GCJ 2008 Round1A A Minimum Scalar Product

//水题
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long ll;
const int MAX_N = 1000;
int n;
int v1[MAX_N], v2[MAX_N];

void solve();

int main()
{
	int k;
	freopen("E:\A-large-practice.txt", "r", stdin); // 输入重定向，输入将从 A.txt 文件中读入
	freopen("E:\A large output.txt", "w", stdout); // 创建文件，输出重定向，输出数据将保存在 A output.txt 文件中 
//	scanf("%d", &k);
	cin >> k;
	for (int kase = 1; kase <= k; kase++)
	{
		cin >> n;
		for (int i = 0; i < n; i++) cin >> v1[i];
		//scanf("%d", v1 + i);
		for (int i = 0; i < n; i++) cin >> v2[i];
		//scanf("%d", v2 + i);
		cout << "Case #" << kase << ": ";
		solve();
	}
	fclose(stdin);
	fclose(stdout);
	return 0;
}

void solve()
{
	sort(v1, v1 + n);
	sort(v2, v2 + n);
	ll ans = 0;
	for (int i = 0; i < n; i++) ans += (ll)v1[i] * v2[n - 1 - i];
//	printf("%ll
", ans);
	cout << ans << endl;
}