Beijing University of Science and Technology Information eleventh Programming Contest (reproduce season) H

H andy and shopping

Topic links: https://ac.nowcoder.com/acm/contest/940/H

Title Description

andy go to the market to buy n pieces of cargo, the price of each package is A _i . Businesses in order to attract customers, to buy the N pieces of cargo each customer a discount list, there is little in the fractional N b 1 on the list _j represents the discount. For each discount b _J , by the user to decide which to use it to make the price of goods into b _J * A _i , and can only be used once.

andy want you to help him count his minimal cost.

Enter a description:

First enter a positive integer t, the number of samples representative of the group. (1≤t≤10) 

for each sample: 

the first line, enter a positive integer n (1≤n≤1000). 

The second line contains n integers, the first integer i a [i] representative of the i-th List item. (1≤a [i] ≤1e9) 

The third row contains a number of small n b [i], the title meaning as described. (0≤b [i] ≤1)

Output Description:

For each sample, the output of a real number s, three decimal places, represents the minimum cost.

Example 1

Entry

1
5
1 2 3 4 5
0.1 0.2 0.3 0.4 0.5

Export

3.500

Ideas:

Big big discount original price multiplied with multiplied after its sort to accumulate

 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1005;
struct node{
    int a;
    double zhe;
}Node[1005];
int main()
{
 
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int x[maxn];
        double y[maxn];
        for(int i=0;i<n;i++)
            cin>>x[i];
        sort(x,x+n);
        for(int i=0;i<n;i++)
            Node[i].a=x[i];
        for(int i=0;i<n;i++)
            cin>>y[i];
        sort(y,y+n,greater<double>());
        for(int i=0;i<n;i++)
            Node[i].zhe=y[i];
        double sum=0.0;
        for(int i=0;i<n;i++)
           sum+=Node[i].a*Node[i].zhe;
        printf("%.3f\n",sum);
    }
    return 0;
}