GCJ 2008 Round 1AA Minimum Scalar Product: Greed

Http://code.google.com/codejam/contest/32016/dashboard

Problem

You are given two vectors v1= (x1,x2,..., xn) and v2= (Y1,y2,..., yn). The scalar product of these vectors are a single number, calculated as X1y1+x2y2+...+xnyn.

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

Input
The ' I ' input file contains integer number t-the number of test cases. For the all test case, the the contains integer number n. The next two lines contain n integers each, giving the coordinates of V1 and v2.

Output

For each test case, output a line

Case #X: Y

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Where X is the "test Case number", starting from 1, and Y are the minimum scalar product of all permutations of the two given Vectors.

Limits

Small DataSet

T = 1000
1≤n≤8
-1000≤xi, yi≤1000

Large DataSet

T = 10
100≤n≤800
-100000≤xi, yi≤100000

Sample

Input

2
3
1 3-5
-2 4 1
5
1 2 3 4 5

1 0 1 0 1

Output

Case #1:-25
Case #2:6
Is the application of the sort inequality. (Reverse sequence and minimum)

Complete code:

#include <cstdio>  
#include <algorithm>  
#include <functional>  
using namespace std;  
    
__int64 v1[805], v2[805];  
    
int main ()  
{  
    freopen ("a-large-practice.in", "R", stdin);  
    Freopen ("A-large-practice.out", "w", stdout);  
    int T, cas = 0, n, i;  
    __int64 ans;  
    scanf ("%d", &t);  
    while (t--)  
    {  
        scanf ("%d", &n);  
        for (i = 0; i < n; ++i) scanf ("%i64d", &v1[i));  
        for (i = 0; i < n; ++i) scanf ("%i64d", &v2[i));  
        Sort (v1, v1 + N);  
        Sort (v2, V2 + N, greater<int> ());  
        Ans = 0;  
        for (i = 0; i < n; ++i)  
            ans + = v1[i] * V2[i];  
        printf ("Case #%d:%i64d\n", ++cas, ans);  
    }  
    return 0;  
}