Largest Point （2015沈阳赛区网络赛水题）

Problem Description

Given the sequence A with n integers t1,t2,⋯,tn. Given the integral coefficients a and b. The fact that select two elements ti and tj of A and i≠j to maximize the value of at2i+btj, becomes the largest point.

Input

An positive integer T, indicating there are T test cases.
For each test case, the first line contains three integers corresponding to n (2≤n≤5×106), a (0≤|a|≤106) and b (0≤|b|≤106). The second line contains n integers t1,t2,⋯,tn where 0≤|ti|≤106 for 1≤i≤n.

The sum of n for all cases would not be larger than 5×106.

 

Output

The output contains exactly T lines.
For each test case, you should output the maximum value of at2i+btj.

 

Sample Input

2

3 2 1

1 2 3

5 -1 0

-3 -3 0 3 3

 

Sample Output

Case #1: 20

Case #2: 0

 

题目大意：给出一系列的数t，给出a、b，找出最大的a*ti²+b*t_j，其中，i<>j。

题目分析：将a*ti²和b*t_j分别存放在两个数组中，排下序并找出最大的。若两个最大的下标不相同，则和即为答案；若相同，再找两个次大的，求来自不同数组的最大的加次大的的和，再取二和之中大的便是答案。

 

 

代码如下：

# include<iostream>
# include<cstdio>
# include<cstring>
# include<vector>
# include<map>
# include<set>
# include<list>
# include<cstdlib>
# include<string>
# include<iomanip>
# include<algorithm>
using namespace std;
# define LL long double
struct arr
{
    LL val;
    int id;
    arr(){}
    arr(LL a,int b):val(a),id(b){}
    bool operator < (const arr& a) const {
        return val<a.val;
    }
};
arr w1[500005],w2[500005];
int main()
{
    int T,a,b,n,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        int k;
        scanf("%d%d%d",&n,&a,&b);
        for(int i=0;i<n;++i){
            scanf("%d",&k);
            w1[i]=arr((LL)a*(LL)k*(LL)k,i);
            w2[i]=arr((LL)b*(LL)k,i);
        }
        sort(w1,w1+n);
        sort(w2,w2+n);
        printf("Case #%d: ",++cas);
        if(w1[n-1].id!=w2[n-1].id)
            cout<<fixed<<setprecision(0)<<w1[n-1].val+w2[n-1].val<<endl;
        else{
            LL ans1=w1[n-1].val+w2[n-2].val;
            LL ans2=w1[n-2].val+w2[n-1].val;
            cout<<fixed<<setprecision(0)<<max(ans1,ans2)<<endl;
        }
    }
    return 0;
}