hdu 5461 Largest Point

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Given the sequence

A A with

n n integers

t 1 ,t 2 ,⋯,t n t1,t2,⋯,tn. Given the integral coefficients

a a and

b b. The fact that select two elements

t i ti and

t j tj of

A A and

i≠j i≠j to maximize the value of

at 2 i +bt j ati2+btj, becomes the largest point.

Input

An positive integer

T T, indicating there are

T T test cases.
For each test case, the first line contains three integers corresponding to

n (2≤n≤5×10 6 ), a (0≤|a|≤10 6 ) n (2≤n≤5×106), a (0≤|a|≤106) and

b (0≤|b|≤10 6 ) b (0≤|b|≤106). The second line contains

n n integers

t 1 ,t 2 ,⋯,t n t1,t2,⋯,tn where

0≤|t i |≤10 6 0≤|ti|≤106 for

1≤i≤n 1≤i≤n.

The sum of

n n for all cases would not be larger than

5×10 6 5×106.

Output

The output contains exactly

T T lines.
For each test case, you should output the maximum value of

at 2 i +bt j ati2+btj.

Sample Input

23 2 11 2 35 -1 0-3 -3 0 3 3

Sample Output

Case #1: 20Case #2: 0
按照ab的符号分好四种情况就可以了
ac代码：
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;long long a1[5000005];int main(){    long long n,a,b,t;    cin>>t;    int p=0;    while(t--)    {       p++;       cin>>n>>a>>b;       cin>>a1[0];       long long max1=abs(a1[0]),min1=abs(a1[0]),max2=a1[0],min2=a1[0];       if(a<=0&&b<=0)       {           int k=0;           for(int i=1;i<n;i++)           {               scanf("%lld",&a1[i]);               if(abs(a1[i])<min1)               {min1=abs(a1[i]);k=i;}           }           int flag=0;           for(int i=1;i<n;i++)           {               if(a1[i]<min2&&i!=k)               {                   min2=a1[i];               }           }           printf("Case #%d: %lld\n",p,a*min1*min1+b*min2);       }       if(a<=0&&b>0)       {           int k=0;          for(int i=1;i<n;i++)           {               scanf("%lld",&a1[i]);               if(abs(a1[i])<min1)               {min1=abs(a1[i]);k=i;}           }           for(int i=1;i<n;i++)           if(a1[i]>max2&&i!=k)           max2=a1[i];           printf("Case #%d: %lld\n",p,a*min1*min1+b*max2);       }       if(a>0&&b<=0)       {           int k=0;           for(int i=1;i<n;i++)           {               scanf("%lld",&a1[i]);               if(abs(a1[i])>max1)               {max1=abs(a1[i]);k=i;}           }           for(int i=1;i<n;i++)           {               if(a1[i]<min2&&i!=k)               min2=a1[i];           }           printf("Case #%d: %lld\n",p,a*max1*max1+b*min2);       }       if(a>0&&b>0)       {           int k=0;           for(int i=1;i<n;i++)           {               scanf("%lld",&a1[i]);               if(abs(a1[i])>max1)               {max1=abs(a1[i]);k=i;}           }           for(int i=1;i<n;i++)           {               if(a1[i]>max2&&i!=k)               max2=a1[i];           }           printf("Case #%d: %lld\n",p,a*max1*max1+b*max2);       }    }    return 0;}

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