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\n", ans);cout << ans << endl;}