mi():
return map(int, input().split())
def lm():
return list(map(int, input().split()))
d={}
n,m=mi()
for i in range(m):
a,b=mi()
if a in d:
d[a].append((b-a)%n)
else:
d[a]=[(b-a)%n]
#print(d)
distances={}
answers=[]
for key in d:
distance=n*(len(d[key])-1)+min(d[key])
distances[key]=distance
for i in range(1,n+1):
md=0
for key in distances:
new=distances[key]+(key-i)%n
md=max(new, md)
answers.append(md)
print(*answers)
``` | instruction | 0 | 28,333 | 1 | 56,666 |
Yes | output | 1 | 28,333 | 1 | 56,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds.
Submitted Solution:
```
from sys import exit, setrecursionlimit
setrecursionlimit(10**6)
inn = lambda: input().strip()
mapp = lambda: map(int, inn().split(" "))
N,M = mapp()
station = [[] for _ in range(N)]
for _ in range(M):
a,b = mapp()
a,b = a-1, b-1
station[a].append(b)
answer = []
for start in range(N):
maxCycle = max(len(a) for a in station)
maxCycleStation = tuple(a for a in range(N) if len(station[a])==maxCycle)
atLastCycle = 0
for here in maxCycleStation:
station[here].sort(key=lambda there: (N + there - here)%N)
what = (N + here - start)%N + (N + station[here][0] - here)%N
atLastCycle = max(atLastCycle, what)
answer.append(N*(maxCycle-1) + atLastCycle)
print(" ".join(map(str, answer)))
``` | instruction | 0 | 28,334 | 1 | 56,668 |
No | output | 1 | 28,334 | 1 | 56,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds.
Submitted Solution:
```
def fastio():
import sys
from io import StringIO
from atexit import register
global input
sys.stdin = StringIO(sys.stdin.read())
input = lambda : sys.stdin.readline().rstrip('\r\n')
sys.stdout = StringIO()
register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
n, m = I()
ans = [0 for i in range(n)]
l = [[] for i in range(5001)]
for i in range(m):
a, b = I()
a -= 1
b -= 1
l[a].append(b)
# print(l[1])
for i in range(n):
l[i].sort(key = lambda x:(x - i)%n, reverse = True)
for i in range(n):
res = 0
k = len(l[i])
if k:
res = (k-1)*n
res += (l[i][-1] - i)%n
ans[i] = res
res = [0 for i in range(n)]
# print(ans)
for i in range(n):
for j in range(n):
# print(j, i)
# print((j - i)%n, i, j, ans[j] + (j - i)%n)
res[i] = max(res[i], ans[j] + (j - i)%n)
print(*res)
``` | instruction | 0 | 28,335 | 1 | 56,670 |
No | output | 1 | 28,335 | 1 | 56,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds.
Submitted Solution:
```
import sys
input = sys.stdin.readline
import bisect
n,m=map(int,input().split())
AB=[list(map(int,input().split())) for i in range(m)]
C=[[] for i in range(n+1)]
for a,b in AB:
C[a].append(b)
C2=[0]*(n+1)#εγγ¦γγ©γηγγεΎγ«γγγζι
for i in range(1,n+1):
if C[i]==[]:
continue
ANS=(len(C[i])-1)*n
rest=[]
for c in C[i]:
if c>i:
rest.append(c-i)
else:
rest.append(n-i+c)
C2[i]=ANS+min(rest)
DIAG=list(range(n))
ANS=[]
for i in range(1,n+1):
K=DIAG[i-1:]+DIAG[:i-1]
ANS.append(max([K[i]+C2[i+1] for i in range(n)]))
print(" ".join([str(a) for a in ANS]))
``` | instruction | 0 | 28,336 | 1 | 56,672 |
No | output | 1 | 28,336 | 1 | 56,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is a simplified version of the task Toy Train. These two versions differ only in the constraints. Hacks for this version are disabled.
Alice received a set of Toy Trainβ’ from Bob. It consists of one train and a connected railway network of n stations, enumerated from 1 through n. The train occupies one station at a time and travels around the network of stations in a circular manner. More precisely, the immediate station that the train will visit after station i is station i+1 if 1 β€ i < n or station 1 if i = n. It takes the train 1 second to travel to its next station as described.
Bob gave Alice a fun task before he left: to deliver m candies that are initially at some stations to their independent destinations using the train. The candies are enumerated from 1 through m. Candy i (1 β€ i β€ m), now at station a_i, should be delivered to station b_i (a_i β b_i).
<image> The blue numbers on the candies correspond to b_i values. The image corresponds to the 1-st example.
The train has infinite capacity, and it is possible to load off any number of candies at a station. However, only at most one candy can be loaded from a station onto the train before it leaves the station. You can choose any candy at this station. The time it takes to move the candies is negligible.
Now, Alice wonders how much time is needed for the train to deliver all candies. Your task is to find, for each station, the minimum time the train would need to deliver all the candies were it to start from there.
Input
The first line contains two space-separated integers n and m (2 β€ n β€ 100; 1 β€ m β€ 200) β the number of stations and the number of candies, respectively.
The i-th of the following m lines contains two space-separated integers a_i and b_i (1 β€ a_i, b_i β€ n; a_i β b_i) β the station that initially contains candy i and the destination station of the candy, respectively.
Output
In the first and only line, print n space-separated integers, the i-th of which is the minimum time, in seconds, the train would need to deliver all the candies were it to start from station i.
Examples
Input
5 7
2 4
5 1
2 3
3 4
4 1
5 3
3 5
Output
10 9 10 10 9
Input
2 3
1 2
1 2
1 2
Output
5 6
Note
Consider the second sample.
If the train started at station 1, the optimal strategy is as follows.
1. Load the first candy onto the train.
2. Proceed to station 2. This step takes 1 second.
3. Deliver the first candy.
4. Proceed to station 1. This step takes 1 second.
5. Load the second candy onto the train.
6. Proceed to station 2. This step takes 1 second.
7. Deliver the second candy.
8. Proceed to station 1. This step takes 1 second.
9. Load the third candy onto the train.
10. Proceed to station 2. This step takes 1 second.
11. Deliver the third candy.
Hence, the train needs 5 seconds to complete the tasks.
If the train were to start at station 2, however, it would need to move to station 1 before it could load the first candy, which would take one additional second. Thus, the answer in this scenario is 5+1 = 6 seconds.
Submitted Solution:
```
n, m=list(map(int, input().split()))
a=[0]*m
b=[0]*m
t=[[-1]]*n
for i in range(m):
a[i], b[i] = map(int, input().split())
if t[a[i]-1]==[-1]:
if b[i]>a[i]: t[a[i]-1]=[b[i]-a[i]]
if b[i]<a[i]: t[a[i]-1]=[n-a[i]+b[i]]
else:
if b[i]>a[i]: t[a[i]-1].append(b[i]-a[i])
if b[i]<a[i]: t[a[i]-1].append(n-a[i]+b[i])
res=[0]*n
mx=0
for i in range(n):
t[i].sort()
if t[i]==[-1]:
res[i]=0
continue
res[i]=t[i][0]+n*(len(t[i])-1)
if res[i]>res[mx]: mx=i
#print(t)
#print(res)
ans=[0]*n
for i in range(n):
mn=res[i]
for j in range(1,n):
mn=max(mn, res[(i+j)%n]+j)
ans[i]=mn
for i in range(n): print(ans[i], end=' ')
``` | instruction | 0 | 28,337 | 1 | 56,674 |
No | output | 1 | 28,337 | 1 | 56,675 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed). | instruction | 0 | 28,692 | 1 | 57,384 |
Tags: implementation
Correct Solution:
```
a, ta = map(int, input().split())
b, tb = map(int, input().split())
time = input().split(":")
mins = int(time[0]) * 60 + int(time[1])
ans = 0
for i in range(300, 1440, b):
if i+tb > mins and i < mins+ta:
ans += 1
print(ans)
``` | output | 1 | 28,692 | 1 | 57,385 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed). | instruction | 0 | 28,693 | 1 | 57,386 |
Tags: implementation
Correct Solution:
```
import math
[a, t_a] = map(int,input().split(" "))
[b, t_b] = map(int,input().split(" "))
[h, m] = map(int,input().split(":"))
mm = h * 60 + m
first_bus = 5*60
start = max(first_bus, mm - t_b + 1) - first_bus
end = min(60*23+59, mm + t_a - 1) - first_bus
# print(start, end)
# print(math.ceil(start/b), int(end/b))
print((int(end/b) - math.ceil(start/b)) + 1)
``` | output | 1 | 28,693 | 1 | 57,387 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed). | instruction | 0 | 28,694 | 1 | 57,388 |
Tags: implementation
Correct Solution:
```
import sys
#import random
from bisect import bisect_right as rb
from collections import deque
#sys.setrecursionlimit(10**8)
from queue import PriorityQueue
from math import *
input_ = lambda: sys.stdin.readline().strip("\r\n")
ii = lambda : int(input_())
il = lambda : list(map(int, input_().split()))
ilf = lambda : list(map(float, input_().split()))
ip = lambda : input_()
fi = lambda : float(input_())
ap = lambda ab,bc,cd : ab[bc].append(cd)
li = lambda : list(input_())
pr = lambda x : print(x)
prinT = lambda x : print(x)
f = lambda : sys.stdout.flush()
mod = 10**9 + 7
a,ta = il()
b,tb = il()
s = ip().split(':')
time = 60*int(s[0]) + int(s[1])
ans = 0
for i in range(5*60,24*60,b) :
if (i+tb > time and time+ta > i) :
ans += 1
print(ans)
``` | output | 1 | 28,694 | 1 | 57,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed). | instruction | 0 | 28,695 | 1 | 57,390 |
Tags: implementation
Correct Solution:
```
a, ta = map(int, input().split())
b, tb = map(int, input().split())
hour, minutes = map(int, input().split(':'))
time_dep = hour * 60 + minutes - 300
count = 0
for i in range(0, 1140, b):
if i + tb > time_dep and i < time_dep + ta:
count += 1
#print(i, i + tb, count)
print(count)
``` | output | 1 | 28,695 | 1 | 57,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed). | instruction | 0 | 28,696 | 1 | 57,392 |
Tags: implementation
Correct Solution:
```
import datetime as DT
def bus(a, t_a, b, t_b, t):
t1 = DT.datetime.strptime(t, '%H:%M')
t2 = DT.datetime(1900, 1, 1)
t_start = ((t1 - t2).total_seconds() / 60.0) - 300
t_end = t_start + t_a
t_elapsed = 0
z = 0
while(t_elapsed <= 23 * 60 + 59 - 300):
start = t_elapsed
end = start + t_b
if(max(t_start, start) < min(t_end, end)):
z += 1
t_elapsed += b
return z
if __name__ == "__main__":
a = input().split(' ')
b = input().split(' ')
t = input()
print(bus(int(a[0]), int(a[1]), int(b[0]), int(b[1]), t))
``` | output | 1 | 28,696 | 1 | 57,393 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed). | instruction | 0 | 28,697 | 1 | 57,394 |
Tags: implementation
Correct Solution:
```
a, ta = map(int, input().split(' '))
b, tb = map(int, input().split(' '))
depart = tuple(map(int, input().split(':')))
depart_t = 60*depart[0] + depart[1]
arrival = depart_t + ta
count = 0
ct = 5*60
while ct < arrival and ct < 24*60:
if ct > depart_t - tb:
count += 1
ct += b
print(count)
# busses must be on the road between depart and depart + ta
``` | output | 1 | 28,697 | 1 | 57,395 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed). | instruction | 0 | 28,698 | 1 | 57,396 |
Tags: implementation
Correct Solution:
```
f1 , t1 = [int(i) for i in input().split(" ")]
f2, t2 = [int(i) for i in input().split(" ")]
time = [str(i) for i in input().split(" ")]
time = time[0]
th, tm = time.split(":")
th =int(th)
tm = int(tm)
#print("th is {}, tm is {}".format(th,tm))
tt = ((th- 5) * 60) + tm
#print(tt)
bs = tt - t2 +1
bf = tt + t1 -1
count = 0
if bs < 0:
bs = 0
#print("start is {}, fin is {}".format(bs,bf))
ct = (18*60) + 59
if bf > ct:
bf = ct
#print("start is {}, fin is {}".format(bs,bf))
time = 0
for i in range(int(((18*60) + 59)/f2)+1):
if bs <= time <= bf:
count +=1
time = time + f2
print(count)
``` | output | 1 | 28,698 | 1 | 57,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed). | instruction | 0 | 28,699 | 1 | 57,398 |
Tags: implementation
Correct Solution:
```
a, t1 = map(int, input().split())
b, t2 = map(int, input().split())
hour, minute = map(int, input().split(':'))
minutef = minute + hour * 60 + t1
minutes = minute + hour * 60 - t2
cnt = 0
time = 300
while time < minutef and time < 240 * 6:
if time > minutes:
cnt += 1
time += b
print(cnt)
``` | output | 1 | 28,699 | 1 | 57,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed).
Submitted Solution:
```
import sys
import bisect
from bisect import bisect_left as lb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**6)
mod=10**9+7
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs(root,par):
if par!=-1:
dp[root]=dp[par]+1
for i in range(1,20):
if kthpar[root][i-1]!=-1:
kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
for child in gp[root]:
if child==par:continue
kthpar[child][0]=root
dfs(child,root)
def hnbhai():
a,ta=sd()
b,tb=sd()
hh,mm=input().split(":")
hh=int(hh)
mm=int(mm)
time=60*hh+mm
tot=0
for i in range(5*60,24*60,b):
if(i+tb>time and time+ta>i):
tot+=1
print(tot)
for _ in range(1):
hnbhai()
``` | instruction | 0 | 28,700 | 1 | 57,400 |
Yes | output | 1 | 28,700 | 1 | 57,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed).
Submitted Solution:
```
freqOfA, minSpentA = [int(x) for x in input().split()]
freqOfB, minSpentB = [int(x) for x in input().split()]
departTime = input().split(":")
departTime, Currdepart, encounters = 60 * int(departTime[0]) + int(departTime[1]), 5 * 60, 0 #departure time of the first bus
while True:
if Currdepart >= departTime + minSpentA or Currdepart > 23*60+59: break
if Currdepart + minSpentB > departTime:
encounters += 1
Currdepart += freqOfB
print(encounters)
``` | instruction | 0 | 28,701 | 1 | 57,402 |
Yes | output | 1 | 28,701 | 1 | 57,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed).
Submitted Solution:
```
s=input().split()
a=int(s[0])
ta=int(s[1])
s=input().split()
b=int(s[0])
tb=int(s[1])
s=input().split(':')
hh=int(s[0])
mm=int(s[1])
starttime=hh*60 + mm
endtime=starttime+ta
i=300
l=[]
while(i<1440):
l.append(i);
i+= b;
length=len(l)
c=0
for i in range(0, length):
if(l[i] < endtime and l[i]+tb >starttime):
c+=1
print(c)
``` | instruction | 0 | 28,702 | 1 | 57,404 |
Yes | output | 1 | 28,702 | 1 | 57,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed).
Submitted Solution:
```
a,ta=map(int,input().split())
b,tb=map(int,input().split())
h,m=map(int,input().split(':'))
sa=h*60+m
ea=sa+ta
c=0
i=0
sb=0
eb=sa
while sb<ea and sb<=1439:
if eb>sa:
c+=1
sb=5*60+b*i
eb=sb+tb
i+=1
print(c)
``` | instruction | 0 | 28,703 | 1 | 57,406 |
Yes | output | 1 | 28,703 | 1 | 57,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed).
Submitted Solution:
```
def solve(a, t1, t2):
l1 = t1 - t1 % a + a
if (l1 < 300):
l1 = 300
l2 = t2 - t2 % a
if (l2 > 1439):
l2 = 1439
return (l2-l1) // a + 1 - (l2 == t2)
def trans(h, m):
return 60 * h + m
data1 = [int(x) for x in input().split()]
data2 = [int(x) for x in input().split()]
data3 = input()
h0 = int(data3[0:2])
m0 = int(data3[-2:])
t0 = trans(h0, m0)
sol = solve(data2[0], t0 - data2[1], t0 + data1[1])
print(sol)
``` | instruction | 0 | 28,704 | 1 | 57,408 |
No | output | 1 | 28,704 | 1 | 57,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.
The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.
You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.
Input
The first line contains two integers a, ta (1 β€ a, ta β€ 120) β the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.
The second line contains two integers b, tb (1 β€ b, tb β€ 120) β the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.
The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.
Output
Print the only integer z β the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.
Examples
Input
10 30
10 35
05:20
Output
5
Input
60 120
24 100
13:00
Output
9
Note
In the first example Simion departs form the city A at 05:20 AM and arrives to the city B at 05:50 AM. He will meet the first 5 buses from the city B that departed in the period [05:00 AM - 05:40 AM]. Also Simion will meet a bus in the city B at 05:50 AM, but he will not count it.
Also note that the first encounter will be between 05:26 AM and 05:27 AM (if we suggest that the buses are go with the sustained speed).
Submitted Solution:
```
a, ta = map(int, input().split())
b, tb = map(int, input().split())
h,m = map(int, input().split(":"))
d = 60*h+m
c = 300
r = 0
while c <= d - tb:
c += b
while c < min(24*60 - 1,d + ta):
c += b
r += 1
print(r)
``` | instruction | 0 | 28,705 | 1 | 57,410 |
No | output | 1 | 28,705 | 1 | 57,411 |
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