GCJ Round 1A 2008 Problem A. Minimum Scalar Product

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题目大意：

现有T个测例，每个测例都给定两个大小为N的一维向量x和y，现可以重新排列向量中元素的顺序，要求出两向量可能的最小的点积值。

small：T = 1000, 1 ≤ N ≤ 8, -1000 ≤ xi, yi ≤ 1000

large：T = 10, 100 ≤ N ≤ 800, -100000 ≤ xi, yi ≤ 100000

题目链接

注释代码：

/*                                                    * Problem ID : GCJ Round 1A 2008 Problem A. Minimum Scalar Product * Author     : Lirx.t.Una                                                    * Language   : G++                                                                                    */#include <algorithm>#include <iostream>#include <cstdio>#defineMAXSIZE800using namespace std;typedef__int64llg;intv1[MAXSIZE];intv2[MAXSIZE];intmain() {intnscn, iscn;intn;//向量的大小inti;//计数变量llgans;//保存向量点乘的最终结果//方法就是先排序，然后让最大的和最小的乘，求和即可#ifdef__SMALL__freopen("A-small-practice.in", "r", stdin);//将标准输入输出流重定向到文件中freopen("A-small-practice.out", "w", stdout);//是的scanf、printf都以文件作为对象#endif#ifdef__LARGE__freopen("A-large-practice.in", "r", stdin);freopen("A-large-practice.out", "w", stdout);#endifscanf("%d", &nscn);iscn = 0;while ( nscn-- ) {scanf("%d", &n);for ( i = 0; i < n; i++ ) scanf("%d", v1 + i);for ( i = 0; i < n; i++ ) scanf("%d", v2 + i);sort(v1, v1 + n);sort(v2, v2 + n);ans = 0;for ( i = 0; i < n; i++ )ans += (llg)v1[i] * (llg)v2[n - i - 1];printf("Case #%d: %I64d\n", ++iscn, ans);}fclose(stdin);fclose(stdout);return 0;}

无注释代码：

#include <algorithm>#include <iostream>#include <cstdio>#defineMAXSIZE800using namespace std;typedef__int64llg;intv1[MAXSIZE];intv2[MAXSIZE];intmain() {intnscn, iscn;intn;inti;llgans;#ifdef__SMALL__freopen("A-small-practice.in", "r", stdin);freopen("A-small-practice.out", "w", stdout);#endif#ifdef__LARGE__freopen("A-large-practice.in", "r", stdin);freopen("A-large-practice.out", "w", stdout);#endifscanf("%d", &nscn);iscn = 0;while ( nscn-- ) {scanf("%d", &n);for ( i = 0; i < n; i++ ) scanf("%d", v1 + i);for ( i = 0; i < n; i++ ) scanf("%d", v2 + i);sort(v1, v1 + n);sort(v2, v2 + n);ans = 0;for ( i = 0; i < n; i++ )ans += (llg)v1[i] * (llg)v2[n - i - 1];printf("Case #%d: %I64d\n", ++iscn, ans);}fclose(stdin);fclose(stdout);return 0;}

单词解释：

permutation：n, 全排列，置换

permute：vt, 交换，排列

scalar：adj, 标量的，数量的

product：n, 乘积

scalar product：n, 数量积，点积

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