. Line to Cashier

最新推荐文章于 2018-10-17 18:34:18 发布

dcba-1357

最新推荐文章于 2018-10-17 18:34:18 发布

阅读量588

点赞数

CC 4.0 BY-SA版权

分类专栏： Codeforces

本文链接：https://blog.youkuaiyun.com/cuizhenling_123/article/details/22599775

Codeforces 专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一种算法，用于解决在超市收银台选择最优队列的问题，通过计算每个队伍的等待时间来帮助顾客快速完成结账。

A. Line to Cashier

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 k_i people. The j-th person standing in the queue to the i-th cashier has m_i, 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.

Input

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, ..., k_n (1 ≤ k_i ≤ 100), where k_i is the number of people in the queue to the i-th cashier.

The i-th of the next n lines contains k_i space-separated integers: m_i, 1, m_i, 2, ..., m_{i, k_i} (1 ≤ m_i, j ≤ 100) — the number of products the j-th person in the queue for the i-th cash has.

Output

Print a single integer — the minimum number of seconds Vasya needs to get to the cashier.

Sample test(s)

Input

1
1
1

Output

Input

Output

Note

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.

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int n,i,j;
    int a[105];
    int b[105];
    int c[105];
    while(cin>>n)
    {

        for(i=0;i<n;i++)
            cin>>a[i];
            for(j=0;j<n;j++)
            {
                int sum=0;
                for(i=0;i<a[j];i++)
                {
                    cin>>b[i];
                    sum=sum+(b[i]*5+15);
                }
                c[j]=sum;
            }
            sort(c,c+n);
            cout<<c[0]<<endl;
    }
    return 0;
}