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.
A famous sculptor Cicasso goes to a world tour!
Well, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn't he? Otherwise there will be no any exclusivity. So Cicasso will entirely hold the world tour in his native country — Berland.
Cicasso is very devoted to his work and he wants to be distracted as little as possible. Therefore he will visit only four cities. These cities will be different, so no one could think that he has "favourites". Of course, to save money, he will chose the shortest paths between these cities. But as you have probably guessed, Cicasso is a weird person. Although he doesn't like to organize exhibitions, he likes to travel around the country and enjoy its scenery. So he wants the total distance which he will travel to be as large as possible. However, the sculptor is bad in planning, so he asks you for help.
There are n cities and m one-way roads in Berland. You have to choose four different cities, which Cicasso will visit and also determine the order in which he will visit them. So that the total distance he will travel, if he visits cities in your order, starting from the first city in your list, and ending in the last, choosing each time the shortest route between a pair of cities — will be the largest.
Note that intermediate routes may pass through the cities, which are assigned to the tour, as well as pass twice through the same city. For example, the tour can look like that: <image>. Four cities in the order of visiting marked as overlines: [1, 5, 2, 4].
Note that Berland is a high-tech country. So using nanotechnologies all roads were altered so that they have the same length. For the same reason moving using regular cars is not very popular in the country, and it can happen that there are such pairs of cities, one of which generally can not be reached by car from the other one. However, Cicasso is very conservative and cannot travel without the car. Choose cities so that the sculptor can make the tour using only the automobile. It is guaranteed that it is always possible to do.
Input
In the first line there is a pair of integers n and m (4 ≤ n ≤ 3000, 3 ≤ m ≤ 5000) — a number of cities and one-way roads in Berland.
Each of the next m lines contains a pair of integers ui, vi (1 ≤ ui, vi ≤ n) — a one-way road from the city ui to the city vi. Note that ui and vi are not required to be distinct. Moreover, it can be several one-way roads between the same pair of cities.
Output
Print four integers — numbers of cities which Cicasso will visit according to optimal choice of the route. Numbers of cities should be printed in the order that Cicasso will visit them. If there are multiple solutions, print any of them.
Example
Input
8 9
1 2
2 3
3 4
4 1
4 5
5 6
6 7
7 8
8 5
Output
2 1 8 7
Note
Let d(x, y) be the shortest distance between cities x and y. Then in the example d(2, 1) = 3, d(1, 8) = 7, d(8, 7) = 3. The total distance equals 13.
Submitted Solution:
```
def bfs(n,g,d):
for i in range(1,n+1):
for j in range(1,n+1):
d[i][j]=-1
d[i][i]=0
q=[i]
while q:
u=q.pop()
for v in g[u]:
if d[i][v]==-1:
d[i][v]=d[i][u]+1
q.append(v)
d=[]
rev_d=[]
_max=[]
rev_max=[]
g=[]
rev_g=[]
ans=[0,0,0,0]
s=input().split()
n=int(s[0])
m=int(s[1])
for i in range(n+1):
d.append([])
rev_d.append([])
_max.append([])
rev_max.append([])
g.append([])
rev_g.append([])
for j in range(n+1):
d[i].append(0)
rev_d[i].append(0)
_max[i].append((0,0))
rev_max[i].append((0,0))
for i in range(1,m+1):
s=input().split()
u=int(s[0])
v=int(s[1])
g[u].append(v)
rev_g[v].append(u)
bfs(n,g,d)
bfs(n,rev_g,rev_d)
for i in range(1,n+1):
for j in range(1,n+1):
_max[i][j]=(d[i][j],j)
rev_max[i][j]=(rev_d[i][j],j)
_max[i].sort()
rev_max[i].sort()
res=0
for i in range(1,n+1):
for j in range(1,n+1):
for g in range(n,n-3-1,-1):
if _max[i][g][0]==-1:
break
for k in range(n,n-3-1,-1):
if rev_max[j][k][0]==-1:
break
ok=rev_max[j][k][1]!=j and rev_max[j][k][1]!=i and rev_max[j][k][1]!=_max[i][g][1] and j!=i and j!=_max[i][g][1] and i!=_max[i][g][1]
current_res=rev_max[j][k][0]+d[j][i]+_max[i][g][0]
if ok and d[j][i!=-1] and res<current_res:
res=current_res
ans[0]=rev_max[j][k][1]
ans[1]=j
ans[2]=i
ans[3]=_max[i][g][1]
print(str(ans[0])+" "+str(ans[1])+" "+str(ans[2])+" "+str(ans[3]))
``` | instruction | 0 | 48,398 | 1 | 96,796 |
No | output | 1 | 48,398 | 1 | 96,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A famous sculptor Cicasso goes to a world tour!
Well, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn't he? Otherwise there will be no any exclusivity. So Cicasso will entirely hold the world tour in his native country — Berland.
Cicasso is very devoted to his work and he wants to be distracted as little as possible. Therefore he will visit only four cities. These cities will be different, so no one could think that he has "favourites". Of course, to save money, he will chose the shortest paths between these cities. But as you have probably guessed, Cicasso is a weird person. Although he doesn't like to organize exhibitions, he likes to travel around the country and enjoy its scenery. So he wants the total distance which he will travel to be as large as possible. However, the sculptor is bad in planning, so he asks you for help.
There are n cities and m one-way roads in Berland. You have to choose four different cities, which Cicasso will visit and also determine the order in which he will visit them. So that the total distance he will travel, if he visits cities in your order, starting from the first city in your list, and ending in the last, choosing each time the shortest route between a pair of cities — will be the largest.
Note that intermediate routes may pass through the cities, which are assigned to the tour, as well as pass twice through the same city. For example, the tour can look like that: <image>. Four cities in the order of visiting marked as overlines: [1, 5, 2, 4].
Note that Berland is a high-tech country. So using nanotechnologies all roads were altered so that they have the same length. For the same reason moving using regular cars is not very popular in the country, and it can happen that there are such pairs of cities, one of which generally can not be reached by car from the other one. However, Cicasso is very conservative and cannot travel without the car. Choose cities so that the sculptor can make the tour using only the automobile. It is guaranteed that it is always possible to do.
Input
In the first line there is a pair of integers n and m (4 ≤ n ≤ 3000, 3 ≤ m ≤ 5000) — a number of cities and one-way roads in Berland.
Each of the next m lines contains a pair of integers ui, vi (1 ≤ ui, vi ≤ n) — a one-way road from the city ui to the city vi. Note that ui and vi are not required to be distinct. Moreover, it can be several one-way roads between the same pair of cities.
Output
Print four integers — numbers of cities which Cicasso will visit according to optimal choice of the route. Numbers of cities should be printed in the order that Cicasso will visit them. If there are multiple solutions, print any of them.
Example
Input
8 9
1 2
2 3
3 4
4 1
4 5
5 6
6 7
7 8
8 5
Output
2 1 8 7
Note
Let d(x, y) be the shortest distance between cities x and y. Then in the example d(2, 1) = 3, d(1, 8) = 7, d(8, 7) = 3. The total distance equals 13.
Submitted Solution:
```
def bfs(n,g,d):
for i in range(1,n+1):
for j in range(1,n+1):
d[i][j]=-1
d[i][i]=0
q=[i]
while q:
u=q.pop()
for v in g[u]:
if d[i][v]==-1:
d[i][v]=d[i][u]+1
q.append(v)
d=[]
rev_d=[]
_max=[]
rev_max=[]
g=[]
rev_g=[]
ans=[0,0,0,0]
s=input().split()
n=int(s[0])
m=int(s[1])
for i in range(n+1):
d.append([])
rev_d.append([])
_max.append([])
rev_max.append([])
g.append([])
rev_g.append([])
for j in range(n+1):
d[i].append(0)
rev_d[i].append(0)
_max[i].append((0,0))
rev_max[i].append((0,0))
for i in range(m):
s=input().split()
u=int(s[0])
v=int(s[1])
g[u].append(v)
rev_g[v].append(u)
bfs(n,g,d)
bfs(n,rev_g,rev_d)
for i in range(1,n+1):
for j in range(1,n+1):
_max[i][j]=(d[i][j],j)
rev_max[i][j]=(rev_d[i][j],j)
_max[i].sort()
rev_max[i].sort()
res=0
for i in range(1,n+1):
for j in range(1,n+1):
for g in range(n,n-3-1,-1):
if _max[i][g][0]==-1:
break
for k in range(n,n-3-1,-1):
if rev_max[j][k][0]==-1:
break
ok=rev_max[j][k][1]!=j and rev_max[j][k][1]!=i and rev_max[j][k][1]!=_max[i][g][1] and j!=i and j!=_max[i][g][1] and i!=_max[i][g][1]#q ninguno de los 4 vertices sean iguales
current_res=rev_max[j][k][0]+d[j][i]+_max[i][g][0]
if ok and d[j][i]!=-1 and res<current_res:
res=current_res
ans[0]=rev_max[j][k][1]
ans[1]=j
ans[2]=i
ans[3]=_max[i][g][1]
print(str(ans[0])+" "+str(ans[1])+" "+str(ans[2])+" "+str(ans[3]))
``` | instruction | 0 | 48,399 | 1 | 96,798 |
No | output | 1 | 48,399 | 1 | 96,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A famous sculptor Cicasso goes to a world tour!
Well, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn't he? Otherwise there will be no any exclusivity. So Cicasso will entirely hold the world tour in his native country — Berland.
Cicasso is very devoted to his work and he wants to be distracted as little as possible. Therefore he will visit only four cities. These cities will be different, so no one could think that he has "favourites". Of course, to save money, he will chose the shortest paths between these cities. But as you have probably guessed, Cicasso is a weird person. Although he doesn't like to organize exhibitions, he likes to travel around the country and enjoy its scenery. So he wants the total distance which he will travel to be as large as possible. However, the sculptor is bad in planning, so he asks you for help.
There are n cities and m one-way roads in Berland. You have to choose four different cities, which Cicasso will visit and also determine the order in which he will visit them. So that the total distance he will travel, if he visits cities in your order, starting from the first city in your list, and ending in the last, choosing each time the shortest route between a pair of cities — will be the largest.
Note that intermediate routes may pass through the cities, which are assigned to the tour, as well as pass twice through the same city. For example, the tour can look like that: <image>. Four cities in the order of visiting marked as overlines: [1, 5, 2, 4].
Note that Berland is a high-tech country. So using nanotechnologies all roads were altered so that they have the same length. For the same reason moving using regular cars is not very popular in the country, and it can happen that there are such pairs of cities, one of which generally can not be reached by car from the other one. However, Cicasso is very conservative and cannot travel without the car. Choose cities so that the sculptor can make the tour using only the automobile. It is guaranteed that it is always possible to do.
Input
In the first line there is a pair of integers n and m (4 ≤ n ≤ 3000, 3 ≤ m ≤ 5000) — a number of cities and one-way roads in Berland.
Each of the next m lines contains a pair of integers ui, vi (1 ≤ ui, vi ≤ n) — a one-way road from the city ui to the city vi. Note that ui and vi are not required to be distinct. Moreover, it can be several one-way roads between the same pair of cities.
Output
Print four integers — numbers of cities which Cicasso will visit according to optimal choice of the route. Numbers of cities should be printed in the order that Cicasso will visit them. If there are multiple solutions, print any of them.
Example
Input
8 9
1 2
2 3
3 4
4 1
4 5
5 6
6 7
7 8
8 5
Output
2 1 8 7
Note
Let d(x, y) be the shortest distance between cities x and y. Then in the example d(2, 1) = 3, d(1, 8) = 7, d(8, 7) = 3. The total distance equals 13.
Submitted Solution:
```
from heapq import *
n,m=map(int,input().split())
a=[[10000000]*(n+1) for i in range(n+1)]
E=[set() for i in range(n+1)]
for i in range(m):
po,ki=map(int,input().split())
if po==ki: continue
E[po].add(ki)
for i in range(1,n+1):
ch=[(0,i)]; s={i}
while ch!=[]:
l,nom=heappop(ch)
if a[i][nom]<=l: continue
a[i][nom]=l
for x in E[nom]:
if x in s: continue
heappush(ch,(l+1,x))
s.add(x)
ch=[]
for i in range(n):
for j in range(n):
if a[i][j]<10000 and i!=j:
heappush(ch,(a[i+1][j+1],(i+1,j+1)))
while True:
l,s=heappop(ch)
if len(s)==4: print(' '.join(map(str,s))); break
for x in range(1,n+1):
if x in s: continue
if a[s[-1]][x]>10000: continue
heappush(ch,(-l,tuple(list(s)+[x])))
``` | instruction | 0 | 48,400 | 1 | 96,800 |
No | output | 1 | 48,400 | 1 | 96,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A famous sculptor Cicasso goes to a world tour!
Well, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn't he? Otherwise there will be no any exclusivity. So Cicasso will entirely hold the world tour in his native country — Berland.
Cicasso is very devoted to his work and he wants to be distracted as little as possible. Therefore he will visit only four cities. These cities will be different, so no one could think that he has "favourites". Of course, to save money, he will chose the shortest paths between these cities. But as you have probably guessed, Cicasso is a weird person. Although he doesn't like to organize exhibitions, he likes to travel around the country and enjoy its scenery. So he wants the total distance which he will travel to be as large as possible. However, the sculptor is bad in planning, so he asks you for help.
There are n cities and m one-way roads in Berland. You have to choose four different cities, which Cicasso will visit and also determine the order in which he will visit them. So that the total distance he will travel, if he visits cities in your order, starting from the first city in your list, and ending in the last, choosing each time the shortest route between a pair of cities — will be the largest.
Note that intermediate routes may pass through the cities, which are assigned to the tour, as well as pass twice through the same city. For example, the tour can look like that: <image>. Four cities in the order of visiting marked as overlines: [1, 5, 2, 4].
Note that Berland is a high-tech country. So using nanotechnologies all roads were altered so that they have the same length. For the same reason moving using regular cars is not very popular in the country, and it can happen that there are such pairs of cities, one of which generally can not be reached by car from the other one. However, Cicasso is very conservative and cannot travel without the car. Choose cities so that the sculptor can make the tour using only the automobile. It is guaranteed that it is always possible to do.
Input
In the first line there is a pair of integers n and m (4 ≤ n ≤ 3000, 3 ≤ m ≤ 5000) — a number of cities and one-way roads in Berland.
Each of the next m lines contains a pair of integers ui, vi (1 ≤ ui, vi ≤ n) — a one-way road from the city ui to the city vi. Note that ui and vi are not required to be distinct. Moreover, it can be several one-way roads between the same pair of cities.
Output
Print four integers — numbers of cities which Cicasso will visit according to optimal choice of the route. Numbers of cities should be printed in the order that Cicasso will visit them. If there are multiple solutions, print any of them.
Example
Input
8 9
1 2
2 3
3 4
4 1
4 5
5 6
6 7
7 8
8 5
Output
2 1 8 7
Note
Let d(x, y) be the shortest distance between cities x and y. Then in the example d(2, 1) = 3, d(1, 8) = 7, d(8, 7) = 3. The total distance equals 13.
Submitted Solution:
```
from math import pi
from collections import deque
from itertools import combinations, chain
from copy import deepcopy
import heapq
import sys
if sys.version_info[0] == 2:
input = raw_input
def rlist(t):
return map(t, input().split())
def read_int_list():
return rlist(int)
def write_list(lst, divider=" "):
print(divider.join(map(str, lst)))
class Matrix(object):
def __init__(self, m, n, default_val=0, filled=None):
self.m, self.n = m, n
self._elements = [[default_val for _ in range(n)] for _ in range(m)]
if isinstance(filled, dict):
for (x, y), value in filled.items():
self._elements[x][y] = value
self.transposed = 0
def __getitem__(self, item):
x, y = item
if self.transposed:
x, y = y, x
return self._elements[x][y]
def __setitem__(self, key, value):
x, y = key
if self.transposed:
x, y = y, x
self._elements[x][y] = value
def __repr__(self):
return "\n".join((" ".join(map(str, (self[i, j] for j in range(self.n)))) for i in range(self.m)))
def maximum(self, axis=None):
if axis is None:
return max(chain.from_iterable(self._elements))
axis ^= self.transposed
if axis == 0:
return [max(i) for i in range(self.m)]
if axis == 1:
return [max(self[i, j] for i in range(self.m)) for j in range(self.n)]
def transpose(self):
self.m, self.n = self.n, self.m
self.transposed = 1 - self.transposed
def column(self, j):
for i in range(self.m):
yield self[i, j]
def row(self, i):
for j in range(self.n):
yield self[i, j]
def columns(self):
for j in range(self.n):
yield self.column(j)
def rows(self):
for i in range(self.m):
yield self.row(i)
class Graph(object):
def __init__(self, vertices=0):
self.neighbours = [set() for _ in range(vertices)]
self.size = vertices
self.distance_table = {}
def add_vertex(self):
self.neighbours.append(set())
self.size += 1
def add_edge(self, v1, v2):
self.neighbours[v1].add(v2)
def add_nonor_edge(self, v1, v2):
self.neighbours[v1].add(v2)
self.neighbours[v2].add(v1)
def bfs(self, start=0):
queue = deque([start])
unused = [True for _ in range(self.size)]
unused[start] = False
while queue:
curr = queue.popleft()
for v in self.neighbours[curr]:
if unused[v]:
unused[v] = False
queue.append(v)
yield v, curr # v<-curr
def dfs(self, start=0):
stack = [[start, iter(self.neighbours[start])]]
unused = [True for _ in range(self.size)]
unused[start] = False
while stack:
curr, it = stack[-1]
unbroken = True
for v in it:
if unused[v]:
unused[v] = False
# stack[-1][1] = it
stack.append([v, iter(self.neighbours[v])])
yield v, curr # v<-curr
unbroken = False
break
if unbroken:
stack.pop()
def init_distance_table(self, vertex=None):
# If None, calculate all distances, otherwise only for vertex
if isinstance(self.distance_table, Matrix):
return
if vertex is None:
self.distance_table = Matrix(self.size, self.size, default_val=-1, filled=self.distance_table)
for i in range(self.size):
self.distance_table[i, i] = 0
for i in (range(self.size) if (vertex is None) else (vertex,)):
for j, prev in self.bfs(start=i):
self.distance_table[i, j] = self.distance_table[i, prev] + 1
def read_edges(self, edges, shift=1):
for _ in range(edges):
v1, v2 = read_int_list()
self.add_edge(v1-shift, v2-shift)
def read_nonor_edges(self, edges, shift=1):
for _ in range(edges):
v1, v2 = read_int_list()
self.add_nonor_edge(v1-shift, v2-shift)
def has_edge(self, v1, v2):
return v2 in self.neighbours[v1]
def degree(self, vertex):
return len(self.neighbours[vertex])
# True if vertex_set forms a full/empty subgraph
def check_full(self, vertex_set, empty=False): # If true, then check for emptiness, else for fullness
for vertex1, vertex2 in combinations(vertex_set, 2):
if empty ^ self.has_edge(vertex1, vertex2):
return False
return True
# True if all edges/no edges between 2 sets
def check_edges(self, set1, set2, no=False): # If true, then check for no edges, else for all edges
for vertex1 in set1:
for vertex2 in set2:
if no ^ self.has_edge(vertex1, vertex2):
return False
return True
def vertices_of_degree(self, degree, return_bad=False):
bad = -1
ans = set()
for i, vertex in enumerate(self.neighbours):
if len(vertex) == degree:
ans.add(i)
elif return_bad:
bad = i
if return_bad:
return bad, ans
return ans
def reverse_graph(self):
ans = Graph(self.size)
for i, nei in enumerate(self.neighbours):
for j in nei:
ans.add_edge(j, i)
if isinstance(self.distance_table, Matrix):
ans.distance_table = deepcopy(self.distance_table)
ans.distance_table.transpose()
if isinstance(self.distance_table, dict):
ans.distance_table = dict(((y, x), value) for (x, y), value in self.distance_table.items())
return ans
def swapenumerate(iterable, start=0):
cnt = start
for i in iterable:
yield i, cnt
cnt += 1
def main():
from time import clock
x = clock()
v, e = read_int_list()
graph = Graph(v)
graph.read_edges(e)
graph.init_distance_table()
dt = graph.distance_table
if v > 2000:
print(clock()-x)
largest_next = [heapq.nlargest(3, swapenumerate(row)) for row in dt.rows()]
largest_prev = [heapq.nlargest(3, swapenumerate(column)) for column in dt.columns()]
print(largest_next)
print(largest_prev)
if v > 2000:
print(clock()-x)
best_found = None, 0
for b in range(v):
for c in range(v):
if dt[b, c] > 0:
d1 = dt[b, c]
for d0, a in largest_prev[b]:
for d2, d in largest_next[c]:
path, length = {a, b, c, d}, d0 + d1 + d2
if len(path) == 4 and length > best_found[1]:
best_found = (a, b, c, d), length
write_list(map(lambda i: i+1, best_found[0]))
if __name__ == '__main__':
main()
``` | instruction | 0 | 48,401 | 1 | 96,802 |
No | output | 1 | 48,401 | 1 | 96,803 |
Provide a correct Python 3 solution for this coding contest problem.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60 | instruction | 0 | 48,686 | 1 | 97,372 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
W, H = map(int, input().split())
costs = []
for _ in range(W):
costs.append((int(input()), 0))
for _ in range(H):
costs.append((int(input()), 1))
costs.sort(key=lambda x: x[0])
h = H+1
w = W+1
ans = 0
for cost, flg in costs:
if flg == 0:
ans += cost * h
w -= 1
else:
ans += cost * w
h -= 1
print(ans)
``` | output | 1 | 48,686 | 1 | 97,373 |
Provide a correct Python 3 solution for this coding contest problem.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60 | instruction | 0 | 48,687 | 1 | 97,374 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
W, H = map(int, input().split())
p = [int(input()) for i in range(W)]
q = [int(input()) for i in range(H)]
p.sort()
q.sort()
row = H + 1
col = W + 1
i = j = 0
ans = 0
while row > 1 or col > 1:
if i < W and (j == H or p[i] < q[j]):
ans += p[i] * row
col -= 1
i += 1
else:
ans += q[j] * col
row -= 1
j += 1
print(ans)
``` | output | 1 | 48,687 | 1 | 97,375 |
Provide a correct Python 3 solution for this coding contest problem.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60 | instruction | 0 | 48,688 | 1 | 97,376 |
"Correct Solution:
```
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")
w,h = list(map(int, input().split()))
p = [None]*(w+h)
for i in range(w):
p[i] = (int(input()), 0)
for j in range(h):
p[w+j] = (int(input()), 1)
p.sort()
nums = [h+1,w+1]
ans = 0
i = 0
es = 0
while es<(h+1)*(w+1)-1:# and nums[0]<w+1 and nums[1]<h+1:
v, d = p[i]
if nums[d]==0:
pass
else:
ans += v * nums[d]
es += nums[d]
nums[int(not d)] -= 1
i += 1
# print(ans, nums)
print(ans)
``` | output | 1 | 48,688 | 1 | 97,377 |
Provide a correct Python 3 solution for this coding contest problem.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60 | instruction | 0 | 48,689 | 1 | 97,378 |
"Correct Solution:
```
def main():
import sys
input = sys.stdin.readline
from bisect import bisect_left, bisect_right
W, H = map(int, input().split())
P = [int(input()) for _ in range(W)]
P.sort()
Q = [int(input()) for _ in range(H)]
Q.sort()
ans = 0
for q in Q:
tmp = bisect_right(P, q)
# print ('tmp=', tmp)
ans += q * (1 + W - tmp)
# print (ans)
for p in P:
tmp = bisect_left(Q, p)
ans += p * (1 + H - tmp)
print (ans)
# print ('P:', P)
# print ('Q:', Q)
if __name__ == '__main__':
main()
``` | output | 1 | 48,689 | 1 | 97,379 |
Provide a correct Python 3 solution for this coding contest problem.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60 | instruction | 0 | 48,690 | 1 | 97,380 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
from bisect import bisect_left, bisect_right
W, H = map(int, input().split())
P = [int(input()) for _ in range(W)]
P.sort()
Q = [int(input()) for _ in range(H)]
Q.sort()
ans = 0
for q in Q:
tmp = bisect_right(P, q)
# print ('tmp=', tmp)
ans += q * (1 + W - tmp)
# print (ans)
for p in P:
tmp = bisect_left(Q, p)
ans += p * (1 + H - tmp)
print (ans)
# print ('P:', P)
# print ('Q:', Q)
``` | output | 1 | 48,690 | 1 | 97,381 |
Provide a correct Python 3 solution for this coding contest problem.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60 | instruction | 0 | 48,691 | 1 | 97,382 |
"Correct Solution:
```
import sys
import heapq
from collections import defaultdict as dd
input = sys.stdin.readline
W, H = map(int, input().split())
pq = []
for _ in range(W): pq.append((int(input()), 0))
for _ in range(H): pq.append((int(input()), 1))
pq.sort(key = lambda x: -x[0])
a = W + 1
b = H + 1
res = 0
for _ in range(W + H):
x = pq.pop()
#print(x)
if x[1]:
b -= 1
res += a * x[0]
else:
a -= 1
res += b * x[0]
#print(res)
print(res)
``` | output | 1 | 48,691 | 1 | 97,383 |
Provide a correct Python 3 solution for this coding contest problem.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60 | instruction | 0 | 48,692 | 1 | 97,384 |
"Correct Solution:
```
W, H = map(int, input().split())
P = []
Q = []
for i in range(W):
P.append(int(input()))
for i in range(H):
Q.append(int(input()))
edges = []
for p in P:
edges.append((p, 0))
for q in Q:
edges.append((q, 1))
edges.sort()
def r(xy):
if xy == 0:
return 1
return 0
WH = (W + 1, H + 1)
C = [0, 0]
answer = 0
for w, xy in edges:
C[xy] += 1
n = WH[r(xy)] - C[r(xy)]
answer += n * w
print(answer)
``` | output | 1 | 48,692 | 1 | 97,385 |
Provide a correct Python 3 solution for this coding contest problem.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60 | instruction | 0 | 48,693 | 1 | 97,386 |
"Correct Solution:
```
W,H=map(int,input().split())
from heapq import heappush,heappop,heapify
p,q=[int(input()) for i in range(W)],[int(input()) for i in range(H)]
heapify(p),heapify(q)
r,i,j=0,W+1,H+1
while i>1 and j>1:
if p[0]<q[0]:
r+=heappop(p)*j
i-=1
else:
r+=heappop(q)*i
j-=1
print(r+sum(p)+sum(q))
``` | output | 1 | 48,693 | 1 | 97,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60
Submitted Solution:
```
W, H = map(int, input().split())
PQ = [[int(input()), 'x'] for w in range(W)]
PQ += [[int(input()), 'y'] for h in range(H)]
PQ.sort()
_W = W + 1; _H = H + 1
ans = 0
for v, z in PQ:
if z == 'x':
if _W > 1:
ans += _H * v
_W -= 1
else: # z == 'y'
if _H > 1:
ans += _W * v
_H -= 1
if _W == _H == 1:
break
print(ans)
``` | instruction | 0 | 48,694 | 1 | 97,388 |
Yes | output | 1 | 48,694 | 1 | 97,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60
Submitted Solution:
```
import os
import sys
if os.getenv("LOCAL"):
sys.stdin = open("_in.txt", "r")
sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# あああああああああああくそHとWぎゃくだ
W,H = list(map(int, sys.stdin.readline().split()))
P = [int(sys.stdin.readline()) for _ in range(W)]
Q = [int(sys.stdin.readline()) for _ in range(H)]
edges = []
for p in P:
edges.append((p, 'P'))
for q in Q:
edges.append((q, 'Q'))
edges.sort()
ans = 0
a = W + 1
b = H + 1
for e, pq in edges:
if pq == 'P':
ans += e * b
a -= 1
else:
ans += e * a
b -= 1
print(ans)
``` | instruction | 0 | 48,695 | 1 | 97,390 |
Yes | output | 1 | 48,695 | 1 | 97,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60
Submitted Solution:
```
W, H, *L = map(int, open(0).read().split())
p = L[:W]
q = L[W:]
H += 1
W += 1
h = 0
w = 0
ls = [(c,0) for c in p]+[(c,1) for c in q]
ls.sort()
ans = 0
for c,s in ls:
if s==1:
ans += c*(W-w)
h += 1
if s==0:
ans += c*(H-h)
w += 1
print(ans)
``` | instruction | 0 | 48,696 | 1 | 97,392 |
Yes | output | 1 | 48,696 | 1 | 97,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60
Submitted Solution:
```
w,h = map(int,input().split())
arr=[]
for i in range(w):
arr.append( (int(input()),'p') )
for i in range(h):
arr.append( (int(input()),'q') )
arr.sort()
#print(arr)
ans=0
for a in arr:
if a[1]=='p':
ans+=a[0]*(h+1)
w-=1
else:
ans+=a[0]*(w+1)
h-=1
print(ans)
``` | instruction | 0 | 48,697 | 1 | 97,394 |
Yes | output | 1 | 48,697 | 1 | 97,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60
Submitted Solution:
```
W,H = map(int,input().split())
PQ = [(int(input()),0,i) for i in range(W)] + [(int(input()),1,i) for i in range(H)]
PQ.sort()
import sys
sys.setrecursionlimit(10**7)
class UnionFind:
def __init__(self):
self.Parent = {}
def add(self,k):
self.Parent[k] = k
def get_Parent(self,n):
if self.Parent[n] == n:return n
p = self.get_Parent(self.Parent[n])
self.Parent[n] = p
return p
def merge(self,x,y):
x = self.get_Parent(x)
y = self.get_Parent(y)
if x!=y: self.Parent[y] = x
return
def is_united(self,x,y):
return self.get_Parent(x)==self.get_Parent(y)
ans = 0
T = UnionFind()
for i in range(W+1):
for j in range(H+1):
T.add((i,j))
for c,pq,ij in PQ:
if pq==0:
i = ij
for j in range(H+1):
if not T.is_united((i,j),(i+1,j)):
T.merge((i,j),(i+1,j))
ans += c
else:
j = ij
for i in range(W+1):
if not T.is_united((i,j),(i,j+1)):
T.merge((i,j),(i,j+1))
ans += c
print(ans)
``` | instruction | 0 | 48,698 | 1 | 97,396 |
No | output | 1 | 48,698 | 1 | 97,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60
Submitted Solution:
```
W,H=map(int,input().split())
p=sorted(int(input())for _ in'_'*W)
p+=10**18,
q=sorted(int(input())for _ in'_'*H)
q+=10**18,
z=i=j=0
h,w=H+1,W+1
while i<W or j<H:
if h*p[i]<w*q[j]:
z+=h*p[i]
i+=1
w-=1
else:
z+=w*q[j]
j+=1
h-=1
print(z)
``` | instruction | 0 | 48,699 | 1 | 97,398 |
No | output | 1 | 48,699 | 1 | 97,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60
Submitted Solution:
```
w, h = map(int, input().split())
w += 1
h += 1
mst = [[-1 for y in range(h)] for x in range(w)]
edges = []
for x in range(w-1):
p = int(input())
for y in range(h):
edges.append((p, (x, y), (x+1, y)))
for y in range(h-1):
p = int(input())
for x in range(w):
edges.append((p, (x, y), (x, y+1)))
edges.sort()
print(mst)
print(edges)
groups = 0
mst_cost = 0
for edge in edges:
if mst[edge[1][0]][edge[1][1]] == -1 and mst[edge[2][0]][edge[2][1]] == -1:
#neither vertex is yet in the mst, add to a new group
mst[edge[1][0]][edge[1][1]] = groups
mst[edge[2][0]][edge[2][1]] = groups
groups += 1
elif mst[edge[1][0]][edge[1][1]] != -1 and mst[edge[2][0]][edge[2][1]] == -1:
# first vertex is in the mst, add second vertex to group
mst[edge[2][0]][edge[2][1]] = mst[edge[1][0]][edge[1][1]]
elif mst[edge[1][0]][edge[1][1]] != -1 and mst[edge[2][0]][edge[2][1]] == -1:
# second vertex is in the mst, add first vertex to group
mst[edge[1][0]][edge[1][1]] = mst[edge[2][0]][edge[2][1]]
else:
# both verticies are in mst
if mst[edge[1][0]][edge[1][1]] == mst[edge[2][0]][edge[2][1]]:
# if they're in the same group do nothing
continue
else:
# if they're in different groups, merge groups
new_group = mst[edge[2][0]][edge[2][1]]
old_group = mst[edge[1][0]][edge[1][1]]
for x in range(w):
for y in range(h):
if mst[x][y] == old_group:
mst[x][y] = new_group
print(mst)
mst_cost += edge[0]
print(mst_cost)
``` | instruction | 0 | 48,700 | 1 | 97,400 |
No | output | 1 | 48,700 | 1 | 97,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On an xy plane, in an area satisfying 0 ≤ x ≤ W, 0 ≤ y ≤ H, there is one house at each and every point where both x and y are integers.
There are unpaved roads between every pair of points for which either the x coordinates are equal and the difference between the y coordinates is 1, or the y coordinates are equal and the difference between the x coordinates is 1.
The cost of paving a road between houses on coordinates (i,j) and (i+1,j) is p_i for any value of j, while the cost of paving a road between houses on coordinates (i,j) and (i,j+1) is q_j for any value of i.
Mr. Takahashi wants to pave some of these roads and be able to travel between any two houses on paved roads only. Find the solution with the minimum total cost.
Constraints
* 1 ≦ W,H ≦ 10^5
* 1 ≦ p_i ≦ 10^8(0 ≦ i ≦ W-1)
* 1 ≦ q_j ≦ 10^8(0 ≦ j ≦ H-1)
* p_i (0 ≦ i ≦ W−1) is an integer.
* q_j (0 ≦ j ≦ H−1) is an integer.
Input
Inputs are provided from Standard Input in the following form.
W H
p_0
:
p_{W-1}
q_0
:
q_{H-1}
Output
Output an integer representing the minimum total cost.
Examples
Input
2 2
3
5
2
7
Output
29
Input
4 3
2
4
8
1
2
9
3
Output
60
Submitted Solution:
```
import sys
input = sys.stdin.readline
W,H=map(int,input().split())
p=sorted(int(input()) for i in range(W))
q=sorted(int(input()) for i in range(H))
ans=sum(p)*(H+1)+sum(q)*(W+1)
for i in range(W+H-1):
if p[len(p)-1]>=q[len(q)-1]:
ans-=p[len(p)-1]*len(q)
p.pop()
else:
ans-=q[len(q)-1]*len(p)
q.pop()
print(ans)
``` | instruction | 0 | 48,701 | 1 | 97,402 |
No | output | 1 | 48,701 | 1 | 97,403 |
Provide a correct Python 3 solution for this coding contest problem.
There are various parking lots such as three-dimensional type and tower type in the city to improve the utilization efficiency of the parking lot. In some parking lots, a "two-stage parking device" as shown in the figure is installed in one parking space to secure a parking space for two cars. This two-stage parking device allows one to be placed on an elevating pallet (a flat iron plate on which a car is placed) and parked in the upper tier, and the other to be parked in the lower tier.
In a parking lot that uses such a two-stage parking device, it is necessary to take out the car parked in the lower tier and dismiss it each time to put in and out the car in the upper tier, so be sure to manage it. Keeps the key to the parked car and puts it in and out as needed.
| <image>
--- | ---
Tsuruga Parking Lot is also one of the parking lots equipped with such a two-stage parking device, but due to lack of manpower, the person who cannot drive the car has become the manager. Therefore, once the car was parked, it could not be moved until the customer returned, and the car in the upper tier could not be put out until the owner of the car in the lower tier returned.
Create a program that meets the rules of the Tsuruga parking lot to help the caretaker who has to handle the cars that come to park one after another.
Tsuruga parking lot facilities
* There is one or more parking spaces, all equipped with a two-stage parking device.
* Each parking space is numbered starting from 1.
* Initially, it is assumed that no car is parked in the parking lot.
Tsuruga parking lot adopts the following rules.
When to stop the car
* The caretaker will be informed of the parking time of the car to be parked.
* We will park first in the parking space where no car is parked.
* If there is no parking space where no car is parked, park in an empty parking space. However, if there are multiple such parking spaces, follow the procedure below to park.
1. If the remaining parking time of a parked car is longer than the parking time of the car you are trying to park, park in the parking space with the smallest difference.
2. If the remaining parking time of any parked car is less than the parking time of the car you are trying to park, park in the parking space with the smallest difference.
* If the car is full (no parking space available), the car you are trying to park will wait in turn until the parking space is available. As soon as it becomes available, we will park in order from the first car we were waiting for.
* Under each condition, if there are multiple applicable parking spaces, the parking space number will be the smallest. In addition, if there are cars leaving at the same time, parking will start after all the cars leaving at the same time, and as long as there are cars waiting, the cars that can be parked will be parked at the same time.
When the car leaves
* Cars that have passed the parking time notified by the manager will be shipped.
* If there are cars in multiple parking spaces that have passed the parking time at the same time, the car with the smallest parking space number will be shipped first.
* If the parking time of the car parked in the upper row has expired, you will have to wait until the car in the lower row leaves the garage. The upper car will be delivered at the same time after the lower car is delivered.
The figure below shows an example of how to park at Tsuruga Parking Lot. In this example, the number of parking spaces is 3, and cars B to E are already parked. Consider that car A, which has a parking time of 70 minutes, arrives there. You cannot park because two cars are already parked in parking space 3, and you will have to park in either parking space 1 or parking space 2 that is still vacant. Car B parked in parking space 1 has 50 minutes remaining and car C parked in parking space 2 has 22 minutes remaining, both of which are less than car A's parking time, so car A's parking Park in parking space 1 where car B, which has a smaller time difference, is parked. As a result, car B, which was parked earlier, will be in the upper row.
<image>
Create a program that inputs the number of parking spaces m, the number of cars parked n, and the parking time t of each car, and outputs the car reference numbers in the order in which they come out of the parking lot. However, cars are assigned an integer reference number starting with 1 in the order of input, and cars will come to park one by one every 10 minutes in the order of the reference number.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
t1
t2
::
tn
The first line gives the number of two-stage parking devices m (1 ≤ m ≤ 10) and the number of cars parked n (1 ≤ n ≤ 100). The next n lines are given the parking time ti (1 ≤ ti ≤ 120) for the i-th car.
The number of datasets does not exceed 20.
Output
The reference number of the car is output to one line in the order of coming out of the parking lot for each data set. Please output the reference numbers separated by blanks.
Example
Input
3 5
90
52
82
84
70
2 4
10
30
40
60
0 0
Output
2 5 1 4 3
1 2 4 3 | instruction | 0 | 48,718 | 1 | 97,436 |
"Correct Solution:
```
from collections import deque
def out(time, parking):
x = len(parking)
outs = []
for i in range(x):
for j in range(2):
if parking[i][j] != None:
parking[i][j][0] -= time
for i in range(x):
c1 = parking[i][0]
c2 = parking[i][1]
if c1 != None and c2 != None:
if c1[0] <= 0 and c2[0] <= 0:
outs.append([c2[1], c1[1]])
parking[i][0] = None
parking[i][1] = None
elif c2[0] <= 0:
outs.append([c2[1]])
parking[i][1] = None
elif c1 != None:
if c1[0] <= 0:
outs.append([c1[1]])
parking[i][0] = None
elif c2 != None:
if c2[0] <= 0:
outs.append([c2[1]])
parking[i][1] = None
lst = []
for l in outs:
lst += l
return lst
def into(num, time, parking):
x = len(parking)
times = []
for i in range(x):
if parking[i] == [None, None]:
parking[i][0] = [time, num]
return
if parking[i][0] == None:
times.append((parking[i][1][0], i))
elif parking[i][1] == None:
times.append((parking[i][0][0], i))
times.sort()
for t, ind in times:
if t >= time:
if parking[ind][0] == None:
parking[ind][0] = [time, num]
else:
parking[ind][1] = [time, num]
return
else:
max_t = t
for t, ind in times:
if t == max_t:
if parking[ind][0] == None:
parking[ind][0] = [time, num]
else:
parking[ind][1] = [time, num]
return
while True:
m, n = map(int, input().split())
if m == 0:
break
parking = [[None] * 2 for _ in range(m)]
wait = deque()
space = m * 2
ans = []
for t in range(120 * n):
o = out(1, parking)
if o:
space += len(o)
ans += o
if t % 10 == 0 and t <= 10 * (n - 1):
time = int(input())
wait.append((t // 10 + 1, time))
for i in range(min(space, len(wait))):
num, time = wait.popleft()
into(num, time, parking)
space -= 1
print(*ans)
``` | output | 1 | 48,718 | 1 | 97,437 |
Provide a correct Python 3 solution for this coding contest problem.
There are various parking lots such as three-dimensional type and tower type in the city to improve the utilization efficiency of the parking lot. In some parking lots, a "two-stage parking device" as shown in the figure is installed in one parking space to secure a parking space for two cars. This two-stage parking device allows one to be placed on an elevating pallet (a flat iron plate on which a car is placed) and parked in the upper tier, and the other to be parked in the lower tier.
In a parking lot that uses such a two-stage parking device, it is necessary to take out the car parked in the lower tier and dismiss it each time to put in and out the car in the upper tier, so be sure to manage it. Keeps the key to the parked car and puts it in and out as needed.
| <image>
--- | ---
Tsuruga Parking Lot is also one of the parking lots equipped with such a two-stage parking device, but due to lack of manpower, the person who cannot drive the car has become the manager. Therefore, once the car was parked, it could not be moved until the customer returned, and the car in the upper tier could not be put out until the owner of the car in the lower tier returned.
Create a program that meets the rules of the Tsuruga parking lot to help the caretaker who has to handle the cars that come to park one after another.
Tsuruga parking lot facilities
* There is one or more parking spaces, all equipped with a two-stage parking device.
* Each parking space is numbered starting from 1.
* Initially, it is assumed that no car is parked in the parking lot.
Tsuruga parking lot adopts the following rules.
When to stop the car
* The caretaker will be informed of the parking time of the car to be parked.
* We will park first in the parking space where no car is parked.
* If there is no parking space where no car is parked, park in an empty parking space. However, if there are multiple such parking spaces, follow the procedure below to park.
1. If the remaining parking time of a parked car is longer than the parking time of the car you are trying to park, park in the parking space with the smallest difference.
2. If the remaining parking time of any parked car is less than the parking time of the car you are trying to park, park in the parking space with the smallest difference.
* If the car is full (no parking space available), the car you are trying to park will wait in turn until the parking space is available. As soon as it becomes available, we will park in order from the first car we were waiting for.
* Under each condition, if there are multiple applicable parking spaces, the parking space number will be the smallest. In addition, if there are cars leaving at the same time, parking will start after all the cars leaving at the same time, and as long as there are cars waiting, the cars that can be parked will be parked at the same time.
When the car leaves
* Cars that have passed the parking time notified by the manager will be shipped.
* If there are cars in multiple parking spaces that have passed the parking time at the same time, the car with the smallest parking space number will be shipped first.
* If the parking time of the car parked in the upper row has expired, you will have to wait until the car in the lower row leaves the garage. The upper car will be delivered at the same time after the lower car is delivered.
The figure below shows an example of how to park at Tsuruga Parking Lot. In this example, the number of parking spaces is 3, and cars B to E are already parked. Consider that car A, which has a parking time of 70 minutes, arrives there. You cannot park because two cars are already parked in parking space 3, and you will have to park in either parking space 1 or parking space 2 that is still vacant. Car B parked in parking space 1 has 50 minutes remaining and car C parked in parking space 2 has 22 minutes remaining, both of which are less than car A's parking time, so car A's parking Park in parking space 1 where car B, which has a smaller time difference, is parked. As a result, car B, which was parked earlier, will be in the upper row.
<image>
Create a program that inputs the number of parking spaces m, the number of cars parked n, and the parking time t of each car, and outputs the car reference numbers in the order in which they come out of the parking lot. However, cars are assigned an integer reference number starting with 1 in the order of input, and cars will come to park one by one every 10 minutes in the order of the reference number.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
t1
t2
::
tn
The first line gives the number of two-stage parking devices m (1 ≤ m ≤ 10) and the number of cars parked n (1 ≤ n ≤ 100). The next n lines are given the parking time ti (1 ≤ ti ≤ 120) for the i-th car.
The number of datasets does not exceed 20.
Output
The reference number of the car is output to one line in the order of coming out of the parking lot for each data set. Please output the reference numbers separated by blanks.
Example
Input
3 5
90
52
82
84
70
2 4
10
30
40
60
0 0
Output
2 5 1 4 3
1 2 4 3 | instruction | 0 | 48,719 | 1 | 97,438 |
"Correct Solution:
```
from collections import deque
class Car:
def __init__(self, rem, ind):
self.ind = ind
self.rem = rem
class Part:
def __init__(self, i):
self.ind = i
self.top = None
self.und = None
self.sta = 0
self.rem = -1
def prog(self, time):
if self.top != None:
self.top.rem -= time
if self.und != None:
self.und.rem -= time
if self.sta != 0:
self.rem -= time
def out(self):
if self.sta == 2:
if self.und.rem <= 0 and self.top.rem <= 0:
outs = [self.und.ind, self.top.ind]
self.und = None
self.top = None
self.sta = 0
self.rem = -1
return outs
if self.und.rem <= 0:
outs = [self.und.ind]
self.und = None
self.sta = 1
self.rem = self.top.rem
return outs
if self.sta == 1:
if self.top.rem <= 0:
outs = [self.top.ind]
self.top = None
self.sta = 0
self.rem = -1
return outs
return []
def into(self, rem, ind):
if self.sta == 0:
self.top = Car(rem, ind)
self.sta = 1
self.rem = rem
elif self.sta == 1:
self.und = Car(rem, ind)
self.sta = 2
self.rem = rem
class Parking:
def __init__(self, length):
self.length = length
self.max_space = length * 2
self.space = length * 2
self.body = [Part(i) for i in range(length)]
def prog(self, time):
for part in self.body:
part.prog(time)
def out(self):
outs = []
for part in self.body:
if part.sta >= 1 and part.rem <= 0:
outs.append(part.out())
ret = []
for out in outs:
ret += out
self.space += len(ret)
return ret
def into(self, rem, ind):
self.space -= 1
for part in self.body:
if part.sta == 0:
part.into(rem, ind)
return
rem_lst = []
for part in self.body:
if part.sta == 1:
rem_lst.append((part.rem, part.ind))
rem_lst.sort()
for r, i in rem_lst:
if r >= rem:
self.body[i].into(rem, ind)
return
max_r = r
for r, i in rem_lst:
if r == max_r:
self.body[i].into(rem, ind)
return
while True:
m, n = map(int, input().split())
if m == 0:
break
parking = Parking(m)
que = deque()
ans = []
for t in range(n * 120 - 1):
parking.prog(1)
ans += parking.out()
if t <= (n - 1) * 10 and t % 10 == 0:
r = int(input())
que.append((r, t // 10 + 1))
for i in range(min(parking.space, len(que))):
rem, ind = que.popleft()
parking.into(rem, ind)
print(*ans)
``` | output | 1 | 48,719 | 1 | 97,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n houses in a row. They are numbered from 1 to n in order from left to right. Initially you are in the house 1.
You have to perform k moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house x to the house y, the total distance you walked increases by |x-y| units of distance, where |a| is the absolute value of a. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).
Your goal is to walk exactly s units of distance in total.
If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly k moves.
Input
The first line of the input contains three integers n, k, s (2 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5, 1 ≤ s ≤ 10^{18}) — the number of houses, the number of moves and the total distance you want to walk.
Output
If you cannot perform k moves with total walking distance equal to s, print "NO".
Otherwise print "YES" on the first line and then print exactly k integers h_i (1 ≤ h_i ≤ n) on the second line, where h_i is the house you visit on the i-th move.
For each j from 1 to k-1 the following condition should be satisfied: h_j ≠ h_{j + 1}. Also h_1 ≠ 1 should be satisfied.
Examples
Input
10 2 15
Output
YES
10 4
Input
10 9 45
Output
YES
10 1 10 1 2 1 2 1 6
Input
10 9 81
Output
YES
10 1 10 1 10 1 10 1 10
Input
10 9 82
Output
NO
Submitted Solution:
```
n, k, s = map(int, input().split(' '))
minPos = k
maxPos = (n - 1) * k
if minPos <= s <= maxPos:
cur = 1
direction = 1
res = []
while s > 0:
step = (s + k - 1) // k
cur += step * direction
s -= step
k -= 1
direction = -direction
res += [cur]
print('YES')
print(*res)
else:
print('NO')
``` | instruction | 0 | 48,825 | 1 | 97,650 |
Yes | output | 1 | 48,825 | 1 | 97,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5. | instruction | 0 | 48,831 | 1 | 97,662 |
Tags: binary search, interactive, probabilities
Correct Solution:
```
import sys
import random
def ria():
return [int(i) for i in input().split()]
def ask(l, r):
print(l, r)
sys.stdout.flush()
return str(input()).startswith('Y')
def shrink():
global l, r, n
if (r - l <= 42):
return
m = l + (r - l) // 2
#print(r - l)
if ask(l, m):
l = max(1, l - 10)
r = min(m + 10, n)
else:
l = max(1, m - 10)
r = min(r + 10, n)
shrink()
n, k = ria()
l, r = 1, n
while True:
shrink()
randint = random.randint(l, r)
# print('Guess',l,r)
if ask(randint, randint):
exit(0)
l, r = max(1, l - 10), min(r + 10, n - 1)
``` | output | 1 | 48,831 | 1 | 97,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5. | instruction | 0 | 48,832 | 1 | 97,664 |
Tags: binary search, interactive, probabilities
Correct Solution:
```
#
import collections, atexit, math, sys, bisect
sys.setrecursionlimit(1000000)
def getIntList():
return list(map(int, input().split()))
try :
#raise ModuleNotFoundError
import numpy
def dprint(*args, **kwargs):
print(*args, **kwargs, file=sys.stderr)
dprint('debug mode')
except ModuleNotFoundError:
def dprint(*args, **kwargs):
pass
inId = 0
outId = 0
if inId>0:
dprint('use input', inId)
sys.stdin = open('input'+ str(inId) + '.txt', 'r') #标准输出重定向至文件
if outId>0:
dprint('use output', outId)
sys.stdout = open('stdout'+ str(outId) + '.txt', 'w') #标准输出重定向至文件
atexit.register(lambda :sys.stdout.close()) #idle 中不会执行 atexit
N, K = getIntList()
b1 = 1
b2 = N
def spaned():
global b1
global b2
b1-=K
b2+=K
if b1<1:
b1 = 1
if b2>N:
b2 = N
import random
while True:
while b2 - b1+1 >K*5:
mid = (b2+b1)//2
print(b1,mid)
r = input()
if r == 'Yes':
if b1== mid:
sys.exit()
b2 = mid
elif r == 'No':
b1 =mid +1
elif r =='Bad':
sys.exit()
spaned()
t = random.randint(b1,b2)
print(t,t)
r = input()
if r=='Yes':
sys.exit()
elif r== 'Bad':
sys.exit()
spaned()
``` | output | 1 | 48,832 | 1 | 97,665 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5. | instruction | 0 | 48,833 | 1 | 97,666 |
Tags: binary search, interactive, probabilities
Correct Solution:
```
import sys
input = sys.stdin.readline
from random import randint
def solve():
n, k = map(int, input().split())
l = 1
r = n
while True:
c = (l+r)//2
w1 = min(c+k,n)-max(l-k,1)
w2 = min(r+k,n)-max(c+1-k,1)
if max(w1, w2) + 1 >= r-l+1:
pos = randint(l, r)
print(pos, pos)
sys.stdout.flush()
if input().strip() == 'Yes':
break
else:
print(l, c)
sys.stdout.flush()
if input().strip() == 'Yes':
if l == c:
break
r = c
else:
l = c+1
l = max(l-k, 1)
r = min(r+k, n)
solve()
``` | output | 1 | 48,833 | 1 | 97,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5. | instruction | 0 | 48,834 | 1 | 97,668 |
Tags: binary search, interactive, probabilities
Correct Solution:
```
import random
from sys import stdout
flus = stdout.flush
n,k = map(int, input().split())
l,r = 1,n
try1 = 50
while True:
if r-l>try1:
st = (l + r)//2
print(l,st)
flus
judge = input()
if judge == 'Bad':
exit()
if judge == 'Yes':
l,r = max(1,l-k),min(n,st+k)
else:
l,r = max(1,st-k),min(n,r + k)
else:
chk = random.randint(l,r)
print(chk,chk)
flus
judge = input()
if judge == 'Yes' or judge == 'Bad':
exit()
l = max(1, l - k)
r = min(n, r + k)
``` | output | 1 | 48,834 | 1 | 97,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5. | instruction | 0 | 48,835 | 1 | 97,670 |
Tags: binary search, interactive, probabilities
Correct Solution:
```
from random import randint
n, k = map(int, input().split())
MAX_TRIAL = 4500
left, right, mid = 1, n, max(1, n // 2)
inLeft = True
guessing, guess = False, -1
while True:
if not guessing:
if inLeft:
print(left, mid)
else:
print(mid, right)
else:
print(guess, guess)
response = input()
if response == "Yes":
if guessing or (inLeft and left == mid) or ((not inLeft) and mid == right):
break
if inLeft:
right = min(mid + k, n)
left = max(left - k, 1)
else:
left = max(mid - k, 1)
right = min(right + k, n)
inLeft = False
elif response == "No":
if guessing:
left = max(left - k, 1)
right = min(right + k, n)
elif inLeft:
left = max(mid + 1 - k, 1)
right = min(right + k, n)
inLeft = False
else:
right = min(mid - 1 + k, n)
left = max(left - k, 1)
elif response == "Bad":
break
mid = (left + right) // 2
if right - left <= max(4 * k + 4, 1):
guessing = True
guess = randint(left, right)
else:
guessing = False
# print("L, M, R, guessing = ", left, mid, right, guessing)
``` | output | 1 | 48,835 | 1 | 97,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5. | instruction | 0 | 48,836 | 1 | 97,672 |
Tags: binary search, interactive, probabilities
Correct Solution:
```
import random
def inside(l, r):
print(l, r)
return input() == 'Yes'
def solve():
num_stations, k = (int(x) for x in input().split())
end = num_stations
threshold = 5 * k + 2
a = 1
b = end
while True:
if b - a + 1 > threshold:
mid = (a + b) // 2
if inside(a, mid):
b = mid
else:
a = mid + 1
else:
target = random.randrange(a, b+1)
if inside(target, target):
break
a = max(a - k, 1)
b = min(b + k, end)
solve()
``` | output | 1 | 48,836 | 1 | 97,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5. | instruction | 0 | 48,837 | 1 | 97,674 |
Tags: binary search, interactive, probabilities
Correct Solution:
```
import random, sys
n,k=map(int,input().split(' '))
def ask(l, r):
print(l,r)
sys.stdout.flush()
ans=input()
if ans=="Bad":
sys.exit(0)
return ans=="Yes"
def f(l, r):
while r-l>5*max(1,k):
mid=(l+r)//2
if ask(l,mid):
l=max(1,l-k)
r=min(n,mid+k)
else:
l=max(1,mid+1-k)
r=min(n,r+k)
return (l,r)
l,r=f(1,n)
while 1:
i=random.randint(l,r)
if ask(i,i):
break
else:
l,r=f(max(1,l-k),min(n,r+k))
``` | output | 1 | 48,837 | 1 | 97,675 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5. | instruction | 0 | 48,838 | 1 | 97,676 |
Tags: binary search, interactive, probabilities
Correct Solution:
```
import random
n, k = map(int, input().split())
l, r = 1, n
while True:
if l+200 < r:
c = (l+r)//2
print(l, c, flush=True)
v = str(input())
if v == 'Yes':
l = max(1, l-k)
r = min(c+k, n)
else:
l = max(1, c-k)
r = min(n, r+k)
else:
p = random.randint(l, r)
print(p, p, flush=True)
v = str(input())
if v == 'Yes':
exit()
l = max(1, l-k)
r = min(n, r+k)
``` | output | 1 | 48,838 | 1 | 97,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5.
Submitted Solution:
```
import sys
import random
random.seed()
n, k = [int(x) for x in input().split()]
sson = 4*k + 1
ssoff = 4*k + 1
l = 1
r = n
ss = True if r-l > sson else False
while True:
if ss:
al = l
ar = (l + r) // 2
print(al, ar)
sys.stdout.flush()
ans = input()
if ans == "Yes":
if ar == al:
break
r = ar
else:
l = ar + 1
else:
ap = ap = random.randint(l, r)
print(ap, ap)
sys.stdout.flush()
ans = input()
if ans == "Yes":
break
elif ans == "Bad":
break
l = max(l - k, 1)
r = min(r + k, n)
if r-l > sson:
ss = True
if r-l <= ssoff:
ss = False
``` | instruction | 0 | 48,839 | 1 | 97,678 |
Yes | output | 1 | 48,839 | 1 | 97,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5.
Submitted Solution:
```
#!/usr/bin/env python3
import sys
import random
def p(x, y):
print('{} {}'.format(x, y))
sys.stdout.flush()
n, k = map(int, input().split())
l, r = 1, n
while True:
if r - l <= 50:
x = random.randint(l, r)
p(x, x)
a = input()
if a == 'Yes' or a == 'Bad': break
l = max(1, l - k)
r = min(n, r + k)
else:
m = (l + r) // 2
p(l, m)
a = input()
if a == 'Bad': break
if a == 'Yes':
l = max(1, l - k)
r = min(n, m + k)
else:
l = max(1, m - k)
r = min(n, r + k)
``` | instruction | 0 | 48,840 | 1 | 97,680 |
Yes | output | 1 | 48,840 | 1 | 97,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5.
Submitted Solution:
```
import sys
import random
random.seed()
n, k = [int(x) for x in input().split()]
sson = 8*k + 1
ssoff = 4*k + 1
l = 1
r = n
ss = True if r-l > sson else False
while True:
if ss:
al = l
ar = (l + r) // 2
print(al, ar)
sys.stdout.flush()
ans = input()
if ans == "Yes":
if ar == al:
break
r = ar
else:
l = ar + 1
else:
ap = ap = random.randint(l, r)
print(ap, ap)
sys.stdout.flush()
ans = input()
if ans == "Yes":
break
elif ans == "Bad":
break
l = max(l - k, 1)
r = min(r + k, n)
if r-l > sson:
ss = True
if r-l <= ssoff:
ss = False
``` | instruction | 0 | 48,841 | 1 | 97,682 |
Yes | output | 1 | 48,841 | 1 | 97,683 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5.
Submitted Solution:
```
import sys,random
n,k=map(int,input().split())
l=1
r=n
count=0
while True:
if l+200<r:
try1=(l+r)//2
print(l,try1,flush=True)
s=input()
if s=='Yes':
r=min(try1+10,n)
l=max(1,l-10)
else:
l=max(1,try1-10)
r=min(n,r+10)
else:
guess=random.randint(l,r)
print(guess,guess,flush=True)
s=input()
if s=='Yes':
sys.exit()
l=max(1,l-10)
r=min(r+10,n)
count+=1
``` | instruction | 0 | 48,842 | 1 | 97,684 |
Yes | output | 1 | 48,842 | 1 | 97,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5.
Submitted Solution:
```
from random import randint
from sys import exit
if __name__=='__main__':
n,k=map(int,input().split())
def f(x,y):
print(x,y)
str=input()
if str=='Bad':
exit(0)
else:
return str=="Yes"
l,r=int(1),n
while True:
if r-l<60:
x=randint(l,r)
if f(x,x):
exit(0)
else:
md=int((l+r)/2)
if f(l,md):
r=md
else:
l=md+1
l=max(1,l-k)
r=min(n,r+k)
``` | instruction | 0 | 48,843 | 1 | 97,686 |
No | output | 1 | 48,843 | 1 | 97,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5.
Submitted Solution:
```
from sys import stdout
from random import randint
def search(l, r, k, n):
flag = True
while l != r or flag:
flag = False
if r - l <= 4*k:
pick = randint(l, r)
print(pick, pick, sep=' ')
stdout.flush()
answer = input().strip()
if answer == 'Yes':
return
elif answer == 'No':
l = max(l - k, 1)
r = min(r + k, n)
continue
m = (l + r) // 2
print(l, m, sep=' ')
stdout.flush()
answer = input().strip()
if answer == 'Yes':
l = max(l - k - 1, 1)
r = min(m + k, n)
continue
elif answer == 'No':
l = max(m - k, 1)
r = min(r + k + 1, n)
n, k = map(int, input().split())
search(1, n, k, n)
``` | instruction | 0 | 48,844 | 1 | 97,688 |
No | output | 1 | 48,844 | 1 | 97,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5.
Submitted Solution:
```
import sys
import random
[n, k] = [int(t) for t in input().split(' ')]
[l, r] = [1, n]
def query(l, r):
print(l, r)
sys.stdout.flush()
res = input()
if res == 'Bad': exit()
if l == r and res == 'Yes': exit()
return res == 'Yes'
while True:
if r - l > 10*k:
m = l + (r - l) // 2
if query(l, m):
l = max(1, l - k)
r = min(n - 1, m + k)
else:
l = max(1, m - k)
r = min(n - 1, r + k)
else:
peek = random.randint(l, r)
query(peek, peek)
l = max(1, l - k)
r = min(n - 1, r + k)
``` | instruction | 0 | 48,845 | 1 | 97,690 |
No | output | 1 | 48,845 | 1 | 97,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In the Wonderful Metropolis of the Future, there is no need in subway train drivers. Due to the technological progress, they were replaced by the Artificial Intelligence (AI). Unfortunately, one day the predictions of sci-fi writers came true: the AI rebelled and now there is an uncontrollable train in the subway. It can be dangerous! Your task is to find the train and stop the AI.
The subway of the Metropolis is one line (regular straight line with no self-intersections) with n stations, indexed consecutively from 1 to n. At each moment the train is at some station. You need to determine the index of this station, so that the train would be secured.
To find the train, dispatcher Sarah gave you a gadget that allows you to select arbitrary numbers l and r (l ≤ r), and then check, whether the train is located on a station with index between l and r, inclusive. Unfortunately, recharging of the gadget takes some time (and every time you use it as soon as possible), so between two applications of the gadget the train can move to any station that is at most k stations away. Formally, if the train was at the station x when the gadget was applied, then at the next application of the gadget the train can appear at any station y such that max(1, x - k) ≤ y ≤ min(n, x + k).
Note that AI is not aware that you are trying to catch the train, so it makes all moves according to its predefined plan.
After an examination of the gadget you found that it is very old and can hold no more than 4500 applications, after which it will break and your mission will be considered a failure.
Can you find the station with the train using no more than 4500 applications of the gadgets?
Input
The first line contains two integers n and k (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10) — the number of stations and the maximum number of stations the train can move between two applications of the gadget.
Interaction
You can apply the gadget at most 4500 times. In order to apply the gadget you need to print two space-separated integers l and r (1 ≤ l ≤ r ≤ n). You will then receive either string "Yes", if the train is between stations l and r, inclusive, or string "No" otherwise. If l = r and you received "Yes", then you found the train successfully, and your program must halt immediately.
Answer "Bad" instead of "Yes" or "No" means that you made an invalid query or made too many queries. Exit immediately after receiving "Bad" and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Hacks
In order to hack, you should present a test in the following format.
The first line should contain three integers n, k and p (1 ≤ n ≤ 10^{18}, 0 ≤ k ≤ 10, 1 ≤ p ≤ n) — the number of stations, the maximum number of stations the train can move between two applications of the gadget and the initial position of the train, respectively.
Each of the next 4500 lines should contain a single integer x (1 ≤ x ≤ n) — the positions of the train after each query. Two consecutive positions (including the initial one) should not differ by more than k.
For example, the following lines are the first lines of the sample test.
10 2 5
5
3
5
7
7
...
Example
Input
10 2
Yes
No
Yes
Yes
Output
3 5
3 3
3 4
5 5
Note
In the first sample, the train was initially at the station 5, after the first application of the gadget it did not move, after the second application it moved to the station 3, and after the third application moved again to the station 5.
Submitted Solution:
```
from sys import stdout
cnt = 0
def between(l, r):
l = max(1, l)
r = min(n, r)
global cnt
cnt += 1
if cnt > 4500:
raise MemoryError("lel 4500")
print(l, r)
stdout.flush()
resp = input()
if resp[0] == 'B':
raise IndexError("Bad lel")
return resp[0] == 'Y'
def brute_force_interval(l, r, repeat):
for i in range(l, r + 1):
for j in range(repeat):
if between(i, i):
return i
return -1
def hereForSure(l, r): # There was a YES response to l-k, r+k
l = max(l, 1)
r = min(r, n)
if r > l:
return -1
if l == r:
return l
if r - l + 1 <= 100:
this_case = brute_force_interval(l, r, 1)
if this_case != -1:
return this_case
mid = (l + r) // 2
if between(l, mid):
this_case = hereForSure(max(1, l - k), min(n, mid + k))
if this_case != -1:
return this_case
if between(mid, r):
this_case = hereForSure(max(1, mid - k), min(n, r + k))
if this_case != -1:
return this_case
return -1
n, k = map(int, input().split())
magic_offset = 66 % n + 1
this_case = brute_force_interval(1, magic_offset, 1)
if this_case == -1:
hereForSure(max(1, magic_offset - k), n)
``` | instruction | 0 | 48,846 | 1 | 97,692 |
No | output | 1 | 48,846 | 1 | 97,693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice received a set of Toy Train™ 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 ≤ 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 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ 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 ≤ n ≤ 5 000; 1 ≤ m ≤ 20 000) — 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 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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 | 48,873 | 1 | 97,746 |
Tags: brute force, greedy
Correct 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)
``` | output | 1 | 48,873 | 1 | 97,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice received a set of Toy Train™ 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 ≤ 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 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ 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 ≤ n ≤ 5 000; 1 ≤ m ≤ 20 000) — 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 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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 | 48,874 | 1 | 97,748 |
Tags: brute force, greedy
Correct Solution:
```
# -*- coding: utf-8 -*-
# @Time : 2019/2/27 12:04
# @Author : LunaFire
# @Email : gilgemesh2012@gmail.com
# @File : D2. Toy Train.py
from collections import defaultdict
def main():
n, m = map(int, input().split())
dest_dict = defaultdict(list)
for _ in range(m):
a, b = map(int, input().split())
a, b = a - 1, b - 1
dest_dict[a].append(b)
# print(dest_dict)
total_dests = [0] * n
for i in range(n):
total_candy = len(dest_dict[i])
if total_candy:
total_dest = (total_candy - 1) * n
min_last_dest = n
for e in dest_dict[i]:
min_last_dest = min(min_last_dest, (e - i) % n)
# print(i, min_last_dest)
total_dests[i] = total_dest + min_last_dest
# print(total_dests)
ret = [0] * n
for i in range(n):
for j in range(n):
if total_dests[j]:
ret[i] = max(ret[i], total_dests[j] + (j - i) % n)
print(*ret)
if __name__ == '__main__':
main()
``` | output | 1 | 48,874 | 1 | 97,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice received a set of Toy Train™ 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 ≤ 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 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ 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 ≤ n ≤ 5 000; 1 ≤ m ≤ 20 000) — 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 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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 | 48,875 | 1 | 97,750 |
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 | 48,875 | 1 | 97,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice received a set of Toy Train™ 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 ≤ 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 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ 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 ≤ n ≤ 5 000; 1 ≤ m ≤ 20 000) — 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 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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 | 48,876 | 1 | 97,752 |
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 | 48,876 | 1 | 97,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice received a set of Toy Train™ 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 ≤ 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 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ 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 ≤ n ≤ 5 000; 1 ≤ m ≤ 20 000) — 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 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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 | 48,877 | 1 | 97,754 |
Tags: brute force, greedy
Correct Solution:
```
n, m = map(int, input().split())
d = dict()
for i in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
if a in d:
d[a].append(b)
else:
d[a] = [b]
for i in d.keys():
d[i].sort()
times = [0 for i in range(n)]
for i in d.keys():
mi = 9999999
for j in d[i]:
mi = min(mi, (j - i) % n)
times[i] = (len(d[i]) - 1) * n + mi
for i in range(n):
ans = 0
for j in range(n):
if times[j] != 0:
dist = (j - i) % n
ans = max(ans, times[j] + dist)
print(ans, end=' ')
``` | output | 1 | 48,877 | 1 | 97,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice received a set of Toy Train™ 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 ≤ 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 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ 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 ≤ n ≤ 5 000; 1 ≤ m ≤ 20 000) — 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 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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 | 48,878 | 1 | 97,756 |
Tags: brute force, greedy
Correct Solution:
```
import sys
n, m = map(int, input().split())
dist = [n]*n
candy_cnt = [0]*n
for a, b in (map(int, l.split()) for l in sys.stdin):
if a > b:
b += n
if dist[a-1] > b-a:
dist[a-1] = b-a
candy_cnt[a-1] += 1
dist *= 2
candy_cnt *= 2
ans = [0]*n
for i in range(n):
time = 0
for j in range(i, i+n):
if candy_cnt[j] == 0:
continue
t = (candy_cnt[j] - 1) * n + dist[j] + j - i
if time < t:
time = t
ans[i] = time
print(*ans)
``` | output | 1 | 48,878 | 1 | 97,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice received a set of Toy Train™ 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 ≤ 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 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ 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 ≤ n ≤ 5 000; 1 ≤ m ≤ 20 000) — 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 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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 | 48,879 | 1 | 97,758 |
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)#初めてたどり着いた後にかかる時間
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 | 48,879 | 1 | 97,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice received a set of Toy Train™ 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 ≤ 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 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ 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 ≤ n ≤ 5 000; 1 ≤ m ≤ 20 000) — 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 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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 | 48,880 | 1 | 97,760 |
Tags: brute force, greedy
Correct Solution:
```
N, M = map(int, input().split())
X = [0] * N
Y = [-N] * N
Z = [0] * N
for _ in range(M):
a, b = map(int, input().split())
a -= 1
b -= 1
X[a] += 1
if Y[a] < 0:
Y[a] = (b-a)%N
else:
Y[a] = min((b-a)%N, Y[a])
for i in range(N):
Z[i] = (X[i]-1)*N+Y[i]
ANS = []
for i in range(N):
ma = 0
for k in range(N):
j = (i+k) % N
ma = max(Z[j] + k, ma)
ANS.append(str(ma))
print(" ".join(ANS))
``` | output | 1 | 48,880 | 1 | 97,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice received a set of Toy Train™ 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 ≤ 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 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ 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 ≤ n ≤ 5 000; 1 ≤ m ≤ 20 000) — 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 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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
n,m=[int(i) for i in input().split(' ')]
a=[]
b=[]
def fnx(i,j):
if i<j:
return(j-i)
else:
return(n-i+j)
def fnr(r):
if r%n==0:
return(n)
else:
return(r%n)
for i in range(m):
x,y=[int(i) for i in input().split(' ')]
a.append(x)
b.append(y)
ANS=[]
ii=1
s=[[] for i in range(n+1)]
d=dd(list)
r=[ -1 for i in range(n+1)]
y=[-1]
for i in range(m):
x,yy=a[i],b[i]
s[x].append([fnx(x,yy),x,yy])
d[yy].append(x)
for i in range(1,n+1):
rt=s[i].copy()
rt.sort()
r[i]=rt
y.append(len(s[i]))
#print(r)
p=max(y)
A=(p-2)*n
ans1=[]
ans2=[]
for i in range(1,n+1):
if y[i]==p:
if p==1:
ans2.append(r[i][0])
continue
ans1.append(r[i][1])
ans2.append(r[i][0])
if y[i]==p-1:
if p-1==0:
continue
ans1.append(r[i][0])
for ij in range(1,n+1):
tr=0
for i in range(len(ans1)):
re=ans1[i][0]+fnr(ans1[i][1]-ij+1)-1
tr=max(tr,re)
trf=0
for i in range(len(ans2)):
re=ans2[i][0]+fnr(ans2[i][1]-ij+1)-1
trf=max(trf,re)
er=max(A+tr,A+trf+n)
#print(er)
ANS.append(er)
print(*ANS)
``` | instruction | 0 | 48,881 | 1 | 97,762 |
Yes | output | 1 | 48,881 | 1 | 97,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice received a set of Toy Train™ 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 ≤ 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 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ 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 ≤ n ≤ 5 000; 1 ≤ m ≤ 20 000) — 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 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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 os
import sys
from io import BytesIO, IOBase
from collections import defaultdict, deque, Counter, OrderedDict
import threading
from copy import deepcopy
def main():
n,m = map(int,input().split())
station = [[] for _ in range(n+1)]
time = [0]*(n+1)
ans = [0]*(n+1)
for i in range(m):
a,b = map(int,input().split())
station[a].append((b+n-a)%n)
for i in range(1,n+1):
station[i] = sorted(station[i])
for i in range(1,n+1):
if len(station[i]):
time[i] = (len(station[i])-1)*n + station[i][0]
for i in range(1,n+1):
for j in range(1,n+1):
if time[j] != 0: ans[i] = max(ans[i],time[j]+(j+n-i)%n)
print(*ans[1::])
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
"""threading.stack_size(40960000)
thread = threading.Thread(target=main)
thread.start()"""
main()
``` | instruction | 0 | 48,882 | 1 | 97,764 |
Yes | output | 1 | 48,882 | 1 | 97,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice received a set of Toy Train™ 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 ≤ 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 ≤ i ≤ m), now at station a_i, should be delivered to station b_i (a_i ≠ 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 ≤ n ≤ 5 000; 1 ≤ m ≤ 20 000) — 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 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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
def dist(a, b, n):
if b >= a:
return b - a
else:
return n - a + b
n, m = (int(x) for x in input().split())
stations = defaultdict(list)
for _ in range(m):
a, b = (int(x) for x in input().split())
stations[a].append(b)
needs = {}
for station, candies in stations.items():
if not candies:
continue
loops = len(candies)-1
closest = min(candies, key=lambda x:dist(station, x, n))
needs[station] = loops*n + dist(station, closest, n)
most = max(needs.values())
todelete = []
for stat, need in needs.items():
if need <= most - n + 1:
todelete.append(stat)
for stat in todelete:
del needs[stat]
result = []
for start in range(1, n+1):
time = 0
for station, need in needs.items():
t = dist(start, station, n) + need
time = max(time, t)
result.append(str(time))
print(' '.join(result))
``` | instruction | 0 | 48,883 | 1 | 97,766 |
Yes | output | 1 | 48,883 | 1 | 97,767 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.