hdu 5461 Largest Point

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

 数据范围不大，直接暴力求解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))
int T;
ll n,a,b;
ll t[1000010];
ll min1,min2,max1,max2,k;//分别存最小的数、第二小的数、最大的数、第二大的数和最接近0的数

int main ()
{
    scanf ("%d",&T);
    int d= 1;
    while(T--)
    {
        scanf ("%lld%lld%lld",&n,&a,&b);
        k=INF;
        for (int i=0; i<n; i++)
            scanf ("%lld",&t[i]);

        printf("Case #%d: ",d++);
        sort(t, t+n);
        for (int i=0; i<n; i++)
        {
            if (t[i]<=0&&t[i+1]>=0)
                k=min(-t[i], t[i+1]);
        }
        min1=t[0];
        min2=t[1];
        max1=t[n-1];
        max2=t[n-2];//找出这五个数

        if (a<0&&b<0)//然后就是苦逼的找最大解了，注意负数的平方为正数
            printf ("%lld\n",a*k*k+b*min1);

        else if (a<0&&b>0)
            printf ("%lld\n",a*k*k+b*max1);

        else if (a>0&&b<0)
            printf ("%lld\n",max(max(a*max1*max1+b*min1, a*min1*min1+b*min2), a*min2*min2+b*min1));

        else if (a>0&&b>0)
            printf ("%lld\n",max(max(a*max1*max1+b*max2, a*max2*max2+b*max1), a*min1*min1+b*max1));

        else printf ("0\n");
    }
    return 0;
}