Algorithm
Little Vasya went to the supermarket to get some groceries. He walked about the supermarket for a long time and got a basket full of products. Now he needs to choose the cashier to pay for the products.
There are n cashiers at the exit from the supermarket. At the moment the queue for the i-th cashier already has ki people. The j-th person standing in the queue to the i-th cashier has mi, j items in the basket. Vasya knows that:
- the cashier needs 5 seconds to scan one item;
- after the cashier scans each item of some customer, he needs 15 seconds to take the customer's money and give him the change.
Of course, Vasya wants to select a queue so that he can leave the supermarket as soon as possible. Help him write a program that displays the minimum number of seconds after which Vasya can get to one of the cashiers.
The first line contains integer n (1 ≤ n ≤ 100) — the number of cashes in the shop. The second line contains n space-separated integers: k1, k2, ..., kn (1 ≤ ki ≤ 100), where ki is the number of people in the queue to the i-th cashier.
The i-th of the next n lines contains ki space-separated integers: mi, 1, mi, 2, ..., mi, ki (1 ≤ mi, j ≤ 100) — the number of products the j-th person in the queue for the i-th cash has.
Print a single integer — the minimum number of seconds Vasya needs to get to the cashier.
1
1
1
20
4
1 4 3 2
100
1 2 2 3
1 9 1
7 8
100
In the second test sample, if Vasya goes to the first queue, he gets to the cashier in 100·5 + 15 = 515 seconds. But if he chooses the second queue, he will need 1·5 + 2·5 + 2·5 + 3·5 + 4·15 = 100 seconds. He will need 1·5 + 9·5 + 1·5 + 3·15 = 100 seconds for the third one and 7·5 + 8·5 + 2·15 = 105 seconds for the fourth one. Thus, Vasya gets to the cashier quicker if he chooses the second or the third queue.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
int const N = 1e2 + 1;
int n, k[N];
int main() {
scanf("%d", &n);
for(int i = 0; i < n; ++i)
scanf("%d", k + i);
int res = 1e9;
for(int i = 0; i < n; ++i) {
int cur = 0;
for(int j = 0, tmp; j < k[i]; ++j) {
scanf("%d", &tmp);
cur += 5 * tmp + 15;
}
res = min(res, cur);
}
printf("%d\n", res);
return 0;
}Input
1
1
Output
Demonstration
Codeforces Solution-Line to Cashier-Solution in C, C++, Java, Python