(code jam)Problem A. Minimum Scalar Product

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 nintegers 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



题意：就是给出两个向量的坐标,比如x(x1,x2,x3...)  y(y1,y2,y3)任意交换两向量之间的值，使x1*y1+x2*y2+x3*y3...的和最小。

题解就是两个排序，一个从小到大，一个从大到小相乘即是最小值。注意的是大数的时候__in64会超范围。所以用double。

#include <iostream>#include <algorithm>#include <cstdio>using namespace std;bool cmp(double a,double b){return a>b;}bool cmp1(double a,double b){return b>a;}double x[1000],y[1000];int main(){int T,n,i,t=0;freopen("A-large-practice.in","r",stdin);freopen("output.out","w",stdout);scanf("%d",&T);while (T--){printf("Case #%d: ",++t);scanf("%d",&n);for (i=0;i<n;i++)scanf("%lf",&x[i]);for (i=0;i<n;i++)scanf("%lf",&y[i]);sort(x,x+n,cmp);sort(y,y+n,cmp1);double s=0;for (i=0;i<n;i++)s+=x[i]*y[i];printf("%.0lf\n",s);}return 0;}