Largest Point

Largest Point

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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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

分析：

给你n个数字，从中取两个出来，使at2i+btj的值最大。

由于考虑时我们只用考虑最大正数，第二大正数，最小负数，第二小负数，绝对值最小的数，所以我们只需要将这5个数提前找出来，然后慢慢分情况讨论即可。

AC代码：

#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))

ll T,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 ("%lld",&T);
for (int cas=1; cas<=T; cas++)
{
scanf ("%lld%lld%lld",&n,&a,&b);
k=INF;
for (int i=0; i<n; i++)
{
scanf ("%lld",&t[i]);
}
cout<<"Case #"<<cas<<": ";
sort(t, t+n);
if(t[0]>0)
k=t[0];
else
{
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;
}