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.
Good old Berland has n cities and m roads. Each road connects a pair of distinct cities and is bidirectional. Between any pair of cities, there is at most one road. For each road, we know its length.
We also know that the President will soon ride along the Berland roads from city s to city t. Naturally, he will choose one of the shortest paths from s to t, but nobody can say for sure which path he will choose.
The Minister for Transport is really afraid that the President might get upset by the state of the roads in the country. That is the reason he is planning to repair the roads in the possible President's path.
Making the budget for such an event is not an easy task. For all possible distinct pairs s, t (s < t) find the number of roads that lie on at least one shortest path from s to t.
Input
The first line of the input contains integers n, m (2 ≤ n ≤ 500, 0 ≤ m ≤ n·(n - 1) / 2) — the number of cities and roads, correspondingly. Then m lines follow, containing the road descriptions, one description per line. Each description contains three integers xi, yi, li (1 ≤ xi, yi ≤ n, xi ≠ yi, 1 ≤ li ≤ 106), where xi, yi are the numbers of the cities connected by the i-th road and li is its length.
Output
Print the sequence of <image> integers c12, c13, ..., c1n, c23, c24, ..., c2n, ..., cn - 1, n, where cst is the number of roads that can lie on the shortest path from s to t. Print the elements of sequence c in the described order. If the pair of cities s and t don't have a path between them, then cst = 0.
Examples
Input
5 6
1 2 1
2 3 1
3 4 1
4 1 1
2 4 2
4 5 4
Output
1 4 1 2 1 5 6 1 2 1
Submitted Solution:
```
INF = float('inf')
n, m = map(int, input().split())
dp = [[INF for _ in range(n+1)] for _ in range(n+1)]
matrix = [[INF for _ in range(n+1)] for _ in range(n+1)]
comingOut = [[0 for _ in range(n+1)] for _ in range(n+1)]
edges = []
def setup():
for i in range(n+1):
for j in range(n+1):
dp[i][j] = matrix[i][j]
dp[i][i] = 0
def floydWarshall():
setup()
for k in range(n+1 ):
for i in range(n+1 ):
if i == k:
continue
for j in range(n+1 ):
if j == k or j == i:
continue
if dp[i][k] + dp[k][j] < dp[i][j]:
dp[i][j] = dp[i][k] + dp[k][j]
handle_negative_cycle()
def handle_negative_cycle():
for k in range(n):
for i in range(n):
if i == k:
continue
for j in range(n):
if j == k or j == i:
continue
if dp[i][k] + dp[k][j] < dp[i][j]:
dp[i][j] = -INF
for i in range(m):
a, b, w = map(int, input().split())
edges.append((a, b, w))
matrix[a][b] = w
matrix[b][a] = w
floydWarshall()
for e in edges:
a, b, w = e
for v in range(1, n+1):
if dp[v][a] + w == dp[v][b]:
comingOut[v][b] += 1
if dp[v][b] + w == dp[v][a]:
comingOut[v][a] += 1
ans =[]
for source in range(1, n+1):
for dest in range(source+1, n+1):
count = 0
for mid in range(1, n+1):
if dp[source][mid] + dp[mid][dest] == dp[source][dest]:
count += comingOut[source][mid]
ans.append(str(count))
print(' '.join(ans))
``` | instruction | 0 | 82,914 | 1 | 165,828 |
No | output | 1 | 82,914 | 1 | 165,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Good old Berland has n cities and m roads. Each road connects a pair of distinct cities and is bidirectional. Between any pair of cities, there is at most one road. For each road, we know its length.
We also know that the President will soon ride along the Berland roads from city s to city t. Naturally, he will choose one of the shortest paths from s to t, but nobody can say for sure which path he will choose.
The Minister for Transport is really afraid that the President might get upset by the state of the roads in the country. That is the reason he is planning to repair the roads in the possible President's path.
Making the budget for such an event is not an easy task. For all possible distinct pairs s, t (s < t) find the number of roads that lie on at least one shortest path from s to t.
Input
The first line of the input contains integers n, m (2 ≤ n ≤ 500, 0 ≤ m ≤ n·(n - 1) / 2) — the number of cities and roads, correspondingly. Then m lines follow, containing the road descriptions, one description per line. Each description contains three integers xi, yi, li (1 ≤ xi, yi ≤ n, xi ≠ yi, 1 ≤ li ≤ 106), where xi, yi are the numbers of the cities connected by the i-th road and li is its length.
Output
Print the sequence of <image> integers c12, c13, ..., c1n, c23, c24, ..., c2n, ..., cn - 1, n, where cst is the number of roads that can lie on the shortest path from s to t. Print the elements of sequence c in the described order. If the pair of cities s and t don't have a path between them, then cst = 0.
Examples
Input
5 6
1 2 1
2 3 1
3 4 1
4 1 1
2 4 2
4 5 4
Output
1 4 1 2 1 5 6 1 2 1
Submitted Solution:
```
from math import inf
def Floyd_Warshall(n: int, E: [[int]]) -> [[int]]:
min_cost = [[inf for _ in range(n)] for _ in range(n)]
for i in range(n):
min_cost[i][i] = 0
for x,y,l in E:
min_cost[x - 1][y - 1] = l
min_cost[y - 1][x - 1] = l
for i in range(n):
for j in range(n):
for k in range(n):
min_cost[i][j] = min(min_cost[i][j], min_cost[i][k] + min_cost[k][j])
return min_cost
def end_edges_count(n: int, min_cost: [[int]], E: [[int]]) -> [[int]]:
end_edges = [[0 for _ in range(n)] for _ in range(n)]
for i in range(n):
for x, y, l in E:
if min_cost[i][x - 1] + l == min_cost[i][y-1]:
end_edges[i][y - 1] += 1
else:
if min_cost[i][y - 1] + l == min_cost[i][x - 1]:
end_edges[i][x - 1] += 1
return end_edges
def shortest_paths_edges_count(n: int, min_cost: [[int]], end_edges: [[int]]) -> [[int]]:
shortest_paths_edges = [[0 for _ in range(n)] for _ in range(n)]
for i in range(n):
for j in range(n):
for k in range(n):
if min_cost[i][k] + min_cost[k][j] == min_cost[i][j]:
shortest_paths_edges[i][j] += end_edges[i][k]
return shortest_paths_edges
inbox: str = input()
inbox: [str] = inbox.split(" ")
n = int(inbox[0])
m = int(inbox[1])
E: [int] = []
for i in range(m):
inbox: str = input()
inbox: [str] = inbox.split(" ")
E.append([int(inbox[0]), int(inbox[1]), int(inbox[2])])
min_cost: [[int]] = Floyd_Warshall(n, E)
end_edges: [[int]] = end_edges_count(n, min_cost, E)
shortest_paths_edges: [[int]] = shortest_paths_edges_count(n, min_cost, end_edges)
for i in range(n):
for j in range(i, n):
if i != j:
print(shortest_paths_edges[i][j], end=" ")
print()
``` | instruction | 0 | 82,915 | 1 | 165,830 |
No | output | 1 | 82,915 | 1 | 165,831 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four. | instruction | 0 | 83,046 | 1 | 166,092 |
Tags: dp, greedy, sortings
Correct Solution:
```
#aawan
n,k=map(int,input().split())
temp=list(map(int,input().split()))
rslt=[]
list=[]
cnt=0
for i in range(n) :
if temp[i] <0:
rslt.append(i)
cnt += 1
if cnt>k :
print(-1)
exit()
if cnt==0 :
print(0)
exit()
k-=cnt
ans=2*cnt
for i in range(cnt-1) :
list.append(rslt[i+1]-rslt[i]-1)
list.sort()
for i in list :
if k<i :
break
k-=i
ans-=2
if k>=(n-rslt[cnt-1]-1) :
ans-=1
print(ans)
``` | output | 1 | 83,046 | 1 | 166,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four. | instruction | 0 | 83,047 | 1 | 166,094 |
Tags: dp, greedy, sortings
Correct Solution:
```
f = lambda: map(int, input().split())
n, k = f()
s, d = [], -1
for q in f():
if q < 0:
k -= 1
if d > 0: s += [d]
d = 0
elif d > -1: d += 1
s.sort()
t = 2 * len(s)
for q in s:
if q > k: break
k -= q
t -= 2
print(-1 if k < 0 else 0 if d < 0 else 2 + t - (k >= d))
``` | output | 1 | 83,047 | 1 | 166,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four. | instruction | 0 | 83,048 | 1 | 166,096 |
Tags: dp, greedy, sortings
Correct Solution:
```
# Copied
n, k = map(int, input().split())
temp = list(map(int, input().split()))
rslt = []
list = []
cnt = 0
for i in range(n):
if temp[i] < 0:
rslt.append(i)
cnt += 1
if cnt > k:
print(-1)
exit()
if cnt == 0:
print(0)
exit()
k -= cnt
ans = 2 * cnt
for i in range(cnt - 1):
list.append(rslt[i + 1] - rslt[i] - 1)
list.sort()
for i in list:
if k < i:
break
k -= i # To not change the tires between to -ve air temp
ans -= 2
if k >= (n - rslt[cnt - 1] - 1):
ans -= 1
print(ans)
``` | output | 1 | 83,048 | 1 | 166,097 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four. | instruction | 0 | 83,049 | 1 | 166,098 |
Tags: dp, greedy, sortings
Correct Solution:
```
n, k = [int(i) for i in input().split()]
t = [int(i) for i in input().split()]
colddays = [i for i in range(n) if t[i]<0 ]
num_colddays = len(colddays)
if num_colddays == 0:
print(0)
elif k >= n:
print(1)
else:
if num_colddays > k:
print(-1)
else:
k -= num_colddays
ans = 0
for i in range(1, len(colddays)):
if colddays[i] - colddays[i-1] > 1:
ans += 1
ans += 1
ans *= 2
gap = [0 for i in range(num_colddays)]
for j in range(1, num_colddays):
gap[j-1] = colddays[j] - colddays[j-1]-1
positivegap = [j for j in gap if j > 0]
positivegap = sorted(positivegap)
for i in positivegap:
if k >= i:
k -= i
ans -= 2
else:
break
remain = n - 1 - colddays[-1]
if k >= remain:
ans -= 1
print(ans)
``` | output | 1 | 83,049 | 1 | 166,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four. | instruction | 0 | 83,050 | 1 | 166,100 |
Tags: dp, greedy, sortings
Correct Solution:
```
n, k = map(int, input().split())
days = list(map(int, input().split()))
a = [0 for i in range(n)] # for < 0
b = [0 for i in range(n)] # for >= 0
asum = 0
ia = 0
ib = 0
inf = n + 199
for x in days:
if x < 0:
a[ia] += 1
asum += 1
if b[ib] != 0:
ib += 1
else:
b[ib] += 1
if a[ia] != 0:
ia += 1
changes = ia + 1 if a[ia] > 0 else ia
changes += changes - 1
if (a[ia] == 0):
ia -= 1
if b[ib] == 0:
ib -= 1
if asum > k:
print(-1)
quit()
for i in range(ia, n):
a[i] = inf if a[i] == 0 else a[i]
for i in range(ib, n):
b[i] = inf if b[i] == 0 else b[i]
if days[0] >= 0:
b[0] = inf
lastb = -1
removedlastb = False
if days[len(days) - 1] >= 0 :
changes += 1
lastb = b[ib]
b.sort()
# print('ia = ' + str(ia))
# print('changes = ' + str(changes))
curb = 0
# take care abount choosing smth in the middle exept for last\
seccurb = curb
secasum = asum
secchanges = changes
secremovedlastb = removedlastb
while (curb <= ib and asum + b[curb] <= k):
asum += b[curb]
if not removedlastb:
if b[curb] == lastb:
changes -= 1
removedlastb = True
else :
changes -= 2
else:
changes -= 2
curb += 1
while (seccurb <= ib and secasum + b[seccurb] <= k):
secasum += b[seccurb]
if not secremovedlastb:
if b[seccurb] == lastb:
secasum -= b[seccurb]
secremovedlastb = True
else :
secchanges -= 2
else:
secchanges -= 2
seccurb += 1
print(max(min(changes, secchanges), 0))
``` | output | 1 | 83,050 | 1 | 166,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four. | instruction | 0 | 83,051 | 1 | 166,102 |
Tags: dp, greedy, sortings
Correct Solution:
```
n,k=map(int,input().split())
d=list(map(int,input().split()))
p=[]
count=0
count1=0
for i in range(len(d)):
if d[i]>=0:
count+=1
if d[i]<0 or i==len(d)-1:
p.append(count)
count=0
count1+=1
if d[len(d)-1]>=0:
count1-=1
if count1==0:
print(0)
exit()
a=p.pop() if d[len(d)-1]>=0 else 0
if len(p)>0:
p.reverse()
p.pop()
p.sort()
count=k-len(p)-1 if count1>1 else k-count1
if count<0:
print(-1)
exit()
count1=0
for i in range(len(p)):
if(count-p[i]>=0):
count1+=1
count-=p[i]
else:
break
ans=2*(len(p)+1-count1)
if count-a>=0:
ans-=1
print(ans)
``` | output | 1 | 83,051 | 1 | 166,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four. | instruction | 0 | 83,052 | 1 | 166,104 |
Tags: dp, greedy, sortings
Correct Solution:
```
import heapq
n,k=[int(i) for i in input().split()]
l=[int(i) for i in input().split()]
cnt=[i for i in l if i<0]
m=0
for a,i in enumerate(l):
if i<0 and m==0:
m=1
elif i<0 and l[a-1]>=0:
m+=1
d={}
q=[]
chk=-1
for i in l:
if i<0:
if chk>0:
if chk in d:
d[chk]+=1
else:
d[chk]=1
heapq.heappush(q,chk)
chk=0
elif chk>=0:
chk+=1
ans=m*2
k-=len(cnt)
while len(q) and k>=q[0]:
k-=q[0]
ans-=2
d[q[0]]-=1
if d[q[0]]==0:
heapq.heappop(q)
if k>=chk and chk>-1:
ans-=1
print(ans if ans>=0 and k>=0 else -1)
``` | output | 1 | 83,052 | 1 | 166,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four. | instruction | 0 | 83,053 | 1 | 166,106 |
Tags: dp, greedy, sortings
Correct Solution:
```
import math
from collections import Counter, defaultdict
from itertools import accumulate
R = lambda: map(int, input().split())
n, k = R()
ts = list(R()) + [-1]
k -= sum(1 if t < 0 else 0 for t in ts[:n])
res = 0 if ts[0] >= 0 else 1
for i in range(1, n):
if ts[i - 1] < 0 and ts[i] >= 0 or ts[i - 1] >= 0 and ts[i] < 0:
res += 1
locs = [i for i, t in enumerate(ts) if t < 0]
ps = sorted((y - x - 1, y) for x, y in zip(locs, locs[1:]) if y - x > 1)
res1 = res
k1 = k
for p in ps:
if k >= p[0]:
res -= (2 if p[1] < n else 1)
k -= p[0]
if k1 >= p[0] and p[1] < n:
res1 -= 2
k1 -= p[0]
res = res if k >= 0 else math.inf
res1 = res1 if k1 >= 0 else math.inf
print(min(res, res1) if min(res, res1) < math.inf else -1)
``` | output | 1 | 83,053 | 1 | 166,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four.
Submitted Solution:
```
import sys
from collections import deque
def weather(temp):
return temp >= 0
INF = int(1e+6)
n, k = (int(x) for x in input().split())
day_to_temp = [int(x) for x in input().split()]
day_to_weather = [weather(temp) for temp in day_to_temp]
first_warm = -1
mid_warms = []
colds = []
curr_state = 0
curr_len = 0
switches_made = 0
for weather in day_to_weather:
if curr_state == 0:
if weather == 1:
curr_len += 1
else:
first_warm = curr_len
curr_state = 2
curr_len = 1
switches_made += 1
elif curr_state == 1:
if weather == 1:
curr_len += 1
else:
mid_warms.append(curr_len)
curr_state = 2
curr_len = 1
switches_made += 1
else:
if weather == 1:
colds.append(curr_len)
curr_state = 1
curr_len = 1
switches_made += 1
else:
curr_len += 1
if curr_state == 1:
last_warm = curr_len
elif curr_state == 2:
colds.append(curr_len)
last_warm = INF
else:
print(0)
sys.exit()
mid_warms_deque = deque(sorted(mid_warms, reverse=True))
cold_days_nr = sum(colds)
free_days = k - cold_days_nr
if free_days < 0:
print(-1)
sys.exit()
elif free_days == 0:
print(switches_made)
sys.exit()
elif len(mid_warms) > 0:
while mid_warms_deque:
curr_warm_len = mid_warms_deque.pop()
if not free_days - curr_warm_len >= 0:
break
switches_made -= 2
free_days -= curr_warm_len
if free_days - last_warm >= 0:
switches_made -= 1
free_days -= last_warm
print(switches_made)
``` | instruction | 0 | 83,054 | 1 | 166,108 |
Yes | output | 1 | 83,054 | 1 | 166,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four.
Submitted Solution:
```
def solve(k, s):
w = [0]
for c in s:
if c < 0:
if w[-1] >= 0:
w.append(-1)
else:
w[-1] -= 1
else:
if w[-1] < 0:
w.append(1)
else:
w[-1] += 1
begin, end = w[0], w[-1]
if len(w) == 1 and w[0] > 0:
return 0
if begin > 0:
w.pop(0)
if end > 0:
w.pop()
neg_seg = [t for t in w if t < 0]
pos_seg = [t for t in w if t > 0]
d = -sum(neg_seg)
if d > k: return -1
changes = 2*len(neg_seg) - (end < 0)
pos_seg.sort()
for seg in pos_seg:
d += seg
if d > k:
if end > 0 and d-seg+end <= k:
return changes-1
else:
return changes
else:
changes -= 2
if end > 0 and d+end <= k:
return changes-1
else:
return changes
n, k = map(int, input().split())
s = [int(x) for x in input().split()]
print(solve(k, s))
``` | instruction | 0 | 83,055 | 1 | 166,110 |
Yes | output | 1 | 83,055 | 1 | 166,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four.
Submitted Solution:
```
n, k = [int(i) for i in input().split(' ')]
w = [int(i) for i in input().split(' ')]
f = 0
while f < n and w[f] >= 0:
f += 1
if f == n:
print(0)
exit()
b = n - 1
while b >= 0 and w[b] >= 0:
b -= 1
# if n == b - f + 1:
# print(1)
# exit()
if f == b:
if k == 0:
print(-1)
exit()
if f+1 == n:
print(1)
exit()
else:
if (k - 1) >= (n - (b + 1)):
print(1)
exit()
else:
print(2)
exit()
inv = []
ps = 0
ns = 0
for i in range(f, b):
if w[i] >= 0:
ps += 1
if w[i+1] < 0:
inv.append(ps)
elif w[i] < 0:
ns += 1
if w[i+1] >= 0:
ps = 0
ns += 1
spare = k - ns
if spare < 0:
print(-1)
exit()
if not inv:
if w[n-1] < 0:
print(1)
exit()
else:
if spare >= (n - (b + 1)):
print(1)
exit()
else:
print(2)
exit()
invs = sorted(inv)
if spare < invs[0]:
ch = (len(invs) + 1) * 2
remainder = spare
else:
use, invsn = invs[0], -1
for i in range(1, len(invs)):
use += invs[i]
if spare < use:
use -= invs[i]
invsn = i
break
if invsn == -1:
invsn = len(invs)
ch = (len(invs) - invsn + 1) * 2
remainder = spare - use
if remainder >= (n - (b + 1)):
ch -= 1
print(ch)
``` | instruction | 0 | 83,056 | 1 | 166,112 |
Yes | output | 1 | 83,056 | 1 | 166,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four.
Submitted Solution:
```
n, k = list(map(int,input().split()))
a = list(map(lambda x:int(x)>=0,input().split()))
for ind_positive in range(n):
if not a[ind_positive]:
break
else:
print(0)
exit()
a = a[ind_positive:]
n = len(a)
for ind_positive in range(n-1, -1, -1):
if not a[ind_positive]:
break
a = a[:ind_positive+1]
len_suf = n-ind_positive-1
n = len(a)
#print(a, len_suf)
b = []
i = 0
while True:
j = 0
while i < n and a[i]:
i += 1
j += 1
if j:
b.append(j)
i += 1
if i >= n:
break
b_sort = b.copy()
b_sort.sort(reverse = True)
s = 0
i = 0
if n-s>k:
for i in range(len(b)):
s += b_sort[i]
if n-s <= k:
i += 1
break
if n-s > k:
ans2 = -1
else:
ans2 = i*2+2
k -= len_suf
if k < 0:
ans1 = -1
else:
b = []
i = 0
while True:
j = 0
while i < n and a[i]:
i += 1
j += 1
if j:
b.append(j)
i += 1
if i >= n:
break
b_sort = b.copy()
b_sort.sort(reverse = True)
s = 0
i = 0
if n-s>k:
for i in range(len(b)):
s += b_sort[i]
if n-s <= k:
i += 1
break
if n-s > k:
ans1 = -1
else:
ans1 = i*2+1
if ans1 == -1:
print(ans2)
elif ans2 == -1:
print(ans1)
else:
print(min(ans1,ans2))
``` | instruction | 0 | 83,057 | 1 | 166,114 |
Yes | output | 1 | 83,057 | 1 | 166,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four.
Submitted Solution:
```
from collections import deque
import heapq
def main():
from sys import stdin
lines = deque(line.strip() for line in stdin.readlines())
# lines will now contain all of the input's lines in a list
n, k = [int(x) for x in lines.popleft().split()]
temps = deque(int(x) for x in lines.popleft().split())
temps.appendleft(0)
streaks = []
count = 0
ispositive = True
numnegative = 0
changes = 0
while temps:
temp = temps.popleft()
positive = temp >= 0
if positive:
count += 1
if not ispositive:
changes += 1
else:
numnegative += 1
if ispositive:
streaks.append(count)
count = 0
changes += 1
ispositive = positive
if numnegative == 0:
print(0)
return
if numnegative > k:
print(-1)
return
if k >= n:
print(1)
return
if count > 0:
streaks.append(count)
k -= numnegative
streaks = streaks[1:]
heapq.heapify(streaks)
while streaks and k >= streaks[0]:
k -= heapq.heappop(streaks)
changes -= 2
if count:
if streaks and streaks[0] > count:
# don't need to change back to summer tires if we end in winter tires
changes += 1
print(changes)
if __name__ == '__main__':
main()
``` | instruction | 0 | 83,058 | 1 | 166,116 |
No | output | 1 | 83,058 | 1 | 166,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four.
Submitted Solution:
```
n, k = list(map(int,input().split()))
a = list(map(lambda x:int(x)>=0,input().split()))
for ind_positive in range(n):
if not a[ind_positive]:
break
if ind_positive == n-1:
print(0)
exit()
a = a[ind_positive:]
n = len(a)
for ind_positive in range(n-1, -1, -1):
if not a[ind_positive]:
break
a = a[:ind_positive+1]
len_suf = n-ind_positive-1
n = len(a)
b = []
i = 0
while True:
j = 0
while i < n and a[i]:
i += 1
j += 1
if j:
b.append(j)
i += 1
if i >= n:
break
b_sort = b.copy()
b_sort.sort(reverse = True)
s = 0
i = 0
if n-s>k:
for i in range(len(b)):
s += b_sort[i]
if n-s <= k:
i += 1
break
if n-s > k:
ans2 = -1
else:
ans2 = i*2+2
k -= len_suf
if k < 0:
ans1 = -1
else:
b = []
i = 0
while True:
j = 0
while i < n and a[i]:
i += 1
j += 1
if j:
b.append(j)
i += 1
if i >= n:
break
b_sort = b.copy()
b_sort.sort(reverse = True)
s = 0
i = 0
if n-s>k:
for i in range(len(b)):
s += b_sort[i]
if n-s <= k:
i += 1
break
if n-s > k:
ans1 = -1
else:
ans1 = i*2+1
if ans1 == -1:
print(ans2)
elif ans2 == -1:
print(ans1)
else:
print(min(ans1,ans2))
``` | instruction | 0 | 83,059 | 1 | 166,118 |
No | output | 1 | 83,059 | 1 | 166,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four.
Submitted Solution:
```
n, k = [int(x) for x in input().strip().split(" ")]
t = [int (x) for x in input().strip().split(" ")]
def yolo():
negatives = 0
summersegments = []
lastOne = 0
prevIndex = -1
winter = False
for i in range(len(t)):
if t[i] < 0:
negatives += 1
if prevIndex == -1:
prevIndex = i
winter = True
continue
if not winter:
summersegments.append(i - prevIndex - 1)
lastOne = i-prevIndex - 1
winter = True
prevIndex = i
else:
winter = False
if i == len(t)-1:
summersegments.append(i - prevIndex)
lastOne = i - prevIndex
if negatives > k:
print(-1)
return
if negatives == 0:
print(0)
return
if k == negatives and k == n:
print(1)
return
summersegments = sorted(summersegments)
#print(summersegments)
for i in range(len(summersegments)):
if negatives + summersegments[i] > k:
if t[-1] < 0:
print(2*(len(summersegments) - i) + 1)
else:
if lastOne <= summersegments[i] and i > 0:
print(2*(len(summersegments) - i) + 1)
else:
print(2*(len(summersegments) - i))
break
else:
negatives += summersegments[i]
#print(negatives)
def yolo2():
global t
negatives = 0
summersegments = []
first = 0
prevIndex = -1
winter = False
for i in range(len(t)):
if t[i] < 0:
negatives += 1
if prevIndex == -1:
first = i
prevIndex = i
winter = True
continue
if not winter:
summersegments.append(i - prevIndex - 1)
winter = True
prevIndex = i
else:
winter = False
if i == len(t)-1:
summersegments.append(i - prevIndex)
if negatives > k:
print(-1)
return
if negatives == 0:
print(0)
return
if k == negatives and k == n:
print(1)
return
t = t[first:]
summersegments = sorted(summersegments)
keep = 0
for i in range(len(summersegments)):
if negatives + summersegments[i] > k:
break
keep += 1
negatives += summersegments[i]
summersegments = summersegments[:keep]
prevIndex = -1
switches = 1
winter = False
#print(summersegments)
for i in range(len(t)):
if t[i] < 0:
if prevIndex == -1:
prevIndex = i
winter = True
continue
if not winter:
if i - prevIndex - 1 in summersegments:
del(summersegments[summersegments.index(i - prevIndex - 1)])
else:
switches += 2
#print(i)
winter = True
prevIndex = i
else:
winter = False
if i == len(t)-1:
if i - prevIndex in summersegments:
pass
else:
switches += 1
print(switches)
yolo2()
``` | instruction | 0 | 83,060 | 1 | 166,120 |
No | output | 1 | 83,060 | 1 | 166,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, ..., tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
Input
4 3
-5 20 -3 0
Output
2
Input
4 2
-5 20 -3 0
Output
4
Input
10 6
2 -5 1 3 0 0 -4 -3 1 0
Output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires' changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires' changes equals four.
Submitted Solution:
```
n, k = map(int, input().split())
inF = False
bad = 0
last = None
stretches = [0]
for d in map(lambda x: int(x) >= 0, input().split()):
last = d
if d and not inF:
continue
else:
inF = True
if d:
stretches[-1] += 1
else:
bad += 1
if stretches[-1] != 0:
stretches.append(0)
if bad > k:
print(-1)
elif bad == 0:
print(0)
else:
k -= bad
num = 1
if last:
num -= 1
if stretches[-1] == 0:
del stretches[-1]
num_ = num
stretches_ = stretches[:-1]
k_ = k
# print(num, stretches, k)
if len(stretches) > 0:
num += (len(stretches)) * 2
last = stretches[-1]
stretches.sort()
while len(stretches) > 0 and stretches[0] <= k:
num -= 2
k -= stretches[0]
del stretches[0]
if not last in stretches:
num += 1
# print(num_, stretches_, k_)
if len(stretches_) > 0:
num_ += (len(stretches_) + 1) * 2
stretches_.sort()
while len(stretches_) > 0 and stretches_[0] <= k_:
num_ -= 2
k_ -= stretches_[0]
del stretches_[0]
print(min(num, num_))
``` | instruction | 0 | 83,061 | 1 | 166,122 |
No | output | 1 | 83,061 | 1 | 166,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oleg the bank client lives in Bankopolia. There are n cities in Bankopolia and some pair of cities are connected directly by bi-directional roads. The cities are numbered from 1 to n. There are a total of m roads in Bankopolia, the i-th road connects cities ui and vi. It is guaranteed that from each city it is possible to travel to any other city using some of the roads.
Oleg wants to give a label to each city. Suppose the label of city i is equal to xi. Then, it must hold that for all pairs of cities (u, v) the condition |xu - xv| ≤ 1 holds if and only if there is a road connecting u and v.
Oleg wonders if such a labeling is possible. Find an example of such labeling if the task is possible and state that it is impossible otherwise.
Input
The first line of input contains two space-separated integers n and m (2 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105) — the number of cities and the number of roads.
Next, m lines follow. The i-th line contains two space-separated integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the cities connected by the i-th road. It is guaranteed that there is at most one road between each pair of cities and it is possible to travel from any city to any other city using some roads.
Output
If the required labeling is not possible, output a single line containing the string "NO" (without quotes).
Otherwise, output the string "YES" (without quotes) on the first line. On the next line, output n space-separated integers, x1, x2, ..., xn. The condition 1 ≤ xi ≤ 109 must hold for all i, and for all pairs of cities (u, v) the condition |xu - xv| ≤ 1 must hold if and only if there is a road connecting u and v.
Examples
Input
4 4
1 2
1 3
1 4
3 4
Output
YES
2 3 1 1
Input
5 10
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
5 4
Output
YES
1 1 1 1 1
Input
4 3
1 2
1 3
1 4
Output
NO
Note
For the first sample, x1 = 2, x2 = 3, x3 = x4 = 1 is a valid labeling. Indeed, (3, 4), (1, 2), (1, 3), (1, 4) are the only pairs of cities with difference of labels not greater than 1, and these are precisely the roads of Bankopolia.
For the second sample, all pairs of cities have difference of labels not greater than 1 and all pairs of cities have a road connecting them.
For the last sample, it is impossible to construct a labeling satisfying the given constraints.
Submitted Solution:
```
read = lambda: map(int, input().split())
``` | instruction | 0 | 83,078 | 1 | 166,156 |
No | output | 1 | 83,078 | 1 | 166,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
All Berland residents are waiting for an unprecedented tour of wizard in his Blue Helicopter over the cities of Berland!
It is well-known that there are n cities in Berland, some pairs of which are connected by bidirectional roads. Each pair of cities is connected by no more than one road. It is not guaranteed that the road network is connected, i.e. it is possible that you can't reach some city from some other.
The tour will contain several episodes. In each of the episodes:
* the wizard will disembark at some city x from the Helicopter;
* he will give a performance and show a movie for free at the city x;
* he will drive to some neighboring city y using a road;
* he will give a performance and show a movie for free at the city y;
* he will drive to some neighboring to y city z;
* he will give a performance and show a movie for free at the city z;
* he will embark the Helicopter and fly away from the city z.
It is known that the wizard doesn't like to use roads, so he agrees to use each road at most once (regardless of direction). In other words, for road between a and b he only can drive once from a to b, or drive once from b to a, or do not use this road at all.
The wizards wants to plan as many episodes as possible without violation the above rules. Help the wizard!
Please note that the wizard can visit the same city multiple times, the restriction is on roads only.
Input
The first line contains two integers n, m (1 ≤ n ≤ 2·105, 0 ≤ m ≤ 2·105) — the number of cities and the number of roads in Berland, respectively.
The roads description follow, one in each line. Each description is a pair of two integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the ids of the cities connected by the i-th road. It is guaranteed that there are no two roads connecting the same pair of cities. Every road is bidirectional. The cities are numbered from 1 to n.
It is possible that the road network in Berland is not connected.
Output
In the first line print w — the maximum possible number of episodes. The next w lines should contain the episodes in format x, y, z — the three integers denoting the ids of the cities in the order of the wizard's visits.
Examples
Input
4 5
1 2
3 2
2 4
3 4
4 1
Output
2
1 4 2
4 3 2
Input
5 8
5 3
1 2
4 5
5 1
2 5
4 3
1 4
3 2
Output
4
1 4 5
2 3 4
1 5 3
5 2 1
Submitted Solution:
```
n, m = map(int, input().split())
A = []
for i in range(n):
A.append([0] * n)
for i in range(m):
p1, p2 = map(int, input().split())
A[p1 - 1][p2-1] = 1
A[p2 - 1][p1 -1] = 1
print(m//2)
for q in range(n):
Q = []
for l in range(n):
if A[q][l] == 1:
Q.append(l)
if len(Q) > 1:
for i in Q:
A[q][i] = 0
A[i][q] = 0
for h in range(len(Q)//2):
print(Q.pop() + 1, q + 1, Q.pop() + 1)
``` | instruction | 0 | 83,087 | 1 | 166,174 |
No | output | 1 | 83,087 | 1 | 166,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
All Berland residents are waiting for an unprecedented tour of wizard in his Blue Helicopter over the cities of Berland!
It is well-known that there are n cities in Berland, some pairs of which are connected by bidirectional roads. Each pair of cities is connected by no more than one road. It is not guaranteed that the road network is connected, i.e. it is possible that you can't reach some city from some other.
The tour will contain several episodes. In each of the episodes:
* the wizard will disembark at some city x from the Helicopter;
* he will give a performance and show a movie for free at the city x;
* he will drive to some neighboring city y using a road;
* he will give a performance and show a movie for free at the city y;
* he will drive to some neighboring to y city z;
* he will give a performance and show a movie for free at the city z;
* he will embark the Helicopter and fly away from the city z.
It is known that the wizard doesn't like to use roads, so he agrees to use each road at most once (regardless of direction). In other words, for road between a and b he only can drive once from a to b, or drive once from b to a, or do not use this road at all.
The wizards wants to plan as many episodes as possible without violation the above rules. Help the wizard!
Please note that the wizard can visit the same city multiple times, the restriction is on roads only.
Input
The first line contains two integers n, m (1 ≤ n ≤ 2·105, 0 ≤ m ≤ 2·105) — the number of cities and the number of roads in Berland, respectively.
The roads description follow, one in each line. Each description is a pair of two integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the ids of the cities connected by the i-th road. It is guaranteed that there are no two roads connecting the same pair of cities. Every road is bidirectional. The cities are numbered from 1 to n.
It is possible that the road network in Berland is not connected.
Output
In the first line print w — the maximum possible number of episodes. The next w lines should contain the episodes in format x, y, z — the three integers denoting the ids of the cities in the order of the wizard's visits.
Examples
Input
4 5
1 2
3 2
2 4
3 4
4 1
Output
2
1 4 2
4 3 2
Input
5 8
5 3
1 2
4 5
5 1
2 5
4 3
1 4
3 2
Output
4
1 4 5
2 3 4
1 5 3
5 2 1
Submitted Solution:
```
from collections import defaultdict
from sys import stdin
def dfs_try(u,vis,count):
global path
for v,e in g[u]:
if not vis[e]:
path.append(v+1)
vis[e]=1
count+=1
if count==2:
return 1
else:
return dfs_try(v,vis,count)
path.pop()
return 0
g=defaultdict(list)
n,m=map(int, stdin.readline().split())
count=[0]*n
for i in range(m):
a,b=map(int, stdin.readline().split())
g[a-1].append([b-1,i])
g[b-1].append([a-1,i])
ans=0
vis=[0]*m
path=[]
final=[]
for x in range(n):
path=[x+1]
if dfs_try(x,vis,0):
ans+=1
final.append(path)
print(ans)
for x in final:
print(" ".join(map(str, x)))
``` | instruction | 0 | 83,088 | 1 | 166,176 |
No | output | 1 | 83,088 | 1 | 166,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
All Berland residents are waiting for an unprecedented tour of wizard in his Blue Helicopter over the cities of Berland!
It is well-known that there are n cities in Berland, some pairs of which are connected by bidirectional roads. Each pair of cities is connected by no more than one road. It is not guaranteed that the road network is connected, i.e. it is possible that you can't reach some city from some other.
The tour will contain several episodes. In each of the episodes:
* the wizard will disembark at some city x from the Helicopter;
* he will give a performance and show a movie for free at the city x;
* he will drive to some neighboring city y using a road;
* he will give a performance and show a movie for free at the city y;
* he will drive to some neighboring to y city z;
* he will give a performance and show a movie for free at the city z;
* he will embark the Helicopter and fly away from the city z.
It is known that the wizard doesn't like to use roads, so he agrees to use each road at most once (regardless of direction). In other words, for road between a and b he only can drive once from a to b, or drive once from b to a, or do not use this road at all.
The wizards wants to plan as many episodes as possible without violation the above rules. Help the wizard!
Please note that the wizard can visit the same city multiple times, the restriction is on roads only.
Input
The first line contains two integers n, m (1 ≤ n ≤ 2·105, 0 ≤ m ≤ 2·105) — the number of cities and the number of roads in Berland, respectively.
The roads description follow, one in each line. Each description is a pair of two integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the ids of the cities connected by the i-th road. It is guaranteed that there are no two roads connecting the same pair of cities. Every road is bidirectional. The cities are numbered from 1 to n.
It is possible that the road network in Berland is not connected.
Output
In the first line print w — the maximum possible number of episodes. The next w lines should contain the episodes in format x, y, z — the three integers denoting the ids of the cities in the order of the wizard's visits.
Examples
Input
4 5
1 2
3 2
2 4
3 4
4 1
Output
2
1 4 2
4 3 2
Input
5 8
5 3
1 2
4 5
5 1
2 5
4 3
1 4
3 2
Output
4
1 4 5
2 3 4
1 5 3
5 2 1
Submitted Solution:
```
from collections import defaultdict
def dfs_try(u,vis,count):
global path
for v,e in g[u]:
if not vis[e]:
path.append(v+1)
vis[e]=1
count+=1
if count==2:
return 1
else:
return dfs_try(v,vis,count)
path.pop()
return 0
g=defaultdict(list)
n,m=map(int, input().split())
count=[0]*n
for i in range(m):
a,b=map(int, input().split())
g[a-1].append([b-1,i])
g[b-1].append([a-1,i])
ans=0
vis=[0]*m
path=[]
final=[]
for x in range(n):
path=[x+1]
if dfs_try(x,vis,0):
ans+=1
final.append(path)
print(ans)
for x in final:
print(x)
``` | instruction | 0 | 83,089 | 1 | 166,178 |
No | output | 1 | 83,089 | 1 | 166,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
All Berland residents are waiting for an unprecedented tour of wizard in his Blue Helicopter over the cities of Berland!
It is well-known that there are n cities in Berland, some pairs of which are connected by bidirectional roads. Each pair of cities is connected by no more than one road. It is not guaranteed that the road network is connected, i.e. it is possible that you can't reach some city from some other.
The tour will contain several episodes. In each of the episodes:
* the wizard will disembark at some city x from the Helicopter;
* he will give a performance and show a movie for free at the city x;
* he will drive to some neighboring city y using a road;
* he will give a performance and show a movie for free at the city y;
* he will drive to some neighboring to y city z;
* he will give a performance and show a movie for free at the city z;
* he will embark the Helicopter and fly away from the city z.
It is known that the wizard doesn't like to use roads, so he agrees to use each road at most once (regardless of direction). In other words, for road between a and b he only can drive once from a to b, or drive once from b to a, or do not use this road at all.
The wizards wants to plan as many episodes as possible without violation the above rules. Help the wizard!
Please note that the wizard can visit the same city multiple times, the restriction is on roads only.
Input
The first line contains two integers n, m (1 ≤ n ≤ 2·105, 0 ≤ m ≤ 2·105) — the number of cities and the number of roads in Berland, respectively.
The roads description follow, one in each line. Each description is a pair of two integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the ids of the cities connected by the i-th road. It is guaranteed that there are no two roads connecting the same pair of cities. Every road is bidirectional. The cities are numbered from 1 to n.
It is possible that the road network in Berland is not connected.
Output
In the first line print w — the maximum possible number of episodes. The next w lines should contain the episodes in format x, y, z — the three integers denoting the ids of the cities in the order of the wizard's visits.
Examples
Input
4 5
1 2
3 2
2 4
3 4
4 1
Output
2
1 4 2
4 3 2
Input
5 8
5 3
1 2
4 5
5 1
2 5
4 3
1 4
3 2
Output
4
1 4 5
2 3 4
1 5 3
5 2 1
Submitted Solution:
```
def delete(n,l,r):
global m
if m[l]==n:
m=m[:l]+m[l+1:]
return
elif m[r]==n:
m=m[:r]+m[r+1:]
return
else:
mid=(r+l)//2
if m[mid]<n:
delete(n,mid,r)
elif m[mid]>n:
delete(n,l,mid)
else:
m=m[:mid]+m[mid+1:]
return
n,m=map(int,input().split())
smezh=[[] for i in range(n)]
for i in range(m):
a,b=map(int,input().split())
a-=1
b-=1
smezh[a].append(b)
smezh[b].append(a)
ans=[]
for i in range(n):
smezh[i].sort()
def dfs(v):
global smezh
global m
global ans
first=-1
i=0
while i<len(smezh[v]):
u=smezh[v][i]
smezh[v]=smezh[v][1:]
m=smezh[u]
delete(v,0,len(m)-1)
smezh[u]=m
findfs=dfs(u)
if findfs!=-1:
if first==-1:
first=findfs
else:
ans.append([first+1,v+1,findfs+1])
first=-1
if first==-1:
first=u
else:
ans.append([first+1,u+1,v+1])
first=-1
return first
for i in range(n):
dfs(i)
print(len(ans))
for i in range(len(ans)):
print(ans[i][0],ans[i][1],ans[i][2])
``` | instruction | 0 | 83,090 | 1 | 166,180 |
No | output | 1 | 83,090 | 1 | 166,181 |
Provide a correct Python 3 solution for this coding contest problem.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998 | instruction | 0 | 83,204 | 1 | 166,408 |
"Correct Solution:
```
from bisect import bisect
a,b,q,*stx = map(int,open(0).read().split())
s = [-1e10] + stx[:a] + [2*1e10]
t = [-1e10] + stx[a:a+b] + [2*1e10]
x = stx[a+b:a+b+q]
for i in x:
pos_s = bisect(s,i)
pos_t = bisect(t,i)
sl = s[pos_s-1]; sr = s[pos_s]
tl = t[pos_t-1]; tr = t[pos_t]
ans = min( i-min(sl,tl), max(sr,tr)-i, (sr-i) + (sr-tl), (tr-i) + (tr-sl), (i-sl) + (tr-sl), (i-tl) + (sr-tl) )
print(ans)
``` | output | 1 | 83,204 | 1 | 166,409 |
Provide a correct Python 3 solution for this coding contest problem.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998 | instruction | 0 | 83,205 | 1 | 166,410 |
"Correct Solution:
```
import bisect
a, b, q = map(int, input().split())
INF = 10 ** 12
S = [-INF] + [int(input()) for i in range(a)] + [INF]
T = [-INF] + [int(input()) for i in range(b)] + [INF]
for i in range(q):
x = int(input())
sid = bisect.bisect_left(S, x)
tid = bisect.bisect_left(T, x)
ans = INF
for s in (S[sid - 1], S[sid]):
for t in (T[tid - 1], T[tid]):
ans = min(ans, abs(x - s) +abs(s - t))
ans = min(ans, abs(x - t) + abs(t - s))
print(ans)
``` | output | 1 | 83,205 | 1 | 166,411 |
Provide a correct Python 3 solution for this coding contest problem.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998 | instruction | 0 | 83,206 | 1 | 166,412 |
"Correct Solution:
```
import bisect
A, B, Q = map(int, input().split())
INF = 10**18
S = [-INF]
for i in range(A):
S.append(int(input()))
S.append(INF)
T = [-INF]
for i in range(B):
T.append(int(input()))
T.append(INF)
X = []
for i in range(Q):
X.append(int(input()))
for x in X:
a, b = bisect.bisect_right(S, x), bisect.bisect_right(T, x)
print(min(abs(s-t) + min(abs(s-x), abs(t-x))
for s in S[a-1:a+1] for t in T[b-1:b+1]))
``` | output | 1 | 83,206 | 1 | 166,413 |
Provide a correct Python 3 solution for this coding contest problem.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998 | instruction | 0 | 83,207 | 1 | 166,414 |
"Correct Solution:
```
from bisect import bisect_right as br
a,b,q=map(int,input().split())
I=2<<60
s,t=[-I]+[int(input()) for _ in range(a)]+[I],[-I]+[int(input()) for _ in range(b)]+[I]
for question in range(q):
x=int(input())
b,d,r=br(s,x),br(t,x),I
for S in [s[b-1],s[b]]:
for T in [t[d-1],t[d]]:
d1,d2=abs(S-x)+abs(T-S),abs(T-x)+abs(S-T)
r=min(r,d1,d2)
print(r)
``` | output | 1 | 83,207 | 1 | 166,415 |
Provide a correct Python 3 solution for this coding contest problem.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998 | instruction | 0 | 83,208 | 1 | 166,416 |
"Correct Solution:
```
import bisect
a,b,q=map(int,input().split())
INF=10**18
s=[-INF]+[int(input()) for _ in range(a)]+[INF]
t=[-INF]+[int(input()) for _ in range(b)]+[INF]
for k in range(q):
x=int(input())
i=bisect.bisect_right(s,x)
j=bisect.bisect_right(t,x)
d=INF
for S in [s[i-1],s[i]]:
for T in [t[j-1],t[j]]:
d1=abs(S-x)+abs(S-T)
d2=abs(T-x)+abs(S-T)
d=min(d,d1,d2)
print(d)
``` | output | 1 | 83,208 | 1 | 166,417 |
Provide a correct Python 3 solution for this coding contest problem.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998 | instruction | 0 | 83,209 | 1 | 166,418 |
"Correct Solution:
```
INF = 1e20
A, B, Q = [int(i) for i in input().split()]
S = [-INF] + [int(input()) for _ in range(A)] + [INF]
T = [-INF] + [int(input()) for _ in range(B)] + [INF]
X = [int(input()) for _ in range(Q)]
from bisect import bisect_left
mins = []
for x in X:
s = bisect_left(S, x)
t = bisect_left(T, x)
mi = INF
for ss in [S[s], S[s - 1]]:
for tt in [T[t], T[t - 1]]:
mi = min(mi, abs(ss - x) + abs(tt - ss), abs(tt - x) + abs(ss - tt))
mins.append(str(mi))
print("\n".join(mins))
``` | output | 1 | 83,209 | 1 | 166,419 |
Provide a correct Python 3 solution for this coding contest problem.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998 | instruction | 0 | 83,210 | 1 | 166,420 |
"Correct Solution:
```
from bisect import bisect
a,b,q=map(int,input().split())
INF=float("inf")
s=[-INF]+[int(input()) for _ in range(a)]+[INF]
t=[-INF]+[int(input()) for _ in range(b)]+[INF]
xx=[int(input()) for _ in range(q)]
for x in xx:
si=bisect(s,x)
ti=bisect(t,x)
sf,sb=x-s[si-1],s[si]-x
tf,tb=x-t[ti-1],t[ti]-x
print(min(2*sf+tb,sf+tb*2,2*sb+tf,sb+tf*2,max(sf,tf),max(sb,tb)))
``` | output | 1 | 83,210 | 1 | 166,421 |
Provide a correct Python 3 solution for this coding contest problem.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998 | instruction | 0 | 83,211 | 1 | 166,422 |
"Correct Solution:
```
import bisect
A, B, Q = map(int, input().split())
s = [int(input()) for i in range(A)]
t = [int(input()) for i in range(B)]
x = [int(input()) for i in range(Q)]
for i in x: # N
idx_i = bisect.bisect(s, i) # log A
z = float('inf')
for j in s[max(0, idx_i - 1): idx_i + 1]: # 2
idx_j = bisect.bisect(t, j) # log B
for k in t[max(0, idx_j - 1): idx_j + 1]: # 2
_z = min(abs(i - j) + abs(j - k), abs(i - k) + abs(k - j))
z = min(z, _z)
print(z)
``` | output | 1 | 83,211 | 1 | 166,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998
Submitted Solution:
```
import bisect
INF = 10 ** 18
A, B, Q = map(int, input().split())
s = [-INF] + [int(input()) for i in range(A)] + [INF]
t = [-INF] + [int(input()) for i in range(B)] + [INF]
for q in range(Q):
x = int(input())
b, d = bisect.bisect_right(s, x), bisect.bisect_right(t, x)
res = INF
for S in [s[b - 1], s[b]]:
for T in [t[d - 1], t[d]]:
d1, d2 = abs(S - x) + abs(T - S), abs(T - x) + abs(S - T)
res = min(res, d1, d2)
print(res)
``` | instruction | 0 | 83,212 | 1 | 166,424 |
Yes | output | 1 | 83,212 | 1 | 166,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998
Submitted Solution:
```
from bisect import bisect
a,b,q=map(int,input().split())
s=[-10**18]+[int(input()) for i in range(a)]+[10**18]
t=[-10**18]+[int(input()) for i in range(b)]+[10**18]
for i in range(q):
x=int(input())
s_inx=bisect(s,x)
t_inx=bisect(t,x)
s_l=s[s_inx-1]
s_r=s[s_inx]
t_l=t[t_inx-1]
t_r=t[t_inx]
ans=10**18
for i in [s_l,s_r]:
for j in [t_l,t_r]:
d1=abs(i-x)+abs(j-i)
d2=abs(j-x)+abs(i-j)
ans=min(ans,d1,d2)
print(ans)
``` | instruction | 0 | 83,213 | 1 | 166,426 |
Yes | output | 1 | 83,213 | 1 | 166,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998
Submitted Solution:
```
import bisect,sys
input=sys.stdin.readline
a,b,Q=map(int,input().split())
INF=10**18
s=[-INF]+[int(input()) for _ in range(a)]+[INF]
t=[-INF]+[int(input()) for _ in range(b)]+[INF]
for q in range(Q):
x=int(input())
b,d=bisect.bisect_right(s,x),bisect.bisect_right(t,x)
res=INF
for S in [s[b-1],s[b]]:
for T in [t[d-1],t[d]]:
d1,d2=abs(S-x)+abs(T-S),abs(T-x)+abs(T-S)
res=min(res,d1,d2)
print(res)
``` | instruction | 0 | 83,214 | 1 | 166,428 |
Yes | output | 1 | 83,214 | 1 | 166,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998
Submitted Solution:
```
import bisect
INF=10**11
A,B,q=map(int,input().split())
a=[-INF]+[int(input()) for _ in range(A)]+[INF]
b=[-INF]+[int(input()) for _ in range(B)]+[INF]
for _ in range(q):
x=int(input())
i=bisect.bisect_left(a,x)
j=bisect.bisect_left(b,x)
ans=min(max(a[i],b[j])-x,x-min(a[i-1],b[j-1]),a[i]-b[j-1]+min(a[i]-x,x-b[j-1]),b[j]-a[i-1]+min(b[j]-x,x-a[i-1]))
print(ans)
``` | instruction | 0 | 83,215 | 1 | 166,430 |
Yes | output | 1 | 83,215 | 1 | 166,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998
Submitted Solution:
```
import sys
from bisect import bisect_left
#inpurt = sys.stdin.readline
a, b, q = map(int, input().split())
INF = 10 ** 11
s = [-INF]
s_append = s.append
for i in range(a):
si = int(input())
s_append(si)
s_append(INF)
t = [-INF]
#t_append = t.append
for i in range(b):
ti = int(input())
t.append(ti)
t.append(INF)
x = []
#x_append = x.append
for i in range(q):
xi = int(input())
x.append(xi)
for xi in x:
index_s = bisect_left(s, xi)
s1 = xi - s[index_s - 1]
s2 = s[index_s] - xi
index_t = bisect_left(t, xi)
t1 = xi - t[index_t - 1]
t2 = t[index_t] -xi
print(min(max(s1, t1), max(s2, t2), min(s1, t2) * 2 + max(s1, t2), min(s2, t1) * 2 + max(s2, t1)))
``` | instruction | 0 | 83,216 | 1 | 166,432 |
No | output | 1 | 83,216 | 1 | 166,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998
Submitted Solution:
```
import bisect
A, B, Q = map(int,input().split())
INF = 30000000000
s = [-1*INF]
t = [-1*INF]
for _ in range(A):
s.append(int(input()))
s.append(INF)
for _ in range(B):
t.append(int(input()))
t.append(INF)
for _ in range(Q):
x = int(input())
s_index = bisect.bisect_left(s,x)
t_index = bisect.bisect_left(t,x)
s_l = x - s[s_index-1]
s_r = s[s_index] - x
t_l = x - t[t_index-1]
t_r = t[t_index] - x
print(min(max(s_l,t_l), s_l + t_r*2, s_r + t_l*2, s_l*2 + t_r, s_r*2 + t_l, max(s_r,t_r)))
``` | instruction | 0 | 83,217 | 1 | 166,434 |
No | output | 1 | 83,217 | 1 | 166,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998
Submitted Solution:
```
import numpy as np
A,B,Q = list(map(int,input().split()))
S = [int(input()) for _ in range(A) ]
T = [int(input()) for _ in range(B) ]
x = [int(input()) for _ in range(Q)]
S = np.array(S)
T = np.array(T)
for a in x:
ans1 = 0
ans2 = 0
S1 = abs(S-a)
T1 = abs(T-a)
ans1 +=np.min(T1)
ans1 += min(abs(S-T[np.argmin(T1)]))
ans2+=np.min(S1)
ans2+= min(abs(T-S[np.argmin(S1)]))
print(min(ans1,ans2))
``` | instruction | 0 | 83,218 | 1 | 166,436 |
No | output | 1 | 83,218 | 1 | 166,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Along a road running in an east-west direction, there are A shrines and B temples. The i-th shrine from the west is located at a distance of s_i meters from the west end of the road, and the i-th temple from the west is located at a distance of t_i meters from the west end of the road.
Answer the following Q queries:
* Query i (1 \leq i \leq Q): If we start from a point at a distance of x_i meters from the west end of the road and freely travel along the road, what is the minimum distance that needs to be traveled in order to visit one shrine and one temple? (It is allowed to pass by more shrines and temples than required.)
Constraints
* 1 \leq A, B \leq 10^5
* 1 \leq Q \leq 10^5
* 1 \leq s_1 < s_2 < ... < s_A \leq 10^{10}
* 1 \leq t_1 < t_2 < ... < t_B \leq 10^{10}
* 1 \leq x_i \leq 10^{10}
* s_1, ..., s_A, t_1, ..., t_B, x_1, ..., x_Q are all different.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B Q
s_1
:
s_A
t_1
:
t_B
x_1
:
x_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Examples
Input
2 3 4
100
600
400
900
1000
150
2000
899
799
Output
350
1400
301
399
Input
1 1 3
1
10000000000
2
9999999999
5000000000
Output
10000000000
10000000000
14999999998
Submitted Solution:
```
num=list(map(int,input().split()))
s_list = [int(input()) for i in range(num[0])]
t_list = [int(input()) for i in range(num[1])]
x_list = [int(input()) for i in range(num[2])]
for x_ in x_list:
min_s=min(list(map(lambda x: abs(x-x_), s_list)))
min_t=min(list(map(lambda x: abs(x-x_), t_list)))
min_index_t=list(map(lambda x: abs(x-x_), t_list)).index(min_t)
minmin_t=min(list(map(lambda x: abs(x-t_list[min_index_t]), s_list)))+min_t
min_index_s=list(map(lambda x: abs(x-x_), s_list)).index(min_s)
minmin_s=min(list(map(lambda x: abs(x-s_list[min_index_s]), t_list)))+min_s
print(min(minminmin_t,minmin_s))
``` | instruction | 0 | 83,219 | 1 | 166,438 |
No | output | 1 | 83,219 | 1 | 166,439 |
Provide a correct Python 3 solution for this coding contest problem.
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33 | instruction | 0 | 83,342 | 1 | 166,684 |
"Correct Solution:
```
while 1:
d,e=map(int,input().split())
mi=((d//2)**2+(d-d//2)**2)**0.5
if d==e==0:break
if d<=e:
print(e-d)
elif mi>=e:
print(f'{mi-e:.5f}')
else:
ans=float("inf")
for i in range(d//2):
mi=abs(((i)**2+(d-i)**2)**0.5-e)
ans=min(ans,mi)
print(f'{ans:.5f}')
``` | output | 1 | 83,342 | 1 | 166,685 |
Provide a correct Python 3 solution for this coding contest problem.
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33 | instruction | 0 | 83,343 | 1 | 166,686 |
"Correct Solution:
```
import math
def get_two_int():
two_int = input().split()
for i in range(2):
two_int[i] = int(two_int[i])
return two_int
def get_diff_set_between_cost_and_budget(walk_distance, budget): #一つの関数にまとめすぎか？
diff_set = set()
max_x = walk_distance // 2 + 1
for x in range(max_x):
y = walk_distance - x
abs_diff = abs(calculate_cost(x, y) - budget)
diff_set.add(abs_diff)
return diff_set
def calculate_cost(x, y):
cost = math.sqrt((x ** 2) + (y ** 2))
return cost
if __name__ == "__main__":
while True:
walk_distance, budget = get_two_int()
if walk_distance == 0:
break
diff_set = get_diff_set_between_cost_and_budget(walk_distance, budget)
min_num = min(diff_set)
print(min_num)
``` | output | 1 | 83,343 | 1 | 166,687 |
Provide a correct Python 3 solution for this coding contest problem.
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33 | instruction | 0 | 83,344 | 1 | 166,688 |
"Correct Solution:
```
#!/usr/bin/env python3
from math import hypot
while True:
d, e = map(int, input().split())
if d == 0:
exit()
ans = 1e5
for x in range(d + 1):
y = d - x
ans = min(ans, abs(hypot(x, y) - e))
print(ans)
``` | output | 1 | 83,344 | 1 | 166,689 |
Provide a correct Python 3 solution for this coding contest problem.
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33 | instruction | 0 | 83,345 | 1 | 166,690 |
"Correct Solution:
```
import math
while True:
d,e = map(int,input().split())
sa = max(d,e)
if d == 0 and e == 0:
break
for i in range(math.floor(d/2) + 1):
if math.fabs(math.sqrt(i ** 2 + (d - i) ** 2) - e) < sa:
sa = math.fabs(math.sqrt(i ** 2 + (d - i) ** 2) - e)
print(sa)
``` | output | 1 | 83,345 | 1 | 166,691 |
Provide a correct Python 3 solution for this coding contest problem.
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33 | instruction | 0 | 83,346 | 1 | 166,692 |
"Correct Solution:
```
while True :
D, E = map(int, input().split())
if(D == 0) :
break
else :
ans = 100
for i in range((D // 2) + 1) :
A = (i ** 2 + (D - i) ** 2) ** (1/2)
a = abs(A - E)
if(a < ans) :
ans = a
else :
pass
print(ans)
``` | output | 1 | 83,346 | 1 | 166,693 |
Provide a correct Python 3 solution for this coding contest problem.
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33 | instruction | 0 | 83,347 | 1 | 166,694 |
"Correct Solution:
```
import math
if __name__ == "__main__":
while 1:
d,e = list(map(int,input().strip().split()))
if d == e == 0:break
diff = 20000
for x in range(d + 1):
y = d-x
if diff > math.fabs(math.sqrt(x**2 + y**2)-e):diff =math.fabs(math.sqrt(x**2 + y**2)-e)
print(diff)
``` | output | 1 | 83,347 | 1 | 166,695 |
Provide a correct Python 3 solution for this coding contest problem.
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33 | instruction | 0 | 83,348 | 1 | 166,696 |
"Correct Solution:
```
import math
import sys
def main():
while True:
d,e = map(int,input().split())
ans = sys.maxsize
if not d and not e:
return
for x in range(d+1):
y = d-x
ans = min(ans, abs(math.sqrt(float(x)**2+float(y)**2)-e))
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 83,348 | 1 | 166,697 |
Provide a correct Python 3 solution for this coding contest problem.
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33 | instruction | 0 | 83,349 | 1 | 166,698 |
"Correct Solution:
```
while True:
d,e = map(int,input().split())
if d==e==0: break
r = []
for x in range(int(d/2+1)):
r += [abs((x**2+(d-x)**2)**0.5-e)]
print(min(r) if min(r)%1!=0 else int(min(r)))
``` | output | 1 | 83,349 | 1 | 166,699 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33
Submitted Solution:
```
while 1:
d,e = map(int,input().split())
if d==0: break
print('%.6g'%(min(abs((x**2+(d-x)**2)**0.5-e)for x in range(d//2+1))))
``` | instruction | 0 | 83,350 | 1 | 166,700 |
Yes | output | 1 | 83,350 | 1 | 166,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33
Submitted Solution:
```
import math
while True:
D, E = map(int, input().split())
if D == 0 and E == 0:
break
ans = float("inf")
for x in range(D + 1):
for y in range(D + 1):
if x + y == D:
# print("x: {}, y: {}, sqrt: {}".format(x, y, math.sqrt(x * x + y * y)))
ans = min(abs(math.sqrt(x * x + y * y) - E), ans)
print(ans)
``` | instruction | 0 | 83,351 | 1 | 166,702 |
Yes | output | 1 | 83,351 | 1 | 166,703 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Koto Municipal Subway
Koto Municipal Subway
Koto City is a famous city whose roads are in a grid pattern, as shown in the figure below. The roads extending from north to south and the roads extending from east to west are lined up at intervals of 1 km each. Let Koto station at the southwestern intersection of Koto city be (0, 0), and mark x km east and y km north from it as (x, y) (0 ≤ x, y). ).
<image>
In anticipation of an increase in tourists due to the Olympic Games to be held five years later, the city has decided to build a new subway line starting from Koto station. Currently, we are planning to lay a rail to Shin-Koto station, which will be newly constructed as the next station after Koto station. The rail will be laid straight from Koto station to Shin-Koto station. Therefore, when the location of Shin-Koto station is (x, y), the length of the rail is √ (x2 + y2). The cost of laying the rail is the same as the length of the laid rail. Even if the rail length is a decimal number such as 1.5km, the cost is also 1.5.
The location (x, y) of Shin-Koto station has not been decided yet, and we plan to make it a location that meets the following conditions.
* It is an intersection. That is, x and y are integers, respectively.
* The shortest distance walked along the road from Koto station is exactly D. That is, x + y = D is satisfied.
Among the above two conditions, select the location of Shin-Koto station so that the difference between the rail budget E and the rail cost set by the city | √ (x2 + y2) --E | is minimized. Here, | A | represents the absolute value of A. Your job is to create a program that outputs the difference between the cost and budget for laying rails when the Shin-Koto station was constructed as described above.
Input
The input consists of multiple data sets, and the number of data sets contained in one input is 100 or less. The format of each data set is as follows.
> D E
D (1 ≤ D ≤ 100) is an integer that represents the shortest distance when walking along the road from Koto station to Shin-Koto station. E (1 ≤ E ≤ 100) is an integer representing the budget for rail construction.
The end of the input is indicated by a line of two zeros separated by a blank.
Output
For each data set, output the difference between the cost and budget when laying the rail so as to satisfy the problem condition in one line. The answer must have an absolute error greater than 10-3. Note that if you do not output a line break at the end of each line, or if you output unnecessary characters, it will be judged as an incorrect answer.
Sample Input
twenty one
7 5
7 6
7 7
76 5
8 41
0 0
Output for Sample Input
0.4142135624
0
0.0827625303
0
48.7401153702
33
Hint
In the first data set, as shown in the figure below, an intersection 2 km along the road from Koto station is a candidate for a place to build Shin-Koto station.
<image>
When Shin-Koto station is constructed at each intersection, the difference between the cost for laying the rail and budget 1 is as follows.
* (2, 0): | √ (22 + 02) --1 | = 1.0
* (1, 1): | √ (12 + 12) --1 | = 0.4142135623 ...
* (0, 2): | √ (02 + 22) --1 | = 1.0
Therefore, the difference between the cost and the budget is minimized when the construction is done in (1, 1).
Example
Input
2 1
7 5
7 6
7 7
76 5
8 41
0 0
Output
0.4142135624
0
0.0827625303
0
48.7401153702
33
Submitted Solution:
```
import math
while True:
d, e = map(int, input().split())
if d == 0:break
print(min([abs(e - math.sqrt(i ** 2 + (d - i) ** 2)) for i in range(d // 2 + 1)]))
``` | instruction | 0 | 83,352 | 1 | 166,704 |
Yes | output | 1 | 83,352 | 1 | 166,705 |
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