CF:Line to Cashier

最新推荐文章于 2018-10-17 18:34:18 发布

yangchangshu

最新推荐文章于 2018-10-17 18:34:18 发布

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分类专栏：解题报告

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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 thej-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

111

output

input

41 4 3 21001 2 2 31 9 17 8

output

100

解题报告：模拟题，大意是有n个收银员,每个收银员有k个顾客,每个顾客买了m[i][j]件商品，收银员每扫描一件商品花5秒钟，结算花15秒。计算哪一队的时间最短。假设时间为t,那么:t=这队的总人数*15+所有商品数*5;最后min(t)即可。

参考的代码：

#include <iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;
int p[100][100],c[100];

int main()
{
//freopen("in.txt","r",stdin);
   int i,j,k,m,n,s;
   while(cin>>n)
   {

    int min_s=999999999;
    for(i=0;i<n;i++)
    cin>>c[i];
    for(i=0;i<n;i++)
    for(j=0;j<c[i];j++)
    {
        cin>>p[i][j];
    }
    for(i=0;i<n;i++)
    {
        s=0;
        for(j=0;j<c[i];j++)
        {
         s+=p[i][j]*5;
        }
        s+=c[i]*15;
        if(s<min_s)
        min_s=s;

    }
    cout<<min_s<<endl;
   }
    return 0;
}