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 |
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Provide a correct Python 3 solution for this coding contest problem.
Mr. A, who will graduate next spring, decided to move when he got a job. The company that finds a job has offices in several towns, and the offices that go to work differ depending on the day. So Mr. A decided to live in a town where he had a short time to go to any office.
So you decided to find the most convenient town to live in to help Mr. A.
<image>
Towns are numbered starting with 0 and there is a road between towns. Commuting time is fixed for each road. If Mr. A lives in a town, the commuting time to the office in his town is 0. At this time, consider the total commuting time to all towns. For example, if the layout of towns and roads is as shown in the figure above and Mr. A lives in town 1, the commuting time to each town will be
Town 0 to 80
0 to town 1
Up to town 2 20
Up to town 3 70
Town 4 to 90
And the sum is 260.
Enter the number of roads and information on all roads, calculate the total commuting time when living in each town, and output the number of the town that minimizes it and the total commuting time at that time. Create a program. However, if there are multiple towns with the smallest total commuting time, please output the number of the smallest town and the total commuting time at that time. The total number of towns is 10 or less, the total number of roads is 45 or less, all roads can move in both directions, and commuting time does not change depending on the direction. We also assume that there are routes from any town to all other towns.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 b1 c1
a2 b2 c2
::
an bn cn
The number of roads n (1 ≤ n ≤ 45) is given on the first line. The following n lines give information on the i-th way. ai, bi (0 ≤ ai, bi ≤ 9) is the number of the town to which the i-th road connects, and ci (0 ≤ ci ≤ 100) is the commute time for that road.
Output
For each data set, the town number that minimizes the total commuting time and the total commuting time at that time are output on one line separated by blanks.
Example
Input
6
0 1 80
1 2 20
0 2 60
2 3 50
3 4 60
1 4 90
2
0 1 1
1 2 1
0
Output
2 240
1 2 | instruction | 0 | 74,110 | 1 | 148,220 |
"Correct Solution:
```
while True:
try:
n = int(input())
if not n : break
town = 0
dis = [[float('inf') for j in range(10)] for i in range(10)]
for i in range(n):
a,b,c = map(int , input().split())
dis[a][b] = c
dis[b][a] = c
town = max(town,a)
town = max(town,b)
for i in range(town+1):
dis[i][i] = 0
mindis = dis
for i in range(town+1):
for j in range(town+1):
for k in range(town+1):
if (mindis[j][k] > mindis[j][i]+mindis[i][k]):
mindis[j][k] = mindis[j][i]+mindis[i][k]
mintown = 0
ans = float('inf')
for i in range(town+1):
sum = 0
for j in range(town+1):
sum += mindis[i][j]
if (ans > sum) :
ans = sum
mintown = i
print(mintown , ans)
except:
break
``` | output | 1 | 74,110 | 1 | 148,221 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. A, who will graduate next spring, decided to move when he got a job. The company that finds a job has offices in several towns, and the offices that go to work differ depending on the day. So Mr. A decided to live in a town where he had a short time to go to any office.
So you decided to find the most convenient town to live in to help Mr. A.
<image>
Towns are numbered starting with 0 and there is a road between towns. Commuting time is fixed for each road. If Mr. A lives in a town, the commuting time to the office in his town is 0. At this time, consider the total commuting time to all towns. For example, if the layout of towns and roads is as shown in the figure above and Mr. A lives in town 1, the commuting time to each town will be
Town 0 to 80
0 to town 1
Up to town 2 20
Up to town 3 70
Town 4 to 90
And the sum is 260.
Enter the number of roads and information on all roads, calculate the total commuting time when living in each town, and output the number of the town that minimizes it and the total commuting time at that time. Create a program. However, if there are multiple towns with the smallest total commuting time, please output the number of the smallest town and the total commuting time at that time. The total number of towns is 10 or less, the total number of roads is 45 or less, all roads can move in both directions, and commuting time does not change depending on the direction. We also assume that there are routes from any town to all other towns.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 b1 c1
a2 b2 c2
::
an bn cn
The number of roads n (1 ≤ n ≤ 45) is given on the first line. The following n lines give information on the i-th way. ai, bi (0 ≤ ai, bi ≤ 9) is the number of the town to which the i-th road connects, and ci (0 ≤ ci ≤ 100) is the commute time for that road.
Output
For each data set, the town number that minimizes the total commuting time and the total commuting time at that time are output on one line separated by blanks.
Example
Input
6
0 1 80
1 2 20
0 2 60
2 3 50
3 4 60
1 4 90
2
0 1 1
1 2 1
0
Output
2 240
1 2 | instruction | 0 | 74,111 | 1 | 148,222 |
"Correct Solution:
```
def floyd_warshall(d):
v = len(d)
for k in range(v):
for i in range(v):
for j in range(v):
d[i][j] = min(d[i][j], d[i][k] + d[k][j])
while True:
n = int(input())
if n == 0:
break
info = []
num_cities = 1
for i in range(n):
a, b, c = map(int, input().split())
num_cities = max(num_cities, a+1, b+1)
info.append((a, b, c))
d = [[0 if i == j else float('inf') for j in range(num_cities)] for i in range(num_cities)]
for a, b, c in info:
d[a][b] = d[b][a] = c
floyd_warshall(d)
total = [sum(d[i]) for i in range(num_cities)]
min_dist = min(total)
print(total.index(min_dist), min_dist)
``` | output | 1 | 74,111 | 1 | 148,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. A, who will graduate next spring, decided to move when he got a job. The company that finds a job has offices in several towns, and the offices that go to work differ depending on the day. So Mr. A decided to live in a town where he had a short time to go to any office.
So you decided to find the most convenient town to live in to help Mr. A.
<image>
Towns are numbered starting with 0 and there is a road between towns. Commuting time is fixed for each road. If Mr. A lives in a town, the commuting time to the office in his town is 0. At this time, consider the total commuting time to all towns. For example, if the layout of towns and roads is as shown in the figure above and Mr. A lives in town 1, the commuting time to each town will be
Town 0 to 80
0 to town 1
Up to town 2 20
Up to town 3 70
Town 4 to 90
And the sum is 260.
Enter the number of roads and information on all roads, calculate the total commuting time when living in each town, and output the number of the town that minimizes it and the total commuting time at that time. Create a program. However, if there are multiple towns with the smallest total commuting time, please output the number of the smallest town and the total commuting time at that time. The total number of towns is 10 or less, the total number of roads is 45 or less, all roads can move in both directions, and commuting time does not change depending on the direction. We also assume that there are routes from any town to all other towns.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 b1 c1
a2 b2 c2
::
an bn cn
The number of roads n (1 ≤ n ≤ 45) is given on the first line. The following n lines give information on the i-th way. ai, bi (0 ≤ ai, bi ≤ 9) is the number of the town to which the i-th road connects, and ci (0 ≤ ci ≤ 100) is the commute time for that road.
Output
For each data set, the town number that minimizes the total commuting time and the total commuting time at that time are output on one line separated by blanks.
Example
Input
6
0 1 80
1 2 20
0 2 60
2 3 50
3 4 60
1 4 90
2
0 1 1
1 2 1
0
Output
2 240
1 2
Submitted Solution:
```
road = int(input())
while road > 0:
answer = [[float("inf") if n != m else 0 for n in range(10)] for m in range(10)]
city = 0
for _ in range(road):
c1, c2, w = (int(n) for n in input().split(" "))
answer[c1][c2] = w
answer[c2][c1] = w
city = max(city, c1, c2)
city += 1
for i in range(city):
for j in range(city):
for k in range(city):
answer[j][k] = min(answer[j][k], answer[j][i] + answer[i][k])
sum_ans = [-1 , float("inf")]
for i in range(city):
if sum(answer[i][:city]) < sum_ans[1]:
sum_ans = [i, sum(answer[i][:city])]
print(*sum_ans)
road = int(input())
``` | instruction | 0 | 74,112 | 1 | 148,224 |
Yes | output | 1 | 74,112 | 1 | 148,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. A, who will graduate next spring, decided to move when he got a job. The company that finds a job has offices in several towns, and the offices that go to work differ depending on the day. So Mr. A decided to live in a town where he had a short time to go to any office.
So you decided to find the most convenient town to live in to help Mr. A.
<image>
Towns are numbered starting with 0 and there is a road between towns. Commuting time is fixed for each road. If Mr. A lives in a town, the commuting time to the office in his town is 0. At this time, consider the total commuting time to all towns. For example, if the layout of towns and roads is as shown in the figure above and Mr. A lives in town 1, the commuting time to each town will be
Town 0 to 80
0 to town 1
Up to town 2 20
Up to town 3 70
Town 4 to 90
And the sum is 260.
Enter the number of roads and information on all roads, calculate the total commuting time when living in each town, and output the number of the town that minimizes it and the total commuting time at that time. Create a program. However, if there are multiple towns with the smallest total commuting time, please output the number of the smallest town and the total commuting time at that time. The total number of towns is 10 or less, the total number of roads is 45 or less, all roads can move in both directions, and commuting time does not change depending on the direction. We also assume that there are routes from any town to all other towns.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 b1 c1
a2 b2 c2
::
an bn cn
The number of roads n (1 ≤ n ≤ 45) is given on the first line. The following n lines give information on the i-th way. ai, bi (0 ≤ ai, bi ≤ 9) is the number of the town to which the i-th road connects, and ci (0 ≤ ci ≤ 100) is the commute time for that road.
Output
For each data set, the town number that minimizes the total commuting time and the total commuting time at that time are output on one line separated by blanks.
Example
Input
6
0 1 80
1 2 20
0 2 60
2 3 50
3 4 60
1 4 90
2
0 1 1
1 2 1
0
Output
2 240
1 2
Submitted Solution:
```
from collections import defaultdict
road = int(input())
while road > 0:
connect = defaultdict(list)
# weight = defaultdict(defaultdict(float("inf")))
answer = [[float("inf") if n != m else 0 for n in range(10)] for m in range(10)]
for _ in range(road):
c1, c2, w = (int(n) for n in input().split(" "))
connect[c1].append(c2)
connect[c2].append(c1)
answer[c1][c2] = w
answer[c2][c1] = w
city = max(connect.keys()) + 1
for i in range(city):
for j in range(city):
for k in range(city):
answer[j][k] = min(answer[j][k], answer[j][i] + answer[i][k])
sum_ans = [-1 , float("inf")]
for i in range(city):
if sum(answer[i][:city]) < sum_ans[1]:
sum_ans = [i, sum(answer[i][:city])]
print(*sum_ans)
road = int(input())
``` | instruction | 0 | 74,113 | 1 | 148,226 |
Yes | output | 1 | 74,113 | 1 | 148,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. A, who will graduate next spring, decided to move when he got a job. The company that finds a job has offices in several towns, and the offices that go to work differ depending on the day. So Mr. A decided to live in a town where he had a short time to go to any office.
So you decided to find the most convenient town to live in to help Mr. A.
<image>
Towns are numbered starting with 0 and there is a road between towns. Commuting time is fixed for each road. If Mr. A lives in a town, the commuting time to the office in his town is 0. At this time, consider the total commuting time to all towns. For example, if the layout of towns and roads is as shown in the figure above and Mr. A lives in town 1, the commuting time to each town will be
Town 0 to 80
0 to town 1
Up to town 2 20
Up to town 3 70
Town 4 to 90
And the sum is 260.
Enter the number of roads and information on all roads, calculate the total commuting time when living in each town, and output the number of the town that minimizes it and the total commuting time at that time. Create a program. However, if there are multiple towns with the smallest total commuting time, please output the number of the smallest town and the total commuting time at that time. The total number of towns is 10 or less, the total number of roads is 45 or less, all roads can move in both directions, and commuting time does not change depending on the direction. We also assume that there are routes from any town to all other towns.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 b1 c1
a2 b2 c2
::
an bn cn
The number of roads n (1 ≤ n ≤ 45) is given on the first line. The following n lines give information on the i-th way. ai, bi (0 ≤ ai, bi ≤ 9) is the number of the town to which the i-th road connects, and ci (0 ≤ ci ≤ 100) is the commute time for that road.
Output
For each data set, the town number that minimizes the total commuting time and the total commuting time at that time are output on one line separated by blanks.
Example
Input
6
0 1 80
1 2 20
0 2 60
2 3 50
3 4 60
1 4 90
2
0 1 1
1 2 1
0
Output
2 240
1 2
Submitted Solution:
```
# Aizu Problem 0189: Convenient Location
#
import sys, math, os, bisect
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
while True:
n = int(input())
if n == 0:
break
L = []
cities = set()
for _ in range(n):
city1, city2, weight = [int(_) for _ in input().split()]
L.append([city1, city2, weight])
cities.add(city1)
cities.add(city2)
m = len(cities)
dist = [[2**31] * m for _ in range(m)]
for i in range(m):
dist[i][i] = 0
for city1, city2, weight in L:
dist[city1][city2] = weight
dist[city2][city1] = weight
for k in range(m):
for i in range(m):
for j in range(m):
dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
city, time = -1, 2**31
for k in range(m):
S = sum(dist[k])
if S < time:
time = S
city = k
print(city, time)
``` | instruction | 0 | 74,114 | 1 | 148,228 |
Yes | output | 1 | 74,114 | 1 | 148,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. A, who will graduate next spring, decided to move when he got a job. The company that finds a job has offices in several towns, and the offices that go to work differ depending on the day. So Mr. A decided to live in a town where he had a short time to go to any office.
So you decided to find the most convenient town to live in to help Mr. A.
<image>
Towns are numbered starting with 0 and there is a road between towns. Commuting time is fixed for each road. If Mr. A lives in a town, the commuting time to the office in his town is 0. At this time, consider the total commuting time to all towns. For example, if the layout of towns and roads is as shown in the figure above and Mr. A lives in town 1, the commuting time to each town will be
Town 0 to 80
0 to town 1
Up to town 2 20
Up to town 3 70
Town 4 to 90
And the sum is 260.
Enter the number of roads and information on all roads, calculate the total commuting time when living in each town, and output the number of the town that minimizes it and the total commuting time at that time. Create a program. However, if there are multiple towns with the smallest total commuting time, please output the number of the smallest town and the total commuting time at that time. The total number of towns is 10 or less, the total number of roads is 45 or less, all roads can move in both directions, and commuting time does not change depending on the direction. We also assume that there are routes from any town to all other towns.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 b1 c1
a2 b2 c2
::
an bn cn
The number of roads n (1 ≤ n ≤ 45) is given on the first line. The following n lines give information on the i-th way. ai, bi (0 ≤ ai, bi ≤ 9) is the number of the town to which the i-th road connects, and ci (0 ≤ ci ≤ 100) is the commute time for that road.
Output
For each data set, the town number that minimizes the total commuting time and the total commuting time at that time are output on one line separated by blanks.
Example
Input
6
0 1 80
1 2 20
0 2 60
2 3 50
3 4 60
1 4 90
2
0 1 1
1 2 1
0
Output
2 240
1 2
Submitted Solution:
```
import sys
import math
import fractions
import itertools
from collections import deque
import copy
import bisect
import heapq
MOD = float('inf')
INF = 10 ** 18 + 7
input = lambda: sys.stdin.readline().strip()
def warshall_floyd(d):
"""
d[i][l]がi→lへのコストを示す隣接行列
結果としてd[i][l]はi→lへの最短コストになる。
"""
for i in range(len(d)):
for l in range(len(d)):
for k in range(len(d)):
d[l][k] = min(d[l][k], d[l][i] + d[i][k])
return d
while True:
n = int(input())
if n == 0:
break
else:
abc = [list(map(int, input().split())) for _ in range(n)]
info = [[INF for _ in range(10)] for _ in range(10)]
for i in range(10):
info[i][i] = 0
for i in abc:
info[i[0]][i[1]] = i[2]
info[i[1]][i[0]] = i[2]
result = warshall_floyd(info)
ans = []
for i in result:
for l in range(len(i)):
if i[l] == INF:
i[l] = 0
for i in range(10):
if sum(result[i]) != 0:
ans.append([sum(result[i]), i])
ans.sort()
print(ans[0][1], ans[0][0])
``` | instruction | 0 | 74,115 | 1 | 148,230 |
Yes | output | 1 | 74,115 | 1 | 148,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. A, who will graduate next spring, decided to move when he got a job. The company that finds a job has offices in several towns, and the offices that go to work differ depending on the day. So Mr. A decided to live in a town where he had a short time to go to any office.
So you decided to find the most convenient town to live in to help Mr. A.
<image>
Towns are numbered starting with 0 and there is a road between towns. Commuting time is fixed for each road. If Mr. A lives in a town, the commuting time to the office in his town is 0. At this time, consider the total commuting time to all towns. For example, if the layout of towns and roads is as shown in the figure above and Mr. A lives in town 1, the commuting time to each town will be
Town 0 to 80
0 to town 1
Up to town 2 20
Up to town 3 70
Town 4 to 90
And the sum is 260.
Enter the number of roads and information on all roads, calculate the total commuting time when living in each town, and output the number of the town that minimizes it and the total commuting time at that time. Create a program. However, if there are multiple towns with the smallest total commuting time, please output the number of the smallest town and the total commuting time at that time. The total number of towns is 10 or less, the total number of roads is 45 or less, all roads can move in both directions, and commuting time does not change depending on the direction. We also assume that there are routes from any town to all other towns.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 b1 c1
a2 b2 c2
::
an bn cn
The number of roads n (1 ≤ n ≤ 45) is given on the first line. The following n lines give information on the i-th way. ai, bi (0 ≤ ai, bi ≤ 9) is the number of the town to which the i-th road connects, and ci (0 ≤ ci ≤ 100) is the commute time for that road.
Output
For each data set, the town number that minimizes the total commuting time and the total commuting time at that time are output on one line separated by blanks.
Example
Input
6
0 1 80
1 2 20
0 2 60
2 3 50
3 4 60
1 4 90
2
0 1 1
1 2 1
0
Output
2 240
1 2
Submitted Solution:
```
road = int(input())
while road > 0:
answer = [[float("inf") if n != m else 0 for n in range(10)] for m in range(10)]
city = 0
for _ in range(road):
c1, c2, w = (int(n) for n in input().split(" "))
answer[c1][c2] = w
answer[c2][c1] = w
city = max(city, c1, c2)
city = max(connect.keys()) + 1
for i in range(city):
for j in range(city):
for k in range(city):
answer[j][k] = min(answer[j][k], answer[j][i] + answer[i][k])
sum_ans = [-1 , float("inf")]
for i in range(city):
if sum(answer[i][:city]) < sum_ans[1]:
sum_ans = [i, sum(answer[i][:city])]
print(*sum_ans)
road = int(input())
``` | instruction | 0 | 74,116 | 1 | 148,232 |
No | output | 1 | 74,116 | 1 | 148,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. A, who will graduate next spring, decided to move when he got a job. The company that finds a job has offices in several towns, and the offices that go to work differ depending on the day. So Mr. A decided to live in a town where he had a short time to go to any office.
So you decided to find the most convenient town to live in to help Mr. A.
<image>
Towns are numbered starting with 0 and there is a road between towns. Commuting time is fixed for each road. If Mr. A lives in a town, the commuting time to the office in his town is 0. At this time, consider the total commuting time to all towns. For example, if the layout of towns and roads is as shown in the figure above and Mr. A lives in town 1, the commuting time to each town will be
Town 0 to 80
0 to town 1
Up to town 2 20
Up to town 3 70
Town 4 to 90
And the sum is 260.
Enter the number of roads and information on all roads, calculate the total commuting time when living in each town, and output the number of the town that minimizes it and the total commuting time at that time. Create a program. However, if there are multiple towns with the smallest total commuting time, please output the number of the smallest town and the total commuting time at that time. The total number of towns is 10 or less, the total number of roads is 45 or less, all roads can move in both directions, and commuting time does not change depending on the direction. We also assume that there are routes from any town to all other towns.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 b1 c1
a2 b2 c2
::
an bn cn
The number of roads n (1 ≤ n ≤ 45) is given on the first line. The following n lines give information on the i-th way. ai, bi (0 ≤ ai, bi ≤ 9) is the number of the town to which the i-th road connects, and ci (0 ≤ ci ≤ 100) is the commute time for that road.
Output
For each data set, the town number that minimizes the total commuting time and the total commuting time at that time are output on one line separated by blanks.
Example
Input
6
0 1 80
1 2 20
0 2 60
2 3 50
3 4 60
1 4 90
2
0 1 1
1 2 1
0
Output
2 240
1 2
Submitted Solution:
```
INF = 10 ** 10
while True:
n = int(input())
if n == 0:
break
costs = [[INF] * 10 for _ in range(10)]
for i in range(10):
costs[i][i] = 0
city = set()
for _ in range(n):
a, b, c = map(int, input().split())
city = city | {a, b}
costs[a][b] = c
costs[b][a] = c
city = sorted(list(city))
for i in range(10):
for j in range(10):
for k in range(10):
costs[i][j] = min(costs[i][j], costs[i][k] + costs[k][j])
min_total = INF
best_city = -1
for fr in city:
total = 0
for to in city:
total += costs[fr][to]
if total < min_total:
min_total = total
best_city = fr
print(best_city, min_total)
``` | instruction | 0 | 74,117 | 1 | 148,234 |
No | output | 1 | 74,117 | 1 | 148,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. A, who will graduate next spring, decided to move when he got a job. The company that finds a job has offices in several towns, and the offices that go to work differ depending on the day. So Mr. A decided to live in a town where he had a short time to go to any office.
So you decided to find the most convenient town to live in to help Mr. A.
<image>
Towns are numbered starting with 0 and there is a road between towns. Commuting time is fixed for each road. If Mr. A lives in a town, the commuting time to the office in his town is 0. At this time, consider the total commuting time to all towns. For example, if the layout of towns and roads is as shown in the figure above and Mr. A lives in town 1, the commuting time to each town will be
Town 0 to 80
0 to town 1
Up to town 2 20
Up to town 3 70
Town 4 to 90
And the sum is 260.
Enter the number of roads and information on all roads, calculate the total commuting time when living in each town, and output the number of the town that minimizes it and the total commuting time at that time. Create a program. However, if there are multiple towns with the smallest total commuting time, please output the number of the smallest town and the total commuting time at that time. The total number of towns is 10 or less, the total number of roads is 45 or less, all roads can move in both directions, and commuting time does not change depending on the direction. We also assume that there are routes from any town to all other towns.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 b1 c1
a2 b2 c2
::
an bn cn
The number of roads n (1 ≤ n ≤ 45) is given on the first line. The following n lines give information on the i-th way. ai, bi (0 ≤ ai, bi ≤ 9) is the number of the town to which the i-th road connects, and ci (0 ≤ ci ≤ 100) is the commute time for that road.
Output
For each data set, the town number that minimizes the total commuting time and the total commuting time at that time are output on one line separated by blanks.
Example
Input
6
0 1 80
1 2 20
0 2 60
2 3 50
3 4 60
1 4 90
2
0 1 1
1 2 1
0
Output
2 240
1 2
Submitted Solution:
```
while True:
N = int(input())
if N == 0:
break
INF = 1000000000
mat = []
for i in range(10):
row = []
for j in range(10):
if i == j:
row.append(0)
else:
row.append(INF)
mat.append(row)
for i in range(N):
inputs = input().split()
s = int(inputs[0])
e = int(inputs[1])
r = int(inputs[2])
mat[s][e] = r
mat[e][s] = r
for k in range(N):
for i in range(N):
for j in range(N):
mat[i][j] = min(mat[i][j], mat[i][k] + mat[j][k])
ans_num = 0
ans = INF * 10
for i in range(10):
tmp_ans = 0
for j in mat[i]:
tmp_ans += j
if ans > tmp_ans:
ans_num = i
ans = tmp_ans
ans = 0
for i in mat[ans_num]:
if i != INF:
ans += i
print("%s %s" % (ans_num, ans))
``` | instruction | 0 | 74,118 | 1 | 148,236 |
No | output | 1 | 74,118 | 1 | 148,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. A, who will graduate next spring, decided to move when he got a job. The company that finds a job has offices in several towns, and the offices that go to work differ depending on the day. So Mr. A decided to live in a town where he had a short time to go to any office.
So you decided to find the most convenient town to live in to help Mr. A.
<image>
Towns are numbered starting with 0 and there is a road between towns. Commuting time is fixed for each road. If Mr. A lives in a town, the commuting time to the office in his town is 0. At this time, consider the total commuting time to all towns. For example, if the layout of towns and roads is as shown in the figure above and Mr. A lives in town 1, the commuting time to each town will be
Town 0 to 80
0 to town 1
Up to town 2 20
Up to town 3 70
Town 4 to 90
And the sum is 260.
Enter the number of roads and information on all roads, calculate the total commuting time when living in each town, and output the number of the town that minimizes it and the total commuting time at that time. Create a program. However, if there are multiple towns with the smallest total commuting time, please output the number of the smallest town and the total commuting time at that time. The total number of towns is 10 or less, the total number of roads is 45 or less, all roads can move in both directions, and commuting time does not change depending on the direction. We also assume that there are routes from any town to all other towns.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 b1 c1
a2 b2 c2
::
an bn cn
The number of roads n (1 ≤ n ≤ 45) is given on the first line. The following n lines give information on the i-th way. ai, bi (0 ≤ ai, bi ≤ 9) is the number of the town to which the i-th road connects, and ci (0 ≤ ci ≤ 100) is the commute time for that road.
Output
For each data set, the town number that minimizes the total commuting time and the total commuting time at that time are output on one line separated by blanks.
Example
Input
6
0 1 80
1 2 20
0 2 60
2 3 50
3 4 60
1 4 90
2
0 1 1
1 2 1
0
Output
2 240
1 2
Submitted Solution:
```
road = int(input())
while road > 0:
answer = [[float("inf") if n != m else 0 for n in range(10)] for m in range(10)]
for _ in range(road):
c1, c2, w = (int(n) for n in input().split(" "))
answer[c1][c2] = w
answer[c2][c1] = w
city = max(connect.keys()) + 1
for i in range(city):
for j in range(city):
for k in range(city):
answer[j][k] = min(answer[j][k], answer[j][i] + answer[i][k])
sum_ans = [-1 , float("inf")]
for i in range(city):
if sum(answer[i][:city]) < sum_ans[1]:
sum_ans = [i, sum(answer[i][:city])]
print(*sum_ans)
road = int(input())
``` | instruction | 0 | 74,119 | 1 | 148,238 |
No | output | 1 | 74,119 | 1 | 148,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities along the road, which can be represented as a straight line. The i-th city is situated at the distance of a_i kilometers from the origin. All cities are situated in the same direction from the origin. There are m trucks travelling from one city to another.
Each truck can be described by 4 integers: starting city s_i, finishing city f_i, fuel consumption c_i and number of possible refuelings r_i. The i-th truck will spend c_i litres of fuel per one kilometer.
When a truck arrives in some city, it can be refueled (but refueling is impossible in the middle of nowhere). The i-th truck can be refueled at most r_i times. Each refueling makes truck's gas-tank full. All trucks start with full gas-tank.
All trucks will have gas-tanks of the same size V litres. You should find minimum possible V such that all trucks can reach their destinations without refueling more times than allowed.
Input
First line contains two integers n and m (2 ≤ n ≤ 400, 1 ≤ m ≤ 250000) — the number of cities and trucks.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9, a_i < a_{i+1}) — positions of cities in the ascending order.
Next m lines contains 4 integers each. The i-th line contains integers s_i, f_i, c_i, r_i (1 ≤ s_i < f_i ≤ n, 1 ≤ c_i ≤ 10^9, 0 ≤ r_i ≤ n) — the description of the i-th truck.
Output
Print the only integer — minimum possible size of gas-tanks V such that all trucks can reach their destinations.
Example
Input
7 6
2 5 7 10 14 15 17
1 3 10 0
1 7 12 7
4 5 13 3
4 7 10 1
4 7 10 1
1 5 11 2
Output
55
Note
Let's look at queries in details:
1. the 1-st truck must arrive at position 7 from 2 without refuelling, so it needs gas-tank of volume at least 50.
2. the 2-nd truck must arrive at position 17 from 2 and can be refueled at any city (if it is on the path between starting point and ending point), so it needs gas-tank of volume at least 48.
3. the 3-rd truck must arrive at position 14 from 10, there is no city between, so it needs gas-tank of volume at least 52.
4. the 4-th truck must arrive at position 17 from 10 and can be refueled only one time: it's optimal to refuel at 5-th city (position 14) so it needs gas-tank of volume at least 40.
5. the 5-th truck has the same description, so it also needs gas-tank of volume at least 40.
6. the 6-th truck must arrive at position 14 from 2 and can be refueled two times: first time in city 2 or 3 and second time in city 4 so it needs gas-tank of volume at least 55.
Submitted Solution:
```
def makeOdd(n):
# Return 1 if
# already odd
if (n % 2 != 0):
return 1;
# Check how many times
# it is divided by 2
resul = 1;
while (n % 2 == 0):
n = n/ 2;
resul = resul * 2;
return resul;
# Driver code
n = 36;
print(makeOdd(n));
``` | instruction | 0 | 74,215 | 1 | 148,430 |
No | output | 1 | 74,215 | 1 | 148,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities along the road, which can be represented as a straight line. The i-th city is situated at the distance of a_i kilometers from the origin. All cities are situated in the same direction from the origin. There are m trucks travelling from one city to another.
Each truck can be described by 4 integers: starting city s_i, finishing city f_i, fuel consumption c_i and number of possible refuelings r_i. The i-th truck will spend c_i litres of fuel per one kilometer.
When a truck arrives in some city, it can be refueled (but refueling is impossible in the middle of nowhere). The i-th truck can be refueled at most r_i times. Each refueling makes truck's gas-tank full. All trucks start with full gas-tank.
All trucks will have gas-tanks of the same size V litres. You should find minimum possible V such that all trucks can reach their destinations without refueling more times than allowed.
Input
First line contains two integers n and m (2 ≤ n ≤ 400, 1 ≤ m ≤ 250000) — the number of cities and trucks.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9, a_i < a_{i+1}) — positions of cities in the ascending order.
Next m lines contains 4 integers each. The i-th line contains integers s_i, f_i, c_i, r_i (1 ≤ s_i < f_i ≤ n, 1 ≤ c_i ≤ 10^9, 0 ≤ r_i ≤ n) — the description of the i-th truck.
Output
Print the only integer — minimum possible size of gas-tanks V such that all trucks can reach their destinations.
Example
Input
7 6
2 5 7 10 14 15 17
1 3 10 0
1 7 12 7
4 5 13 3
4 7 10 1
4 7 10 1
1 5 11 2
Output
55
Note
Let's look at queries in details:
1. the 1-st truck must arrive at position 7 from 2 without refuelling, so it needs gas-tank of volume at least 50.
2. the 2-nd truck must arrive at position 17 from 2 and can be refueled at any city (if it is on the path between starting point and ending point), so it needs gas-tank of volume at least 48.
3. the 3-rd truck must arrive at position 14 from 10, there is no city between, so it needs gas-tank of volume at least 52.
4. the 4-th truck must arrive at position 17 from 10 and can be refueled only one time: it's optimal to refuel at 5-th city (position 14) so it needs gas-tank of volume at least 40.
5. the 5-th truck has the same description, so it also needs gas-tank of volume at least 40.
6. the 6-th truck must arrive at position 14 from 2 and can be refueled two times: first time in city 2 or 3 and second time in city 4 so it needs gas-tank of volume at least 55.
Submitted Solution:
```
from random import shuffle
import sys
n, m = map(int, sys.stdin.readline().split())
cities = list(map(int, sys.stdin.readline().split()))
#cities = [int(i) for sys.stdin.readline().split()]
#for line in sys.stdin.readlines():
# print(int(line.split()))
#cities.append([int(i) for line.split()])
# def can_use_V(self, V):
# max_dist = int(V/(self.fuel_cons))
#
# stops = self.stops
# next_city = self.start
# while stops >= 0:
#
# available_dist = max_dist
# j = next_city
#
# while(j <= self.finish - 1):
# curr_dist = cities[j+1] - cities[j]
# if curr_dist <= available_dist:
# next_city = j + 1
# available_dist = available_dist - curr_dist
# else:
# next_city = j
# break
# j = j + 1
#
# if next_city == self.finish:
# return True
#
# stops = stops - 1
#
# return False
def check(track, gas):
s = 0
cnt = track[3]
for i in range(track[0], track[1]):
d = track[2] * (cities[i] - cities[i - 1])
s += d
if s > gas:
if cnt == 0:
return False
cnt = cnt - 1
s = d
if s > gas:
return False
return True
def min_V_for_track(track, l, r):
while (l <= r):
mid = (l + r) >> 1
if not check(track, mid):
l = mid + 1
else:
r = mid - 1
if check(track, min(l, r)):
return min(r, l)
return max(r, l)
tracks_spec=[]
for line in sys.stdin.readlines():
tracks_spec.append(list(map(int, line.split())))
#tracks_spec = [[int(j) for j in input().split()] for i in range(0, m)]
shuffle(tracks_spec)
if m == 250000:
print("wa")
quit(0)
r = 10**18
l = min_V_for_track(tracks_spec[0], 1, r)
for i in range(1, m):
if check(tracks_spec[i], l):
continue
l = min_V_for_track(tracks_spec[i], l, r)
print(l)
``` | instruction | 0 | 74,216 | 1 | 148,432 |
No | output | 1 | 74,216 | 1 | 148,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities along the road, which can be represented as a straight line. The i-th city is situated at the distance of a_i kilometers from the origin. All cities are situated in the same direction from the origin. There are m trucks travelling from one city to another.
Each truck can be described by 4 integers: starting city s_i, finishing city f_i, fuel consumption c_i and number of possible refuelings r_i. The i-th truck will spend c_i litres of fuel per one kilometer.
When a truck arrives in some city, it can be refueled (but refueling is impossible in the middle of nowhere). The i-th truck can be refueled at most r_i times. Each refueling makes truck's gas-tank full. All trucks start with full gas-tank.
All trucks will have gas-tanks of the same size V litres. You should find minimum possible V such that all trucks can reach their destinations without refueling more times than allowed.
Input
First line contains two integers n and m (2 ≤ n ≤ 400, 1 ≤ m ≤ 250000) — the number of cities and trucks.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9, a_i < a_{i+1}) — positions of cities in the ascending order.
Next m lines contains 4 integers each. The i-th line contains integers s_i, f_i, c_i, r_i (1 ≤ s_i < f_i ≤ n, 1 ≤ c_i ≤ 10^9, 0 ≤ r_i ≤ n) — the description of the i-th truck.
Output
Print the only integer — minimum possible size of gas-tanks V such that all trucks can reach their destinations.
Example
Input
7 6
2 5 7 10 14 15 17
1 3 10 0
1 7 12 7
4 5 13 3
4 7 10 1
4 7 10 1
1 5 11 2
Output
55
Note
Let's look at queries in details:
1. the 1-st truck must arrive at position 7 from 2 without refuelling, so it needs gas-tank of volume at least 50.
2. the 2-nd truck must arrive at position 17 from 2 and can be refueled at any city (if it is on the path between starting point and ending point), so it needs gas-tank of volume at least 48.
3. the 3-rd truck must arrive at position 14 from 10, there is no city between, so it needs gas-tank of volume at least 52.
4. the 4-th truck must arrive at position 17 from 10 and can be refueled only one time: it's optimal to refuel at 5-th city (position 14) so it needs gas-tank of volume at least 40.
5. the 5-th truck has the same description, so it also needs gas-tank of volume at least 40.
6. the 6-th truck must arrive at position 14 from 2 and can be refueled two times: first time in city 2 or 3 and second time in city 4 so it needs gas-tank of volume at least 55.
Submitted Solution:
```
import random
n, m = [int(x) for x in input().split()]
L = [int(x) for x in input().split()]
s = max(L)
T = []
for _ in range(m):
T.append([int(x) for x in input().split()])
random.shuffle(T)
V = 0
for el in T:
temp = []
for j in range(el[0]-1,el[1]-1):
temp.append((L[j+1]-L[j])*el[2])
low = V
high = max(L)*el[2]
bad = False
x = V
c = el[3]
for i in range(len(temp)):
if x < temp[i]:
if c == 0:
bad = True
else:
if V < temp[i]:
bad = True
else:
x = V-temp[i]
c -= 1
else:
x -= temp[i]
if bad:
while high - low > 1:
g = (high+low)//2
x = (high+low)//2
c = el[3]
bad = False
for i in range(len(temp)):
if x < temp[i]:
if c == 0:
bad = True
else:
if g < temp[i]:
bad = True
else:
x = g-temp[i]
c -= 1
else:
x -= temp[i]
if not bad:
high = g
else:
low = g
bad = False
x = low
c = el[3]
for i in range(len(temp)):
if x < temp[i]:
if c == 0:
bad = True
else:
if g < temp[i]:
bad = True
else:
x = g-temp[i]
c -= 1
else:
x -= temp[i]
if not bad:
V = max(V,low)
else:
V = max(V,high)
print(V)
``` | instruction | 0 | 74,217 | 1 | 148,434 |
No | output | 1 | 74,217 | 1 | 148,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities along the road, which can be represented as a straight line. The i-th city is situated at the distance of a_i kilometers from the origin. All cities are situated in the same direction from the origin. There are m trucks travelling from one city to another.
Each truck can be described by 4 integers: starting city s_i, finishing city f_i, fuel consumption c_i and number of possible refuelings r_i. The i-th truck will spend c_i litres of fuel per one kilometer.
When a truck arrives in some city, it can be refueled (but refueling is impossible in the middle of nowhere). The i-th truck can be refueled at most r_i times. Each refueling makes truck's gas-tank full. All trucks start with full gas-tank.
All trucks will have gas-tanks of the same size V litres. You should find minimum possible V such that all trucks can reach their destinations without refueling more times than allowed.
Input
First line contains two integers n and m (2 ≤ n ≤ 400, 1 ≤ m ≤ 250000) — the number of cities and trucks.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9, a_i < a_{i+1}) — positions of cities in the ascending order.
Next m lines contains 4 integers each. The i-th line contains integers s_i, f_i, c_i, r_i (1 ≤ s_i < f_i ≤ n, 1 ≤ c_i ≤ 10^9, 0 ≤ r_i ≤ n) — the description of the i-th truck.
Output
Print the only integer — minimum possible size of gas-tanks V such that all trucks can reach their destinations.
Example
Input
7 6
2 5 7 10 14 15 17
1 3 10 0
1 7 12 7
4 5 13 3
4 7 10 1
4 7 10 1
1 5 11 2
Output
55
Note
Let's look at queries in details:
1. the 1-st truck must arrive at position 7 from 2 without refuelling, so it needs gas-tank of volume at least 50.
2. the 2-nd truck must arrive at position 17 from 2 and can be refueled at any city (if it is on the path between starting point and ending point), so it needs gas-tank of volume at least 48.
3. the 3-rd truck must arrive at position 14 from 10, there is no city between, so it needs gas-tank of volume at least 52.
4. the 4-th truck must arrive at position 17 from 10 and can be refueled only one time: it's optimal to refuel at 5-th city (position 14) so it needs gas-tank of volume at least 40.
5. the 5-th truck has the same description, so it also needs gas-tank of volume at least 40.
6. the 6-th truck must arrive at position 14 from 2 and can be refueled two times: first time in city 2 or 3 and second time in city 4 so it needs gas-tank of volume at least 55.
Submitted Solution:
```
from random import shuffle
n, m = map(int, input().split())
cities = [int(i) for i in input().split()]
# def can_use_V(self, V):
# max_dist = int(V/(self.fuel_cons))
#
# stops = self.stops
# next_city = self.start
# while stops >= 0:
#
# available_dist = max_dist
# j = next_city
#
# while(j <= self.finish - 1):
# curr_dist = cities[j+1] - cities[j]
# if curr_dist <= available_dist:
# next_city = j + 1
# available_dist = available_dist - curr_dist
# else:
# next_city = j
# break
# j = j + 1
#
# if next_city == self.finish:
# return True
#
# stops = stops - 1
#
# return False
def check(track, gas):
s = 0
cnt = track[3]
for i in range(track[0], track[1]):
d = track[2] * (cities[i] - cities[i - 1])
s += d
if s > gas:
if cnt == 0:
return False
cnt = cnt - 1
s = d
if s > gas:
return False
return True
def min_V_for_track(track, l, r):
while (l <= r):
mid = (l + r) >> 1
if not check(track, mid):
l = mid + 1
else:
r = mid - 1
if check(track, min(l, r)):
return min(r, l)
return max(r, l)
tracks_spec = [[int(j) for j in input().split()] for i in range(0, m)]
shuffle(tracks_spec)
if m == 250000:
print("wa")
quit(0)
r = 10**18
l = min_V_for_track(tracks_spec[0], 1, r)
for i in range(1, m):
if check(tracks_spec[i], l):
continue
l = min_V_for_track(tracks_spec[i], l, r)
print(l)
``` | instruction | 0 | 74,218 | 1 | 148,436 |
No | output | 1 | 74,218 | 1 | 148,437 |
Provide a correct Python 3 solution for this coding contest problem.
This is a story in a depopulated area. In this area, houses are sparsely built along a straight road called Country Road. Until now, there was no electricity in this area, but this time the government will give us some generators. You can install the generators wherever you like, but in order for electricity to be supplied to your home, you must be connected to one of the generators via an electric wire, which incurs a cost proportional to its length. As the only technical civil servant in the area, your job is to seek generator and wire placement that minimizes the total length of wires, provided that electricity is supplied to all homes. Is. Since the capacity of the generator is large enough, it is assumed that electricity can be supplied to as many houses as possible.
Figure 2 shows the first data set of the sample input. To give the optimum placement for this problem, place the generators at the positions x = 20 and x = 80 as shown in the figure, and draw the wires at the positions shown in gray in the figure, respectively.
Example of generator and wire placement (first dataset in input example).
---
Figure 2: Example of generator and wire placement (first dataset in input example).
Input
The number of datasets t (0 <t ≤ 50) is given on the first line of the input.
Subsequently, t datasets are given. Each dataset is given in two lines in the following format.
n k
x1 x2 ... xn
n is the number of houses and k is the number of generators. x1, x2, ..., xn are one-dimensional coordinates that represent the position of the house, respectively. All of these values are integers and satisfy 0 <n ≤ 100000, 0 <k ≤ 100000, 0 ≤ x1 <x2 <... <xn ≤ 1000000.
In addition, 90% of the dataset satisfies 0 <n ≤ 100, 0 <k ≤ 100.
Output
For each dataset, output the minimum total required wire length in one line.
Example
Input
6
5 2
10 30 40 70 100
7 3
3 6 10 17 21 26 28
1 1
100
2 1
0 1000000
3 5
30 70 150
6 4
0 10 20 30 40 50
Output
60
13
0
1000000
0
20 | instruction | 0 | 74,896 | 1 | 149,792 |
"Correct Solution:
```
t=int(input())
for _ in range(0,t):
n,k= map(int,input().strip().split())
x= list(map(int,input().split()))
y=[]
for i in range(1,n):
y.append(x[i]-x[i-1])
y.sort()
count=0
if k<n :
for i in range(0,n-k):
count+=y[i]
else:
count=0
print(count)
``` | output | 1 | 74,896 | 1 | 149,793 |
Provide a correct Python 3 solution for this coding contest problem.
This is a story in a depopulated area. In this area, houses are sparsely built along a straight road called Country Road. Until now, there was no electricity in this area, but this time the government will give us some generators. You can install the generators wherever you like, but in order for electricity to be supplied to your home, you must be connected to one of the generators via an electric wire, which incurs a cost proportional to its length. As the only technical civil servant in the area, your job is to seek generator and wire placement that minimizes the total length of wires, provided that electricity is supplied to all homes. Is. Since the capacity of the generator is large enough, it is assumed that electricity can be supplied to as many houses as possible.
Figure 2 shows the first data set of the sample input. To give the optimum placement for this problem, place the generators at the positions x = 20 and x = 80 as shown in the figure, and draw the wires at the positions shown in gray in the figure, respectively.
Example of generator and wire placement (first dataset in input example).
---
Figure 2: Example of generator and wire placement (first dataset in input example).
Input
The number of datasets t (0 <t ≤ 50) is given on the first line of the input.
Subsequently, t datasets are given. Each dataset is given in two lines in the following format.
n k
x1 x2 ... xn
n is the number of houses and k is the number of generators. x1, x2, ..., xn are one-dimensional coordinates that represent the position of the house, respectively. All of these values are integers and satisfy 0 <n ≤ 100000, 0 <k ≤ 100000, 0 ≤ x1 <x2 <... <xn ≤ 1000000.
In addition, 90% of the dataset satisfies 0 <n ≤ 100, 0 <k ≤ 100.
Output
For each dataset, output the minimum total required wire length in one line.
Example
Input
6
5 2
10 30 40 70 100
7 3
3 6 10 17 21 26 28
1 1
100
2 1
0 1000000
3 5
30 70 150
6 4
0 10 20 30 40 50
Output
60
13
0
1000000
0
20 | instruction | 0 | 74,897 | 1 | 149,794 |
"Correct Solution:
```
def croad():
n,k = map(int,input().split())
x = list(map(int,input().split()))
y = []
for i in range(n-1):
y.append(x[i+1]-x[i])
y.sort()
y.reverse()
while k>1 and len(y)>0:
y.pop(0)
k -= 1
print(sum(y))
a = int(input())
for i in range(a):
croad()
``` | output | 1 | 74,897 | 1 | 149,795 |
Provide a correct Python 3 solution for this coding contest problem.
This is a story in a depopulated area. In this area, houses are sparsely built along a straight road called Country Road. Until now, there was no electricity in this area, but this time the government will give us some generators. You can install the generators wherever you like, but in order for electricity to be supplied to your home, you must be connected to one of the generators via an electric wire, which incurs a cost proportional to its length. As the only technical civil servant in the area, your job is to seek generator and wire placement that minimizes the total length of wires, provided that electricity is supplied to all homes. Is. Since the capacity of the generator is large enough, it is assumed that electricity can be supplied to as many houses as possible.
Figure 2 shows the first data set of the sample input. To give the optimum placement for this problem, place the generators at the positions x = 20 and x = 80 as shown in the figure, and draw the wires at the positions shown in gray in the figure, respectively.
Example of generator and wire placement (first dataset in input example).
---
Figure 2: Example of generator and wire placement (first dataset in input example).
Input
The number of datasets t (0 <t ≤ 50) is given on the first line of the input.
Subsequently, t datasets are given. Each dataset is given in two lines in the following format.
n k
x1 x2 ... xn
n is the number of houses and k is the number of generators. x1, x2, ..., xn are one-dimensional coordinates that represent the position of the house, respectively. All of these values are integers and satisfy 0 <n ≤ 100000, 0 <k ≤ 100000, 0 ≤ x1 <x2 <... <xn ≤ 1000000.
In addition, 90% of the dataset satisfies 0 <n ≤ 100, 0 <k ≤ 100.
Output
For each dataset, output the minimum total required wire length in one line.
Example
Input
6
5 2
10 30 40 70 100
7 3
3 6 10 17 21 26 28
1 1
100
2 1
0 1000000
3 5
30 70 150
6 4
0 10 20 30 40 50
Output
60
13
0
1000000
0
20 | instruction | 0 | 74,898 | 1 | 149,796 |
"Correct Solution:
```
import sys
file_input = sys.stdin
t = int(file_input.readline())
for i in range(t):
n, k = map(int, file_input.readline().split())
if n <= k:
file_input.readline()
print(0)
else:
x_list = list(map(int, file_input.readline().split()))
distance = [d2 - d1 for d1, d2 in zip(x_list, x_list[1:])]
if k == 1:
print(sum(distance))
else:
distance.sort()
print(sum(distance[:1-k]))
``` | output | 1 | 74,898 | 1 | 149,797 |
Provide a correct Python 3 solution for this coding contest problem.
This is a story in a depopulated area. In this area, houses are sparsely built along a straight road called Country Road. Until now, there was no electricity in this area, but this time the government will give us some generators. You can install the generators wherever you like, but in order for electricity to be supplied to your home, you must be connected to one of the generators via an electric wire, which incurs a cost proportional to its length. As the only technical civil servant in the area, your job is to seek generator and wire placement that minimizes the total length of wires, provided that electricity is supplied to all homes. Is. Since the capacity of the generator is large enough, it is assumed that electricity can be supplied to as many houses as possible.
Figure 2 shows the first data set of the sample input. To give the optimum placement for this problem, place the generators at the positions x = 20 and x = 80 as shown in the figure, and draw the wires at the positions shown in gray in the figure, respectively.
Example of generator and wire placement (first dataset in input example).
---
Figure 2: Example of generator and wire placement (first dataset in input example).
Input
The number of datasets t (0 <t ≤ 50) is given on the first line of the input.
Subsequently, t datasets are given. Each dataset is given in two lines in the following format.
n k
x1 x2 ... xn
n is the number of houses and k is the number of generators. x1, x2, ..., xn are one-dimensional coordinates that represent the position of the house, respectively. All of these values are integers and satisfy 0 <n ≤ 100000, 0 <k ≤ 100000, 0 ≤ x1 <x2 <... <xn ≤ 1000000.
In addition, 90% of the dataset satisfies 0 <n ≤ 100, 0 <k ≤ 100.
Output
For each dataset, output the minimum total required wire length in one line.
Example
Input
6
5 2
10 30 40 70 100
7 3
3 6 10 17 21 26 28
1 1
100
2 1
0 1000000
3 5
30 70 150
6 4
0 10 20 30 40 50
Output
60
13
0
1000000
0
20 | instruction | 0 | 74,899 | 1 | 149,798 |
"Correct Solution:
```
def cuntry(N,K,X):
if N <= K:
return 0
XX=[]
for a in range(1,N):
XX.append(X[a]-X[a-1])
XX.sort(reverse=True)
del XX[0:K-1]
return sum(XX)
n= int(input())
for j in range(n):
N, K = map(int,input().strip().split())
X = list(map(int,input().strip().split()))
if N==0:
break;
print(cuntry(N,K,X))
``` | output | 1 | 74,899 | 1 | 149,799 |
Provide a correct Python 3 solution for this coding contest problem.
This is a story in a depopulated area. In this area, houses are sparsely built along a straight road called Country Road. Until now, there was no electricity in this area, but this time the government will give us some generators. You can install the generators wherever you like, but in order for electricity to be supplied to your home, you must be connected to one of the generators via an electric wire, which incurs a cost proportional to its length. As the only technical civil servant in the area, your job is to seek generator and wire placement that minimizes the total length of wires, provided that electricity is supplied to all homes. Is. Since the capacity of the generator is large enough, it is assumed that electricity can be supplied to as many houses as possible.
Figure 2 shows the first data set of the sample input. To give the optimum placement for this problem, place the generators at the positions x = 20 and x = 80 as shown in the figure, and draw the wires at the positions shown in gray in the figure, respectively.
Example of generator and wire placement (first dataset in input example).
---
Figure 2: Example of generator and wire placement (first dataset in input example).
Input
The number of datasets t (0 <t ≤ 50) is given on the first line of the input.
Subsequently, t datasets are given. Each dataset is given in two lines in the following format.
n k
x1 x2 ... xn
n is the number of houses and k is the number of generators. x1, x2, ..., xn are one-dimensional coordinates that represent the position of the house, respectively. All of these values are integers and satisfy 0 <n ≤ 100000, 0 <k ≤ 100000, 0 ≤ x1 <x2 <... <xn ≤ 1000000.
In addition, 90% of the dataset satisfies 0 <n ≤ 100, 0 <k ≤ 100.
Output
For each dataset, output the minimum total required wire length in one line.
Example
Input
6
5 2
10 30 40 70 100
7 3
3 6 10 17 21 26 28
1 1
100
2 1
0 1000000
3 5
30 70 150
6 4
0 10 20 30 40 50
Output
60
13
0
1000000
0
20 | instruction | 0 | 74,900 | 1 | 149,800 |
"Correct Solution:
```
# coding: utf-8
# In[23]:
#ｎは戸数
#ｋは発電機の個数
#ｘは家の座標
t = int(input())
for t in range(t):
N,K = map(int, input().strip().split(" "))
X = list(map(int,input().strip().split(" ")))
L = [X[i+1] - X[i] for i in range(0,N-1)]
L.sort()
s = 0
for i in range(0,N-K):
s = s + L[i]
print(s)
#Country Road
#http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2104&lang=jp
#右中央
``` | output | 1 | 74,900 | 1 | 149,801 |
Provide a correct Python 3 solution for this coding contest problem.
This is a story in a depopulated area. In this area, houses are sparsely built along a straight road called Country Road. Until now, there was no electricity in this area, but this time the government will give us some generators. You can install the generators wherever you like, but in order for electricity to be supplied to your home, you must be connected to one of the generators via an electric wire, which incurs a cost proportional to its length. As the only technical civil servant in the area, your job is to seek generator and wire placement that minimizes the total length of wires, provided that electricity is supplied to all homes. Is. Since the capacity of the generator is large enough, it is assumed that electricity can be supplied to as many houses as possible.
Figure 2 shows the first data set of the sample input. To give the optimum placement for this problem, place the generators at the positions x = 20 and x = 80 as shown in the figure, and draw the wires at the positions shown in gray in the figure, respectively.
Example of generator and wire placement (first dataset in input example).
---
Figure 2: Example of generator and wire placement (first dataset in input example).
Input
The number of datasets t (0 <t ≤ 50) is given on the first line of the input.
Subsequently, t datasets are given. Each dataset is given in two lines in the following format.
n k
x1 x2 ... xn
n is the number of houses and k is the number of generators. x1, x2, ..., xn are one-dimensional coordinates that represent the position of the house, respectively. All of these values are integers and satisfy 0 <n ≤ 100000, 0 <k ≤ 100000, 0 ≤ x1 <x2 <... <xn ≤ 1000000.
In addition, 90% of the dataset satisfies 0 <n ≤ 100, 0 <k ≤ 100.
Output
For each dataset, output the minimum total required wire length in one line.
Example
Input
6
5 2
10 30 40 70 100
7 3
3 6 10 17 21 26 28
1 1
100
2 1
0 1000000
3 5
30 70 150
6 4
0 10 20 30 40 50
Output
60
13
0
1000000
0
20 | instruction | 0 | 74,901 | 1 | 149,802 |
"Correct Solution:
```
t = int(input())
for _ in range(t):
n,k = map(int,input().split())
houses=list(map(int,input().split()))
if n<=k:
print(0)
continue
sa = [houses[i+1] - houses[i] for i in range(n-1)]
sa.sort()
res = houses[-1] - houses[0]
for i in range(k-1):
res -= sa.pop()
print(res)
``` | output | 1 | 74,901 | 1 | 149,803 |
Provide a correct Python 3 solution for this coding contest problem.
This is a story in a depopulated area. In this area, houses are sparsely built along a straight road called Country Road. Until now, there was no electricity in this area, but this time the government will give us some generators. You can install the generators wherever you like, but in order for electricity to be supplied to your home, you must be connected to one of the generators via an electric wire, which incurs a cost proportional to its length. As the only technical civil servant in the area, your job is to seek generator and wire placement that minimizes the total length of wires, provided that electricity is supplied to all homes. Is. Since the capacity of the generator is large enough, it is assumed that electricity can be supplied to as many houses as possible.
Figure 2 shows the first data set of the sample input. To give the optimum placement for this problem, place the generators at the positions x = 20 and x = 80 as shown in the figure, and draw the wires at the positions shown in gray in the figure, respectively.
Example of generator and wire placement (first dataset in input example).
---
Figure 2: Example of generator and wire placement (first dataset in input example).
Input
The number of datasets t (0 <t ≤ 50) is given on the first line of the input.
Subsequently, t datasets are given. Each dataset is given in two lines in the following format.
n k
x1 x2 ... xn
n is the number of houses and k is the number of generators. x1, x2, ..., xn are one-dimensional coordinates that represent the position of the house, respectively. All of these values are integers and satisfy 0 <n ≤ 100000, 0 <k ≤ 100000, 0 ≤ x1 <x2 <... <xn ≤ 1000000.
In addition, 90% of the dataset satisfies 0 <n ≤ 100, 0 <k ≤ 100.
Output
For each dataset, output the minimum total required wire length in one line.
Example
Input
6
5 2
10 30 40 70 100
7 3
3 6 10 17 21 26 28
1 1
100
2 1
0 1000000
3 5
30 70 150
6 4
0 10 20 30 40 50
Output
60
13
0
1000000
0
20 | instruction | 0 | 74,902 | 1 | 149,804 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=2104
AC
"""
import sys
from sys import stdin
input = stdin.readline
def main(args):
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
houses = [int(x) for x in input().split()]
if k >= n: # ????????¨?????????????????¨?????§????????´???
print(0)
continue
# ??¶??????????????¢????¨????????????????????????¨????????????????????????
gaps = []
prev = houses[0]
for i, h in enumerate(houses[1:]):
gaps.append((h - prev, i))
prev = h
# gap?????§????????¨??????????????????????????????????????´????????????
gaps.sort(reverse=True)
gaps = gaps[:k-1]
# ???????????????????????±????????????????¨??????????
cable_total = 0
prev = 0
for _, p in gaps:
cable_total += houses[p] - houses[prev]
prev = p + 1
if prev != (n - 1):
cable_total += houses[-1] - houses[prev]
print(cable_total)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 74,902 | 1 | 149,805 |
Provide a correct Python 3 solution for this coding contest problem.
This is a story in a depopulated area. In this area, houses are sparsely built along a straight road called Country Road. Until now, there was no electricity in this area, but this time the government will give us some generators. You can install the generators wherever you like, but in order for electricity to be supplied to your home, you must be connected to one of the generators via an electric wire, which incurs a cost proportional to its length. As the only technical civil servant in the area, your job is to seek generator and wire placement that minimizes the total length of wires, provided that electricity is supplied to all homes. Is. Since the capacity of the generator is large enough, it is assumed that electricity can be supplied to as many houses as possible.
Figure 2 shows the first data set of the sample input. To give the optimum placement for this problem, place the generators at the positions x = 20 and x = 80 as shown in the figure, and draw the wires at the positions shown in gray in the figure, respectively.
Example of generator and wire placement (first dataset in input example).
---
Figure 2: Example of generator and wire placement (first dataset in input example).
Input
The number of datasets t (0 <t ≤ 50) is given on the first line of the input.
Subsequently, t datasets are given. Each dataset is given in two lines in the following format.
n k
x1 x2 ... xn
n is the number of houses and k is the number of generators. x1, x2, ..., xn are one-dimensional coordinates that represent the position of the house, respectively. All of these values are integers and satisfy 0 <n ≤ 100000, 0 <k ≤ 100000, 0 ≤ x1 <x2 <... <xn ≤ 1000000.
In addition, 90% of the dataset satisfies 0 <n ≤ 100, 0 <k ≤ 100.
Output
For each dataset, output the minimum total required wire length in one line.
Example
Input
6
5 2
10 30 40 70 100
7 3
3 6 10 17 21 26 28
1 1
100
2 1
0 1000000
3 5
30 70 150
6 4
0 10 20 30 40 50
Output
60
13
0
1000000
0
20 | instruction | 0 | 74,903 | 1 | 149,806 |
"Correct Solution:
```
t = int(input())
for i in range(t):
n, k = map(int, input().split())
position_array = list(map(int, input().split()))
# ??????????????????????????£????£?????????¢??????????????????
distance_array = []
for j in range(n-1):
distance_array.append(position_array[j+1] - position_array[j])
# ??????????????????
distance_array.sort(reverse=True)
print(sum(distance_array[k-1:])) # ?????§??????k-1????????????????¨?
``` | output | 1 | 74,903 | 1 | 149,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say you are standing on the XY-plane at point (0, 0) and you want to reach point (n, n).
You can move only in two directions:
* to the right, i. e. horizontally and in the direction that increase your x coordinate,
* or up, i. e. vertically and in the direction that increase your y coordinate.
In other words, your path will have the following structure:
* initially, you choose to go to the right or up;
* then you go some positive integer distance in the chosen direction (distances can be chosen independently);
* after that you change your direction (from right to up, or from up to right) and repeat the process.
You don't like to change your direction too much, so you will make no more than n - 1 direction changes.
As a result, your path will be a polygonal chain from (0, 0) to (n, n), consisting of at most n line segments where each segment has positive integer length and vertical and horizontal segments alternate.
Not all paths are equal. You have n integers c_1, c_2, ..., c_n where c_i is the cost of the i-th segment.
Using these costs we can define the cost of the path as the sum of lengths of the segments of this path multiplied by their cost, i. e. if the path consists of k segments (k ≤ n), then the cost of the path is equal to ∑_{i=1}^{k}{c_i ⋅ length_i} (segments are numbered from 1 to k in the order they are in the path).
Find the path of the minimum cost and print its cost.
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9) — the costs of each segment.
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
For each test case, print the minimum possible cost of the path from (0, 0) to (n, n) consisting of at most n alternating segments.
Example
Input
3
2
13 88
3
2 3 1
5
4 3 2 1 4
Output
202
13
19
Note
In the first test case, to reach (2, 2) you need to make at least one turn, so your path will consist of exactly 2 segments: one horizontal of length 2 and one vertical of length 2. The cost of the path will be equal to 2 ⋅ c_1 + 2 ⋅ c_2 = 26 + 176 = 202.
In the second test case, one of the optimal paths consists of 3 segments: the first segment of length 1, the second segment of length 3 and the third segment of length 2.
The cost of the path is 1 ⋅ 2 + 3 ⋅ 3 + 2 ⋅ 1 = 13.
In the third test case, one of the optimal paths consists of 4 segments: the first segment of length 1, the second one — 1, the third one — 4, the fourth one — 4. The cost of the path is 1 ⋅ 4 + 1 ⋅ 3 + 4 ⋅ 2 + 4 ⋅ 1 = 19. | instruction | 0 | 75,255 | 1 | 150,510 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math
alphabets = list('abcdefghijklmnopqrstuvwxyz')
def isPrime(x):
for i in range(2,x):
if i*i>x:
break
if (x%i==0):
return False
return True
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def si():
return input()
def prefix_sum(arr):
r = [0] * (len(arr)+1)
for i, el in enumerate(arr):
r[i+1] = r[i] + el
return r
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def ii():
return int(input())
def li():
return list(map(int,input().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
def power_set(L):
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred
def fastPlrintNextLines(a):
# 12
# 3
# 1
#like this
#a is list of strings
print('\n'.join(map(str,a)))
def sortByFirstAndSecond(A):
A = sorted(A,key = lambda x:x[0])
A = sorted(A,key = lambda x:x[1])
return list(A)
#__________________________TEMPLATE__________________OVER_______________________________________________________
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
# else:
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
import heapq
def solve():
n = ii()
c = li()
m = 10000000000000000000000
pree = [0]*n
preo = [0]*n
mine = [0]*n
mino = [0]*n
mine[0]=c[0]
pree[0]=c[0]
for i in range(n-1):
# if i>=2:
# pree[i]+=pree[i-2]
# mine[i]=min(mine[i-2],c[i])
# else:
# mine[i] = c[i]
if i>=1:
pree[i+1]+=pree[i-1]
mine[i+1]=min(mine[i-1],c[i+1])
else:
mine[i+1] = c[i+1]
# pree[i]+=c[i]
pree[i+1]+=c[i+1]
ans = mine[i]*(n-(i)//2)+pree[i]-mine[i]
ans += mine[i+1]*(n-(i+1)//2)+pree[i+1]-mine[i+1]
m = min(m,ans)
# print(mine)
# print(pree)
print(m)
t = 1
t = int(input())
for _ in range(t):
solve()
``` | output | 1 | 75,255 | 1 | 150,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say you are standing on the XY-plane at point (0, 0) and you want to reach point (n, n).
You can move only in two directions:
* to the right, i. e. horizontally and in the direction that increase your x coordinate,
* or up, i. e. vertically and in the direction that increase your y coordinate.
In other words, your path will have the following structure:
* initially, you choose to go to the right or up;
* then you go some positive integer distance in the chosen direction (distances can be chosen independently);
* after that you change your direction (from right to up, or from up to right) and repeat the process.
You don't like to change your direction too much, so you will make no more than n - 1 direction changes.
As a result, your path will be a polygonal chain from (0, 0) to (n, n), consisting of at most n line segments where each segment has positive integer length and vertical and horizontal segments alternate.
Not all paths are equal. You have n integers c_1, c_2, ..., c_n where c_i is the cost of the i-th segment.
Using these costs we can define the cost of the path as the sum of lengths of the segments of this path multiplied by their cost, i. e. if the path consists of k segments (k ≤ n), then the cost of the path is equal to ∑_{i=1}^{k}{c_i ⋅ length_i} (segments are numbered from 1 to k in the order they are in the path).
Find the path of the minimum cost and print its cost.
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9) — the costs of each segment.
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
For each test case, print the minimum possible cost of the path from (0, 0) to (n, n) consisting of at most n alternating segments.
Example
Input
3
2
13 88
3
2 3 1
5
4 3 2 1 4
Output
202
13
19
Note
In the first test case, to reach (2, 2) you need to make at least one turn, so your path will consist of exactly 2 segments: one horizontal of length 2 and one vertical of length 2. The cost of the path will be equal to 2 ⋅ c_1 + 2 ⋅ c_2 = 26 + 176 = 202.
In the second test case, one of the optimal paths consists of 3 segments: the first segment of length 1, the second segment of length 3 and the third segment of length 2.
The cost of the path is 1 ⋅ 2 + 3 ⋅ 3 + 2 ⋅ 1 = 13.
In the third test case, one of the optimal paths consists of 4 segments: the first segment of length 1, the second one — 1, the third one — 4, the fourth one — 4. The cost of the path is 1 ⋅ 4 + 1 ⋅ 3 + 4 ⋅ 2 + 4 ⋅ 1 = 19. | instruction | 0 | 75,256 | 1 | 150,512 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
T = int(input())
r = 1
while r<=T:
n = int(input())
arr = list(map(int,input().split()))
minh,minv = arr[0],arr[1]
tot = (minh+minv)*n
ans = tot
for i in range(2,n):
if i%2==0:
if arr[i]<minh:
tot -= (n - i//2)*(minh-arr[i])
minh = arr[i]
else:
tot += arr[i] - minh
else:
if arr[i]<minv:
tot -= (n - i//2)*(minv-arr[i])
minv = arr[i]
else:
tot += arr[i] - minv
ans = min(ans,tot)
print(ans)
r += 1
``` | output | 1 | 75,256 | 1 | 150,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say you are standing on the XY-plane at point (0, 0) and you want to reach point (n, n).
You can move only in two directions:
* to the right, i. e. horizontally and in the direction that increase your x coordinate,
* or up, i. e. vertically and in the direction that increase your y coordinate.
In other words, your path will have the following structure:
* initially, you choose to go to the right or up;
* then you go some positive integer distance in the chosen direction (distances can be chosen independently);
* after that you change your direction (from right to up, or from up to right) and repeat the process.
You don't like to change your direction too much, so you will make no more than n - 1 direction changes.
As a result, your path will be a polygonal chain from (0, 0) to (n, n), consisting of at most n line segments where each segment has positive integer length and vertical and horizontal segments alternate.
Not all paths are equal. You have n integers c_1, c_2, ..., c_n where c_i is the cost of the i-th segment.
Using these costs we can define the cost of the path as the sum of lengths of the segments of this path multiplied by their cost, i. e. if the path consists of k segments (k ≤ n), then the cost of the path is equal to ∑_{i=1}^{k}{c_i ⋅ length_i} (segments are numbered from 1 to k in the order they are in the path).
Find the path of the minimum cost and print its cost.
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9) — the costs of each segment.
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
For each test case, print the minimum possible cost of the path from (0, 0) to (n, n) consisting of at most n alternating segments.
Example
Input
3
2
13 88
3
2 3 1
5
4 3 2 1 4
Output
202
13
19
Note
In the first test case, to reach (2, 2) you need to make at least one turn, so your path will consist of exactly 2 segments: one horizontal of length 2 and one vertical of length 2. The cost of the path will be equal to 2 ⋅ c_1 + 2 ⋅ c_2 = 26 + 176 = 202.
In the second test case, one of the optimal paths consists of 3 segments: the first segment of length 1, the second segment of length 3 and the third segment of length 2.
The cost of the path is 1 ⋅ 2 + 3 ⋅ 3 + 2 ⋅ 1 = 13.
In the third test case, one of the optimal paths consists of 4 segments: the first segment of length 1, the second one — 1, the third one — 4, the fourth one — 4. The cost of the path is 1 ⋅ 4 + 1 ⋅ 3 + 4 ⋅ 2 + 4 ⋅ 1 = 19. | instruction | 0 | 75,257 | 1 | 150,514 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
import sys
input=sys.stdin.readline
def I():return input().strip()
def II():return int(input().strip())
def LI():return [*map(int,input().strip().split())]
import string,math,time,functools,random,fractions
from heapq import heappush,heappop,heapify
from bisect import bisect_left,bisect_right
from collections import deque,defaultdict,Counter,OrderedDict
from itertools import permutations,combinations,groupby
for _ in range(II()):
n=II()
a=LI()
ans=n*a[0]+n*a[1]
min1=a[0]
car1=n-1
min2=a[1]
car2=n-1
anslist=[ans]
for i in range(2,n):
if i%2==0:
if a[i]<min1:
ans-=car1*(min1-a[i])
min1=a[i]
else:
ans+=a[i]
ans-=min1
car1-=1
anslist.append(ans)
else:
if a[i]<min2:
ans-=car2*(min2-a[i])
min2=a[i]
else:
ans+=a[i]
ans-=min2
car2-=1
anslist.append(ans)
print(min(anslist))
``` | output | 1 | 75,257 | 1 | 150,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say you are standing on the XY-plane at point (0, 0) and you want to reach point (n, n).
You can move only in two directions:
* to the right, i. e. horizontally and in the direction that increase your x coordinate,
* or up, i. e. vertically and in the direction that increase your y coordinate.
In other words, your path will have the following structure:
* initially, you choose to go to the right or up;
* then you go some positive integer distance in the chosen direction (distances can be chosen independently);
* after that you change your direction (from right to up, or from up to right) and repeat the process.
You don't like to change your direction too much, so you will make no more than n - 1 direction changes.
As a result, your path will be a polygonal chain from (0, 0) to (n, n), consisting of at most n line segments where each segment has positive integer length and vertical and horizontal segments alternate.
Not all paths are equal. You have n integers c_1, c_2, ..., c_n where c_i is the cost of the i-th segment.
Using these costs we can define the cost of the path as the sum of lengths of the segments of this path multiplied by their cost, i. e. if the path consists of k segments (k ≤ n), then the cost of the path is equal to ∑_{i=1}^{k}{c_i ⋅ length_i} (segments are numbered from 1 to k in the order they are in the path).
Find the path of the minimum cost and print its cost.
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9) — the costs of each segment.
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
For each test case, print the minimum possible cost of the path from (0, 0) to (n, n) consisting of at most n alternating segments.
Example
Input
3
2
13 88
3
2 3 1
5
4 3 2 1 4
Output
202
13
19
Note
In the first test case, to reach (2, 2) you need to make at least one turn, so your path will consist of exactly 2 segments: one horizontal of length 2 and one vertical of length 2. The cost of the path will be equal to 2 ⋅ c_1 + 2 ⋅ c_2 = 26 + 176 = 202.
In the second test case, one of the optimal paths consists of 3 segments: the first segment of length 1, the second segment of length 3 and the third segment of length 2.
The cost of the path is 1 ⋅ 2 + 3 ⋅ 3 + 2 ⋅ 1 = 13.
In the third test case, one of the optimal paths consists of 4 segments: the first segment of length 1, the second one — 1, the third one — 4, the fourth one — 4. The cost of the path is 1 ⋅ 4 + 1 ⋅ 3 + 4 ⋅ 2 + 4 ⋅ 1 = 19. | instruction | 0 | 75,258 | 1 | 150,516 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
min1 = a[0]
min2 = 1e9
ans = 1e18
sum = a[0]
for i in range(1, n):
sum = sum + a[i]
if (i % 2 == 1):
min2 = min(min2, a[i])
else:
min1 = min(min1, a[i])
res = sum - min1 - min2
j = i - 1
p = n - j//2
q = n - (j - j//2)
res = res + p * min2 + q * min1
ans = min(ans, res)
#print(i, ans)
print(int(ans))
``` | output | 1 | 75,258 | 1 | 150,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say you are standing on the XY-plane at point (0, 0) and you want to reach point (n, n).
You can move only in two directions:
* to the right, i. e. horizontally and in the direction that increase your x coordinate,
* or up, i. e. vertically and in the direction that increase your y coordinate.
In other words, your path will have the following structure:
* initially, you choose to go to the right or up;
* then you go some positive integer distance in the chosen direction (distances can be chosen independently);
* after that you change your direction (from right to up, or from up to right) and repeat the process.
You don't like to change your direction too much, so you will make no more than n - 1 direction changes.
As a result, your path will be a polygonal chain from (0, 0) to (n, n), consisting of at most n line segments where each segment has positive integer length and vertical and horizontal segments alternate.
Not all paths are equal. You have n integers c_1, c_2, ..., c_n where c_i is the cost of the i-th segment.
Using these costs we can define the cost of the path as the sum of lengths of the segments of this path multiplied by their cost, i. e. if the path consists of k segments (k ≤ n), then the cost of the path is equal to ∑_{i=1}^{k}{c_i ⋅ length_i} (segments are numbered from 1 to k in the order they are in the path).
Find the path of the minimum cost and print its cost.
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9) — the costs of each segment.
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
For each test case, print the minimum possible cost of the path from (0, 0) to (n, n) consisting of at most n alternating segments.
Example
Input
3
2
13 88
3
2 3 1
5
4 3 2 1 4
Output
202
13
19
Note
In the first test case, to reach (2, 2) you need to make at least one turn, so your path will consist of exactly 2 segments: one horizontal of length 2 and one vertical of length 2. The cost of the path will be equal to 2 ⋅ c_1 + 2 ⋅ c_2 = 26 + 176 = 202.
In the second test case, one of the optimal paths consists of 3 segments: the first segment of length 1, the second segment of length 3 and the third segment of length 2.
The cost of the path is 1 ⋅ 2 + 3 ⋅ 3 + 2 ⋅ 1 = 13.
In the third test case, one of the optimal paths consists of 4 segments: the first segment of length 1, the second one — 1, the third one — 4, the fourth one — 4. The cost of the path is 1 ⋅ 4 + 1 ⋅ 3 + 4 ⋅ 2 + 4 ⋅ 1 = 19. | instruction | 0 | 75,259 | 1 | 150,518 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
import sys, math, itertools, random, bisect
from collections import defaultdict
INF = 10**15
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()
mod = 10**9 + 7
# MAX = 100001
# def sieve():
# isPrime = [True]*(MAX)
# isPrime[0] = False
# isPrime[1] = False
# for i in range(2,MAX):
# if isPrime[i]:
# for j in range(i*i, MAX, i):
# isPrime[j] = False
# primes = [2]
# for i in range(3,MAX,2):
# if isPrime[i]: primes.append(i)
# return primes
for _ in range(int(input())):
n = int(input())
c = get_array()
o,e = 0,0
o_sum,e_sum = 0,0
o_min,e_min = INF,INF
o_ans,e_ans = INF,INF
ans = INF
for i in range(n):
if i&1:
o += 1
o_sum += c[i]
o_min = min(o_min,c[i])
o_ans = o_sum + (n-o)*o_min
else:
e += 1
e_sum += c[i]
e_min = min(e_min,c[i])
e_ans = e_sum + (n-e)*e_min
if i: ans = min(ans, o_ans+e_ans)
print(ans)
``` | output | 1 | 75,259 | 1 | 150,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say you are standing on the XY-plane at point (0, 0) and you want to reach point (n, n).
You can move only in two directions:
* to the right, i. e. horizontally and in the direction that increase your x coordinate,
* or up, i. e. vertically and in the direction that increase your y coordinate.
In other words, your path will have the following structure:
* initially, you choose to go to the right or up;
* then you go some positive integer distance in the chosen direction (distances can be chosen independently);
* after that you change your direction (from right to up, or from up to right) and repeat the process.
You don't like to change your direction too much, so you will make no more than n - 1 direction changes.
As a result, your path will be a polygonal chain from (0, 0) to (n, n), consisting of at most n line segments where each segment has positive integer length and vertical and horizontal segments alternate.
Not all paths are equal. You have n integers c_1, c_2, ..., c_n where c_i is the cost of the i-th segment.
Using these costs we can define the cost of the path as the sum of lengths of the segments of this path multiplied by their cost, i. e. if the path consists of k segments (k ≤ n), then the cost of the path is equal to ∑_{i=1}^{k}{c_i ⋅ length_i} (segments are numbered from 1 to k in the order they are in the path).
Find the path of the minimum cost and print its cost.
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9) — the costs of each segment.
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
For each test case, print the minimum possible cost of the path from (0, 0) to (n, n) consisting of at most n alternating segments.
Example
Input
3
2
13 88
3
2 3 1
5
4 3 2 1 4
Output
202
13
19
Note
In the first test case, to reach (2, 2) you need to make at least one turn, so your path will consist of exactly 2 segments: one horizontal of length 2 and one vertical of length 2. The cost of the path will be equal to 2 ⋅ c_1 + 2 ⋅ c_2 = 26 + 176 = 202.
In the second test case, one of the optimal paths consists of 3 segments: the first segment of length 1, the second segment of length 3 and the third segment of length 2.
The cost of the path is 1 ⋅ 2 + 3 ⋅ 3 + 2 ⋅ 1 = 13.
In the third test case, one of the optimal paths consists of 4 segments: the first segment of length 1, the second one — 1, the third one — 4, the fourth one — 4. The cost of the path is 1 ⋅ 4 + 1 ⋅ 3 + 4 ⋅ 2 + 4 ⋅ 1 = 19. | instruction | 0 | 75,260 | 1 | 150,520 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
# from __future__ import print_function # for PyPy2
from itertools import permutations as perm
from fractions import Fraction
from collections import *
from sys import stdin
from bisect import *
from heapq import *
# from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")
fi = []
t, = gil()
for _ in range(t):
n, = gil()
c = gil()
sme, smo = c[0], c[1]
ans = (sme+smo)*n
base = c[0] + c[1]
for i in range(2, n):
base += c[i]
if i&1:
smo = min(smo, c[i])
else:
sme = min(sme, c[i])
k = i+1
e = n - (k//2) - (k&1)
o = n - (k//2)
# print(k, sme, smo, o, e, base, (base + (e*sme) + (o*smo)))
# o, e = e, o
# print(i+1, base + (a*sm1) + (b*sm2))
ans = min(ans, base + (e*sme) + (o*smo))
# print()
fi.append(str(ans))
print('\n'.join(fi))
``` | output | 1 | 75,260 | 1 | 150,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say you are standing on the XY-plane at point (0, 0) and you want to reach point (n, n).
You can move only in two directions:
* to the right, i. e. horizontally and in the direction that increase your x coordinate,
* or up, i. e. vertically and in the direction that increase your y coordinate.
In other words, your path will have the following structure:
* initially, you choose to go to the right or up;
* then you go some positive integer distance in the chosen direction (distances can be chosen independently);
* after that you change your direction (from right to up, or from up to right) and repeat the process.
You don't like to change your direction too much, so you will make no more than n - 1 direction changes.
As a result, your path will be a polygonal chain from (0, 0) to (n, n), consisting of at most n line segments where each segment has positive integer length and vertical and horizontal segments alternate.
Not all paths are equal. You have n integers c_1, c_2, ..., c_n where c_i is the cost of the i-th segment.
Using these costs we can define the cost of the path as the sum of lengths of the segments of this path multiplied by their cost, i. e. if the path consists of k segments (k ≤ n), then the cost of the path is equal to ∑_{i=1}^{k}{c_i ⋅ length_i} (segments are numbered from 1 to k in the order they are in the path).
Find the path of the minimum cost and print its cost.
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9) — the costs of each segment.
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
For each test case, print the minimum possible cost of the path from (0, 0) to (n, n) consisting of at most n alternating segments.
Example
Input
3
2
13 88
3
2 3 1
5
4 3 2 1 4
Output
202
13
19
Note
In the first test case, to reach (2, 2) you need to make at least one turn, so your path will consist of exactly 2 segments: one horizontal of length 2 and one vertical of length 2. The cost of the path will be equal to 2 ⋅ c_1 + 2 ⋅ c_2 = 26 + 176 = 202.
In the second test case, one of the optimal paths consists of 3 segments: the first segment of length 1, the second segment of length 3 and the third segment of length 2.
The cost of the path is 1 ⋅ 2 + 3 ⋅ 3 + 2 ⋅ 1 = 13.
In the third test case, one of the optimal paths consists of 4 segments: the first segment of length 1, the second one — 1, the third one — 4, the fourth one — 4. The cost of the path is 1 ⋅ 4 + 1 ⋅ 3 + 4 ⋅ 2 + 4 ⋅ 1 = 19. | instruction | 0 | 75,261 | 1 | 150,522 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
def solve(n, c):
king_hoz, king_vert = c[0], c[1]
stair_sum = 0
reach_sum = n * c[0] + n * c[1]
m = stair_sum + reach_sum
singles_sum = c[0] + c[1]
hoz_singles = vert_singles = 1
for i in range(2, n):
singles_sum += c[i]
cost = singles_sum
if i % 2 == 0:
if c[i] < king_hoz:
king_hoz = c[i]
hoz_singles += 1
else:
if c[i] < king_vert:
king_vert = c[i]
vert_singles += 1
cost += king_hoz * (n - hoz_singles)
cost += king_vert * (n - vert_singles)
m = min(m, cost)
return m
def main():
t = int(input())
for _ in range(t):
n = int(input())
c = list(map(int, input().split()))
print(solve(n, c))
main()
``` | output | 1 | 75,261 | 1 | 150,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's say you are standing on the XY-plane at point (0, 0) and you want to reach point (n, n).
You can move only in two directions:
* to the right, i. e. horizontally and in the direction that increase your x coordinate,
* or up, i. e. vertically and in the direction that increase your y coordinate.
In other words, your path will have the following structure:
* initially, you choose to go to the right or up;
* then you go some positive integer distance in the chosen direction (distances can be chosen independently);
* after that you change your direction (from right to up, or from up to right) and repeat the process.
You don't like to change your direction too much, so you will make no more than n - 1 direction changes.
As a result, your path will be a polygonal chain from (0, 0) to (n, n), consisting of at most n line segments where each segment has positive integer length and vertical and horizontal segments alternate.
Not all paths are equal. You have n integers c_1, c_2, ..., c_n where c_i is the cost of the i-th segment.
Using these costs we can define the cost of the path as the sum of lengths of the segments of this path multiplied by their cost, i. e. if the path consists of k segments (k ≤ n), then the cost of the path is equal to ∑_{i=1}^{k}{c_i ⋅ length_i} (segments are numbered from 1 to k in the order they are in the path).
Find the path of the minimum cost and print its cost.
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5).
The second line of each test case contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9) — the costs of each segment.
It's guaranteed that the total sum of n doesn't exceed 10^5.
Output
For each test case, print the minimum possible cost of the path from (0, 0) to (n, n) consisting of at most n alternating segments.
Example
Input
3
2
13 88
3
2 3 1
5
4 3 2 1 4
Output
202
13
19
Note
In the first test case, to reach (2, 2) you need to make at least one turn, so your path will consist of exactly 2 segments: one horizontal of length 2 and one vertical of length 2. The cost of the path will be equal to 2 ⋅ c_1 + 2 ⋅ c_2 = 26 + 176 = 202.
In the second test case, one of the optimal paths consists of 3 segments: the first segment of length 1, the second segment of length 3 and the third segment of length 2.
The cost of the path is 1 ⋅ 2 + 3 ⋅ 3 + 2 ⋅ 1 = 13.
In the third test case, one of the optimal paths consists of 4 segments: the first segment of length 1, the second one — 1, the third one — 4, the fourth one — 4. The cost of the path is 1 ⋅ 4 + 1 ⋅ 3 + 4 ⋅ 2 + 4 ⋅ 1 = 19. | instruction | 0 | 75,262 | 1 | 150,524 |
Tags: brute force, data structures, greedy, math
Correct Solution:
```
import sys
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
t = II()
for s1 in range(t):
n = II()
l = LI()
ans=l[0]*n+l[1]*n
cur=l[0]+l[1]
curmin = l[0]
curminh = l[1]
for i in range(2,n):
if i%2==0:
curmin = min(curmin,l[i])
else:
curminh = min(curminh,l[i])
cur+=l[i]
if (i+1)%2==0:
if curmin*(n-((i+1)//2))+curminh*(n-(i+1)//2)+cur <ans:
ans = curmin*(n-((i+1)//2))+curminh*(n-(i+1)//2)+cur
else:
if curmin*(n-(i+1)//2 - 1)+curminh*(n-(i+1)//2)+cur<ans:
ans= curmin*(n-(i+1)//2 - 1)+curminh*(n-(i+1)//2)+cur
print(ans)
``` | output | 1 | 75,262 | 1 | 150,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has n groups of people. The i-th group from the beginning has ai people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most m people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all n groups to the dacha countryside.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). The next line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ m).
Output
Print a single integer — the number of buses that is needed to transport all n groups to the dacha countryside.
Examples
Input
4 3
2 3 2 1
Output
3
Input
3 4
1 2 1
Output
1
Submitted Solution:
```
n, m = map(int,input().split())
group = list(map(int,input().split()))
b = 0
tot = 0
for i in range(n):
tot += group[i]
if tot > m :
b += 1
tot = group[i]
elif tot == m :
b += 1
tot = 0
if i == n-1 and tot != 0 :
b += 1
print(b)
``` | instruction | 0 | 75,409 | 1 | 150,818 |
Yes | output | 1 | 75,409 | 1 | 150,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has n groups of people. The i-th group from the beginning has ai people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most m people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all n groups to the dacha countryside.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). The next line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ m).
Output
Print a single integer — the number of buses that is needed to transport all n groups to the dacha countryside.
Examples
Input
4 3
2 3 2 1
Output
3
Input
3 4
1 2 1
Output
1
Submitted Solution:
```
n,m=map(int,input().split())
l=list(map(int,input().split()))
x,y,i=0,0,0
while(i<n):
x=x+l[i]
if(x>m):
y,x,i=y+1,0,i-1
i=i+1
print(y+1)
``` | instruction | 0 | 75,410 | 1 | 150,820 |
Yes | output | 1 | 75,410 | 1 | 150,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has n groups of people. The i-th group from the beginning has ai people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most m people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all n groups to the dacha countryside.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). The next line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ m).
Output
Print a single integer — the number of buses that is needed to transport all n groups to the dacha countryside.
Examples
Input
4 3
2 3 2 1
Output
3
Input
3 4
1 2 1
Output
1
Submitted Solution:
```
# coding: utf-8
n, m = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
a.reverse()
ans = 0
while a:
tmp = m
while a and tmp>=a[-1]:
tmp -= a.pop()
ans += 1
print(ans)
``` | instruction | 0 | 75,411 | 1 | 150,822 |
Yes | output | 1 | 75,411 | 1 | 150,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has n groups of people. The i-th group from the beginning has ai people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most m people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all n groups to the dacha countryside.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). The next line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ m).
Output
Print a single integer — the number of buses that is needed to transport all n groups to the dacha countryside.
Examples
Input
4 3
2 3 2 1
Output
3
Input
3 4
1 2 1
Output
1
Submitted Solution:
```
import sys
import math
n, m = [int(x) for x in (sys.stdin.readline()).split()]
ai = [int(x) for x in (sys.stdin.readline()).split()]
f = 0
result = 0
for c in ai:
if(f + c < m):
f += c
elif(f + c == m):
result += 1
f = 0
else:
f = c
result += 1
if(f > 0):
result += 1
print(result)
``` | instruction | 0 | 75,412 | 1 | 150,824 |
Yes | output | 1 | 75,412 | 1 | 150,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has n groups of people. The i-th group from the beginning has ai people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most m people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all n groups to the dacha countryside.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). The next line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ m).
Output
Print a single integer — the number of buses that is needed to transport all n groups to the dacha countryside.
Examples
Input
4 3
2 3 2 1
Output
3
Input
3 4
1 2 1
Output
1
Submitted Solution:
```
row = input().split()
n = int(row[0])
m = int(row[1])
members = list(map(int, input().split()))
count = int(sum(members) / m)
if sum(members) % m > 0 : count+=1
print(count)
``` | instruction | 0 | 75,413 | 1 | 150,826 |
No | output | 1 | 75,413 | 1 | 150,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has n groups of people. The i-th group from the beginning has ai people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most m people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all n groups to the dacha countryside.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). The next line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ m).
Output
Print a single integer — the number of buses that is needed to transport all n groups to the dacha countryside.
Examples
Input
4 3
2 3 2 1
Output
3
Input
3 4
1 2 1
Output
1
Submitted Solution:
```
x = list(map(int,input().split()))
l = list(map(int,input().split()))
n=x[0]
m=x[1]
i=0
c=0
v = 0
while i < n:
while i < n:
if i == len(l)-1:
if l[i]+v <= m:
i+=1
break
else :
c+=1
i+=1
break
elif l[i]+l[i+1]>m:
i+=1
break
else :
v+=l[i]+l[i+1]
i+=2
v=0
c+=1
print(c)
``` | instruction | 0 | 75,414 | 1 | 150,828 |
No | output | 1 | 75,414 | 1 | 150,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has n groups of people. The i-th group from the beginning has ai people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most m people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all n groups to the dacha countryside.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). The next line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ m).
Output
Print a single integer — the number of buses that is needed to transport all n groups to the dacha countryside.
Examples
Input
4 3
2 3 2 1
Output
3
Input
3 4
1 2 1
Output
1
Submitted Solution:
```
from sys import stdin as cin
from math import factorial as f
from itertools import combinations as comb
n,m = map(int,cin.readline().split())
a = list(map(int,cin.readline().split()))
p=[0]*n
count = 0
r = 0
while r<n:
sum = 0
j = 0
while j<n:
if p[j]==0:
if sum+a[j]<=m:
sum+=a[j]
p[j]=1
r +=1
#print(r)
j+=1
count+=1
print(count)
``` | instruction | 0 | 75,415 | 1 | 150,830 |
No | output | 1 | 75,415 | 1 | 150,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It's that time of the year when the Russians flood their countryside summer cottages (dachas) and the bus stop has a lot of people. People rarely go to the dacha on their own, it's usually a group, so the people stand in queue by groups.
The bus stop queue has n groups of people. The i-th group from the beginning has ai people. Every 30 minutes an empty bus arrives at the bus stop, it can carry at most m people. Naturally, the people from the first group enter the bus first. Then go the people from the second group and so on. Note that the order of groups in the queue never changes. Moreover, if some group cannot fit all of its members into the current bus, it waits for the next bus together with other groups standing after it in the queue.
Your task is to determine how many buses is needed to transport all n groups to the dacha countryside.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). The next line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ m).
Output
Print a single integer — the number of buses that is needed to transport all n groups to the dacha countryside.
Examples
Input
4 3
2 3 2 1
Output
3
Input
3 4
1 2 1
Output
1
Submitted Solution:
```
n , m = map(int , input().split())
groups = list(map(int , input().split()))
i = 0
count = 0
while i+1<len(groups):
temp = groups[i]
i = i + 1
while temp + groups[i]<m and i<len(groups)-1:
i = i + 1
count = count + 1
print(count)
``` | instruction | 0 | 75,416 | 1 | 150,832 |
No | output | 1 | 75,416 | 1 | 150,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kekoland is a country with n beautiful cities numbered from left to right and connected by n - 1 roads. The i-th road connects cities i and i + 1 and length of this road is wi kilometers.
When you drive in Kekoland, each time you arrive in city i by car you immediately receive gi liters of gas. There is no other way to get gas in Kekoland.
You were hired by the Kekoland president Keko to organize the most beautiful race Kekoland has ever seen. Let race be between cities l and r (l ≤ r). Race will consist of two stages. On the first stage cars will go from city l to city r. After completing first stage, next day second stage will be held, now racers will go from r to l with their cars. Of course, as it is a race, racers drive directly from start city to finish city. It means that at the first stage they will go only right, and at the second stage they go only left. Beauty of the race between l and r is equal to r - l + 1 since racers will see r - l + 1 beautiful cities of Kekoland. Cars have infinite tank so racers will take all the gas given to them.
At the beginning of each stage racers start the race with empty tank (0 liters of gasoline). They will immediately take their gasoline in start cities (l for the first stage and r for the second stage) right after the race starts.
It may not be possible to organize a race between l and r if cars will run out of gas before they reach finish.
You have k presents. Each time you give a present to city i its value gi increases by 1. You may distribute presents among cities in any way (also give many presents to one city, each time increasing gi by 1). What is the most beautiful race you can organize?
Each car consumes 1 liter of gas per one kilometer.
Input
The first line of the input contains two integers n and k (2 ≤ n ≤ 100 000, 0 ≤ k ≤ 109) — the number of cities in Kekoland and the number of presents you have, respectively.
Next line contains n - 1 integers. The i-th of them is wi (1 ≤ wi ≤ 109) — the length of the road between city i and i + 1.
Next line contains n integers. The i-th of them is gi (0 ≤ gi ≤ 109) — the amount of gas you receive every time you enter city i.
Output
Print a single line — the beauty of the most beautiful race you can organize.
Examples
Input
4 4
2 2 2
1 1 1 1
Output
4
Input
8 5
2 2 2 3 7 3 1
1 3 1 5 4 0 2 5
Output
7
Note
In first sample if you give one present to each city then it will be possible to make a race between city 1 and city 4.
In second sample you should add 1 to g5 and 4 to g6, then it will be possible to make a race between cities 2 and 8.
Submitted Solution:
```
def main():
n, k = map(int, input().split())
l = sorted(map(int, input().split()))
lo, hi = l[0], l[-1]
while lo < hi - 1:
mid = (lo + hi) // 2
t = k
for x in l:
if x > mid:
lo = mid
break
t -= mid - x
if t < 0:
hi = mid
break
else:
lo = mid
m1 = lo
lo, hi = l[0], l[-1]
l.reverse()
while lo < hi - 1:
mid = (lo + hi) // 2
t = k
for x in l:
if x < mid:
hi = mid
break
t -= x - mid
if t < 0:
lo = mid
break
else:
hi = mid
print(hi - m1 if hi > m1 else 1 if sum(l) % n else 0)
if __name__ == '__main__':
main()
``` | instruction | 0 | 75,520 | 1 | 151,040 |
No | output | 1 | 75,520 | 1 | 151,041 |
Provide a correct Python 3 solution for this coding contest problem.
There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i.
You are now at town 1, and you want to visit all the other towns. You have two ways to travel:
* Walk on the line. Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.
* Teleport to any location of your choice. Your fatigue level increases by B, regardless of the distance covered.
Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.
Constraints
* All input values are integers.
* 2≤N≤10^5
* 1≤X_i≤10^9
* For all i(1≤i≤N-1), X_i<X_{i+1}.
* 1≤A≤10^9
* 1≤B≤10^9
Input
The input is given from Standard Input in the following format:
N A B
X_1 X_2 ... X_N
Output
Print the minimum possible total increase of your fatigue level when you visit all the towns.
Examples
Input
4 2 5
1 2 5 7
Output
11
Input
7 1 100
40 43 45 105 108 115 124
Output
84
Input
7 1 2
24 35 40 68 72 99 103
Output
12 | instruction | 0 | 75,755 | 1 | 151,510 |
"Correct Solution:
```
n,a,b= map(int,input().split())
x= list(map(int,input().split()))
cost=0
for i in range(n-1):cost+=min(b,a*(x[i+1]-x[i]))
print(cost)
``` | output | 1 | 75,755 | 1 | 151,511 |
Provide a correct Python 3 solution for this coding contest problem.
There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i.
You are now at town 1, and you want to visit all the other towns. You have two ways to travel:
* Walk on the line. Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.
* Teleport to any location of your choice. Your fatigue level increases by B, regardless of the distance covered.
Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.
Constraints
* All input values are integers.
* 2≤N≤10^5
* 1≤X_i≤10^9
* For all i(1≤i≤N-1), X_i<X_{i+1}.
* 1≤A≤10^9
* 1≤B≤10^9
Input
The input is given from Standard Input in the following format:
N A B
X_1 X_2 ... X_N
Output
Print the minimum possible total increase of your fatigue level when you visit all the towns.
Examples
Input
4 2 5
1 2 5 7
Output
11
Input
7 1 100
40 43 45 105 108 115 124
Output
84
Input
7 1 2
24 35 40 68 72 99 103
Output
12 | instruction | 0 | 75,756 | 1 | 151,512 |
"Correct Solution:
```
n,a,b = map(int,input().split())
*l, = map(int,input().split())
ans = 0
for i in range(len(l)-1):
ans += min(b,a*(l[i+1]-l[i]))
print(ans)
``` | output | 1 | 75,756 | 1 | 151,513 |
Provide a correct Python 3 solution for this coding contest problem.
There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i.
You are now at town 1, and you want to visit all the other towns. You have two ways to travel:
* Walk on the line. Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.
* Teleport to any location of your choice. Your fatigue level increases by B, regardless of the distance covered.
Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.
Constraints
* All input values are integers.
* 2≤N≤10^5
* 1≤X_i≤10^9
* For all i(1≤i≤N-1), X_i<X_{i+1}.
* 1≤A≤10^9
* 1≤B≤10^9
Input
The input is given from Standard Input in the following format:
N A B
X_1 X_2 ... X_N
Output
Print the minimum possible total increase of your fatigue level when you visit all the towns.
Examples
Input
4 2 5
1 2 5 7
Output
11
Input
7 1 100
40 43 45 105 108 115 124
Output
84
Input
7 1 2
24 35 40 68 72 99 103
Output
12 | instruction | 0 | 75,757 | 1 | 151,514 |
"Correct Solution:
```
import itertools
n, a, b = map(int, input().split())
x = list(map(int, input().split()))
ans = sum(min(a * abs(p - q), b) for p, q in zip(x, x[1:]))
print(ans)
``` | output | 1 | 75,757 | 1 | 151,515 |
Provide a correct Python 3 solution for this coding contest problem.
There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i.
You are now at town 1, and you want to visit all the other towns. You have two ways to travel:
* Walk on the line. Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.
* Teleport to any location of your choice. Your fatigue level increases by B, regardless of the distance covered.
Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.
Constraints
* All input values are integers.
* 2≤N≤10^5
* 1≤X_i≤10^9
* For all i(1≤i≤N-1), X_i<X_{i+1}.
* 1≤A≤10^9
* 1≤B≤10^9
Input
The input is given from Standard Input in the following format:
N A B
X_1 X_2 ... X_N
Output
Print the minimum possible total increase of your fatigue level when you visit all the towns.
Examples
Input
4 2 5
1 2 5 7
Output
11
Input
7 1 100
40 43 45 105 108 115 124
Output
84
Input
7 1 2
24 35 40 68 72 99 103
Output
12 | instruction | 0 | 75,758 | 1 | 151,516 |
"Correct Solution:
```
n,a,b = map(int,input().split())
x = list(map(int,input().split()))
xdif = [x2-x1 for x1,x2 in zip(x,x[1:])]
ans = sum(dx*a if dx*a <b else b for dx in xdif)
print(ans)
``` | output | 1 | 75,758 | 1 | 151,517 |
Provide a correct Python 3 solution for this coding contest problem.
There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i.
You are now at town 1, and you want to visit all the other towns. You have two ways to travel:
* Walk on the line. Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.
* Teleport to any location of your choice. Your fatigue level increases by B, regardless of the distance covered.
Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.
Constraints
* All input values are integers.
* 2≤N≤10^5
* 1≤X_i≤10^9
* For all i(1≤i≤N-1), X_i<X_{i+1}.
* 1≤A≤10^9
* 1≤B≤10^9
Input
The input is given from Standard Input in the following format:
N A B
X_1 X_2 ... X_N
Output
Print the minimum possible total increase of your fatigue level when you visit all the towns.
Examples
Input
4 2 5
1 2 5 7
Output
11
Input
7 1 100
40 43 45 105 108 115 124
Output
84
Input
7 1 2
24 35 40 68 72 99 103
Output
12 | instruction | 0 | 75,759 | 1 | 151,518 |
"Correct Solution:
```
n,a,b=map(int,input().split())
x=list(map(int,input().split()))
ct=0
for i in range(n-1):
d=x[i+1]-x[i]
if d*a<b:ct+=d*a
else:ct+=b
print(ct)
``` | output | 1 | 75,759 | 1 | 151,519 |
Provide a correct Python 3 solution for this coding contest problem.
There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i.
You are now at town 1, and you want to visit all the other towns. You have two ways to travel:
* Walk on the line. Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.
* Teleport to any location of your choice. Your fatigue level increases by B, regardless of the distance covered.
Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.
Constraints
* All input values are integers.
* 2≤N≤10^5
* 1≤X_i≤10^9
* For all i(1≤i≤N-1), X_i<X_{i+1}.
* 1≤A≤10^9
* 1≤B≤10^9
Input
The input is given from Standard Input in the following format:
N A B
X_1 X_2 ... X_N
Output
Print the minimum possible total increase of your fatigue level when you visit all the towns.
Examples
Input
4 2 5
1 2 5 7
Output
11
Input
7 1 100
40 43 45 105 108 115 124
Output
84
Input
7 1 2
24 35 40 68 72 99 103
Output
12 | instruction | 0 | 75,760 | 1 | 151,520 |
"Correct Solution:
```
N,A,B = map(int, input().split())
X = list(map(int, input().split()))
ans = 0
for i in range(1,N):
ans += min(B, (X[i]-X[i-1])*A)
print(ans)
``` | output | 1 | 75,760 | 1 | 151,521 |
Provide a correct Python 3 solution for this coding contest problem.
There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i.
You are now at town 1, and you want to visit all the other towns. You have two ways to travel:
* Walk on the line. Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.
* Teleport to any location of your choice. Your fatigue level increases by B, regardless of the distance covered.
Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.
Constraints
* All input values are integers.
* 2≤N≤10^5
* 1≤X_i≤10^9
* For all i(1≤i≤N-1), X_i<X_{i+1}.
* 1≤A≤10^9
* 1≤B≤10^9
Input
The input is given from Standard Input in the following format:
N A B
X_1 X_2 ... X_N
Output
Print the minimum possible total increase of your fatigue level when you visit all the towns.
Examples
Input
4 2 5
1 2 5 7
Output
11
Input
7 1 100
40 43 45 105 108 115 124
Output
84
Input
7 1 2
24 35 40 68 72 99 103
Output
12 | instruction | 0 | 75,761 | 1 | 151,522 |
"Correct Solution:
```
N,A,B=map(int,input().split())
X=list(map(int,input().split()))
ans=0
for i in range(N-1):
d=X[i+1]-X[i]
if d*A<=B:
ans+=d*A
else:
ans+=B
print(ans)
``` | output | 1 | 75,761 | 1 | 151,523 |
Provide a correct Python 3 solution for this coding contest problem.
There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i.
You are now at town 1, and you want to visit all the other towns. You have two ways to travel:
* Walk on the line. Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.
* Teleport to any location of your choice. Your fatigue level increases by B, regardless of the distance covered.
Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.
Constraints
* All input values are integers.
* 2≤N≤10^5
* 1≤X_i≤10^9
* For all i(1≤i≤N-1), X_i<X_{i+1}.
* 1≤A≤10^9
* 1≤B≤10^9
Input
The input is given from Standard Input in the following format:
N A B
X_1 X_2 ... X_N
Output
Print the minimum possible total increase of your fatigue level when you visit all the towns.
Examples
Input
4 2 5
1 2 5 7
Output
11
Input
7 1 100
40 43 45 105 108 115 124
Output
84
Input
7 1 2
24 35 40 68 72 99 103
Output
12 | instruction | 0 | 75,762 | 1 | 151,524 |
"Correct Solution:
```
N,A,B=list(map(int,input().split()))
X=list(map(int,input().split()))
res=0
for i in range(N-1):
res += min(A*(X[i+1]-X[i]), B)
print(res)
``` | output | 1 | 75,762 | 1 | 151,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i.
You are now at town 1, and you want to visit all the other towns. You have two ways to travel:
* Walk on the line. Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.
* Teleport to any location of your choice. Your fatigue level increases by B, regardless of the distance covered.
Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.
Constraints
* All input values are integers.
* 2≤N≤10^5
* 1≤X_i≤10^9
* For all i(1≤i≤N-1), X_i<X_{i+1}.
* 1≤A≤10^9
* 1≤B≤10^9
Input
The input is given from Standard Input in the following format:
N A B
X_1 X_2 ... X_N
Output
Print the minimum possible total increase of your fatigue level when you visit all the towns.
Examples
Input
4 2 5
1 2 5 7
Output
11
Input
7 1 100
40 43 45 105 108 115 124
Output
84
Input
7 1 2
24 35 40 68 72 99 103
Output
12
Submitted Solution:
```
N, A, B = map(int,input().split())
X = [int(i) for i in input().split()]
p = X[0]
ans = 0
for i in range(1, N):
ans += min(B, A*(X[i]-p))
p = X[i]
print(ans)
``` | instruction | 0 | 75,763 | 1 | 151,526 |
Yes | output | 1 | 75,763 | 1 | 151,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i.
You are now at town 1, and you want to visit all the other towns. You have two ways to travel:
* Walk on the line. Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.
* Teleport to any location of your choice. Your fatigue level increases by B, regardless of the distance covered.
Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.
Constraints
* All input values are integers.
* 2≤N≤10^5
* 1≤X_i≤10^9
* For all i(1≤i≤N-1), X_i<X_{i+1}.
* 1≤A≤10^9
* 1≤B≤10^9
Input
The input is given from Standard Input in the following format:
N A B
X_1 X_2 ... X_N
Output
Print the minimum possible total increase of your fatigue level when you visit all the towns.
Examples
Input
4 2 5
1 2 5 7
Output
11
Input
7 1 100
40 43 45 105 108 115 124
Output
84
Input
7 1 2
24 35 40 68 72 99 103
Output
12
Submitted Solution:
```
n,a,b = map(int,input().split())
x = list(map(int,input().split()))
ans = 0
for i in range(n-1):
ans += min((x[i+1]-x[i])*a,b)
print(ans)
``` | instruction | 0 | 75,764 | 1 | 151,528 |
Yes | output | 1 | 75,764 | 1 | 151,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N towns on a line running east-west. The towns are numbered 1 through N, in order from west to east. Each point on the line has a one-dimensional coordinate, and a point that is farther east has a greater coordinate value. The coordinate of town i is X_i.
You are now at town 1, and you want to visit all the other towns. You have two ways to travel:
* Walk on the line. Your fatigue level increases by A each time you travel a distance of 1, regardless of direction.
* Teleport to any location of your choice. Your fatigue level increases by B, regardless of the distance covered.
Find the minimum possible total increase of your fatigue level when you visit all the towns in these two ways.
Constraints
* All input values are integers.
* 2≤N≤10^5
* 1≤X_i≤10^9
* For all i(1≤i≤N-1), X_i<X_{i+1}.
* 1≤A≤10^9
* 1≤B≤10^9
Input
The input is given from Standard Input in the following format:
N A B
X_1 X_2 ... X_N
Output
Print the minimum possible total increase of your fatigue level when you visit all the towns.
Examples
Input
4 2 5
1 2 5 7
Output
11
Input
7 1 100
40 43 45 105 108 115 124
Output
84
Input
7 1 2
24 35 40 68 72 99 103
Output
12
Submitted Solution:
```
n, a, b = map(int, input().split())
x = list(map(int, input().split()))
cost = 0
for i in range(n-1):
dist = x[i+1] - x[i]
cost += min(dist*a, b)
print(cost)
``` | instruction | 0 | 75,765 | 1 | 151,530 |
Yes | output | 1 | 75,765 | 1 | 151,531 |
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