Category: Tips

Problem

You are given two vectors v₁=(x₁,x₂,…,x_n) and v₂=(y₁,y₂,…,y_n). The scalar product of these vectors is a single number, calculated as x₁y₁+x₂y₂+…+x_ny_n.

Suppose you are allowed to permute the coordinates of each vector as you wish. Choose two permutations such that the scalar product of your two new vectors is the smallest possible, and output that minimum scalar product.

Input

The first line of the input file contains integer number T – the number of test cases. For each test case, the first line contains integer number n. The next two lines contain n integers each, giving the coordinates of v₁ and v₂ respectively.

Output

For each test case, output a line

Case #X: Y

where X is the test case number, starting from 1, and Y is the minimum scalar product of all permutations of the two given vectors.

Limits

 

Small dataset

T = 1000
1 ≤ n ≤ 8
-1000 ≤ x_i, y_i ≤ 1000

Large dataset

T = 10
100 ≤ n ≤ 800
-100000 ≤ x_i, y_i ≤ 100000

Sample

Input	Output
2 3 1 3 -5 -2 4 1 5 1 2 3 4 5 1 0 1 0 1	Case #1: -25 Case #2: 6

C++ Solution:

/*
    Author      :   Sreejith Sreekantan
    Description :   Problem A. Minimum Scalar Product
                    (https://code.google.com/codejam/contest/32016/dashboard#s=p0)

*/

#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

int main(int argc, char const *argv[])
{
    // freopen("as.in", "rt", stdin);
    // freopen("as.out", "wt", stdout);
    int n;
    cin >> n;
    int c;
    int tmp;
    vector v1, v2;
    int a, b;
    for(int a=0; a<n; a++)=""    ="" {=""        ="" v1.clear();="" v2.clear();="" cin="">> c;
        for(int b=0; b<c; b++)=""        ="" {=""            ="" cin="">> tmp;
            v1.push_back(tmp);
        }
        for(int b=0; b<c; b++)=""        ="" {=""            ="" cin="">> tmp;
            v2.push_back(tmp);
        }
        sort(v1.begin(),v1.end());
        sort(v2.rbegin, v2.rend());

        long long sumofprod = inner_product(v1.begin(), v1.end(), v2.begin(), (long long)0);
        cout << "Case #" << a + 1 << ": " << sumofprod << endl;
    }
    return 0;
}