GCJ Round 1A 2008 Problem A. Minimum Scalar Product(YY)
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题目链接:https://code.google.com/codejam/contest/32016/dashboard
题意:给两个整型数组,两个数组各项相乘,数组内的数可以任意调换位置,求乘积最小
思路:先对这两个数组进行排序,然后用数组1的最小值乘以数组2的最大值,然后加上数组1的次小值乘以数组2的次大值,依此类推,得到的最终值就是那个内积的最小值。
#include<cstdio>#include<algorithm>#include<iostream>using namespace std;typedef long long ll;ll a[1000],b[1000];bool cmp(int a,int b){ return a>b;}int main(){ freopen("A-large-practice.in","r",stdin); freopen("A-large-practice.out","w",stdout); int n,m; scanf("%d",&n); for(int cas=1;cas<=n;cas++){ ll ans=0; scanf("%d",&m); for(int i=1;i<=m;i++) scanf("%I64d",&a[i]); for(int i=1;i<=m;i++) scanf("%I64d",&b[i]); sort(a+1,a+m+1); sort(b+1,b+m+1,cmp); for(int i=1;i<=m;i++){ ans+=a[i]*b[i]; } printf("Case #%d: %I64d\n",cas,ans); } return 0;}
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