The minimum number of plot 1403

This question yaozhaoguilv

Subject description:

There are two vectors v1 = (x1, x2, x3 , ..., xn), v2 = (y1, y2, y3, ..., yn). Sequence allows any switching component v1 and v2, calculate the inner product of x1 v1 and v2 X2 + Y1 Y2 minimum value xn * yn + ... + a. (1 <= n <= 800 , -1000 <= xi <= 1000, -1000 <= yi <= 1000)

Enter a description:

The first input line of a positive integer n, the number of components of each vector.

The second line of the input vector v1 n integers components, separated by a space between each component.

The third line of input vector v2 n integers components, between each component separated by a space.

Output Description:

An integer, per line, the minimum value of the product of the two vectors.

Sample input:

3

1 3 -5

-2 4 1

Sample output:

-25

Idea: a ascending order, a descending order

Minimum product.

 1 #include<iostream>
 2 using namespace std;
 3 int main(){
 4     int n,i,j;
 5     while(cin>>n){
 6         long v1[n];
 7         long v2[n];
 8         long sum=0,p,q;
 9         for(i=0;i<n;i++){
10             cin>>v1[i];
11         }
12         for(i=0;i<n;i++){
13             cin>>v2[i];
14         }
15         for(i=0;i<n-1;i++){
16             for(j=0;j<n-i-1;j++){
17                 if(v1[j]>v1[j+1]){
18                     p=v1[j];
19                     v1[j]=v1[j+1];
20                     v1[j+1]=p;
21                 }
22                 if(v2[j]<v2[j+1]){
23                     q=v2[j];
24                     v2[j]=v2[j+1];
25                     v2[j+1]=q;
26                 }
27             }
28         }
29         for(i=0;i<n;i++){
30             sum=sum+v1[i]*v2[i];
31         }
32         cout<<sum<<endl;
33     }
34     return 0;
35 }