挑战2.7.1Round 1A 2008A. Minimum Scalar Product贪心

题目链接:

https://code.google.com/codejam/contest/32016/dashboard#s=p0

题意:

题解:

代码:

#include <bits/stdc++.h> using namespace std; typedef long long ll; #define MS(a) memset(a,0,sizeof(a)) #define MP make_pair #define PB push_back const int INF = 0x3f3f3f3f; const ll INFLL = 0x3f3f3f3f3f3f3f3fLL; inline ll read(){ ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } ////////////////////////////////////////////////////////////////////////// const int maxn = 800+10; ll a[maxn],b[maxn]; int main(){ freopen("2.7.1_A-large-practice.in","r",stdin); freopen("2.7.1_A-large-practice.out","w",stdout); int T=read(); for(int cas=1; cas<=T; cas++){ int n = read(); for(int i=1; i<=n; i++) a[i] = read(); for(int i=1; i<=n; i++) b[i] = read(); sort(a+1,a+1+n); sort(b+1,b+1+n); ll ans = 0; for(int i=1; i<=n; i++){ ans += a[i]*b[n-i+1]; } cout << "Case #" << cas << ": " << ans << endl; } return 0; }