GCJ 2008 Round1A Problem A. Minimum Scalar Product

                                        Problem A. Minimum Scalar Product
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Small input
5 points
Solve A-small
Large input
10 points

Solve A-large

Problem
You are given two vectors v1=(x1,x2,...,xn) and v2=(y1,y2,...,yn). The scalar product of these vectors is 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 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 ¡Ü n ¡Ü 8
-1000 ¡Ü xi, yi ¡Ü 1000

Large dataset

T = 10
100 ¡Ü n ¡Ü 800
-100000 ¡Ü xi, yi ¡Ü 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

题意就是两行对应相乘的和的最小值

#include <cstdio>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 1000 + 10;int t;int n;int v1[maxn], v2[maxn];int main(){    scanf("%d", &t);    for (int cas = 1; cas <= t; cas++){        scanf("%d", &n);        for (int i = 0; i < n; i++)            scanf("%d", &v1[i]);        for (int i = 0; i < n; i++)            scanf("%d", &v2[i]);        sort(v1, v1 + n);        sort(v2, v2 + n);        ll ans = 0;        for (int i = 0; i < n; i++)            ans += (ll)v1[i] * v2[n - i - 1];        printf("Case #%d: %lld\n", cas, ans);    }    return 0;}