HDU 5461 Largest point

Problem Descriptiongiven the sequenceAWithNIntegersT1,T2,?,tn . Given the integral coefficientsaandb. The fact that select the elementsTi andTJ OfAandi≠j To maximize the value of at2i+btJ, becomes the Largest point.

Inputan positive integerT, indicating there isTTest cases.
For each test case, the first line contains three integers corresponding toN(2≤N≤5X106) ,  a  ( 0≤ | a| ≤106) andb (0≤| b| ≤6) . The second line containsNIntegersT1,T2,?,tn where0≤| TI| ≤6 For1≤i≤n .

The sum of n For all cases would is larger than 5x6.

Outputthe output contains exactlyt  lines.
For each test case, you should output the maximum value Of  at2i+b tj .

Sample Input23 2 2 35-1 0-3-3 0 3 3Sample outputcase #1: Case #2:0 Data range is small, direct violence solution

#include <iostream>#include<cstdio>#include<cstring>#include<stack>#include<queue>#include<map>#include<Set>#include<vector>#include<cmath>#include<algorithm>using namespacestd;Const DoubleEPS = 1e-6;Const DoublePI = ACOs (-1.0);Const intINF =0x3f3f3f3f;Const intMOD =1000000007;#definell Long Long#defineCL (a) memset (A,0,sizeof (a))intt;ll n,a,b;ll t[1000010];ll min1,min2,max1,max2,k;//The smallest number, the second small number, the largest number, the second largest number, and the nearest 0 number, respectively.intMain () {scanf ("%d",&T); intD=1;  while(t--) {scanf ("%lld%lld%lld",&n,&a,&b); K=INF;  for(intI=0; i<n; i++) scanf ("%lld",&T[i]); printf ("Case #%d:", d++); Sort (T, T+N);  for(intI=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];//find these five numbers .        if(a<0&&b<0)//then the bitter force to find the biggest solution, notice the negative square is a positive numberprintf ("%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)); Elseprintf ("0\n"); }    return 0;}

HDU 5461 Largest point