Largest Point

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

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

简单题  求a*ti*ti+b*tj(i!=j)的最大值

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;

int tz[5000050],tf[5000050];
bool compare(int a,int b)
{
    return a<b;
}
int main()
{
    int T,tt=1;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        long long a,b;
        int tzlen=0,tflen=0;
        scanf("%d%lld%lld",&n,&a,&b);
        for(int i=0; i<n; i++)
        {
            int num;
            scanf("%d",&num);
            if(num>=0)
                tz[tzlen++]=num;
            else tf[tflen++]=-num;
        }
        sort(tz,tz+tzlen,compare);
        sort(tf,tf+tflen,compare);
        long long Max;
        if(a>=0&&b>=0)
        {
            Max=a*tz[tzlen-1]*tz[tzlen-1]+b*tz[tzlen-2];
            Max=max(Max,a*tz[tzlen-2]*tz[tzlen-2]+b*tz[tzlen-1]);
            Max=max(Max,a*tf[tflen-1]*tf[tflen-1]+b*tz[tzlen-1]);
        }else if(a>=0&&b<0)
        {
            Max=-b*tf[tflen-1]+a*tz[tzlen-1]*tz[tzlen-1];
            Max=max(Max,-b*tf[tflen-1]+a*tf[tflen-2]*tf[tflen-2]);
            Max=max(Max,-b*tf[tflen-2]+a*tf[tflen-1]*tf[tflen-1]);
        }else if(a<0&&b>=0)
        {
            Max=a*tz[0]*tz[0]+b*tz[tzlen-1];
            Max=max(Max,a*tf[0]*tf[0]+b*tz[tzlen-1]);
        }else
        {
            Max=-b*tf[tflen-1]+a*tz[0]*tz[0];
            Max=max(Max,-b*tf[tflen-1]+a*tf[0]*tf[0]);
        }

        printf("Case #%d: %lld\n",tt++,Max);
    }
}

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