考虑两种情况

Largest Point

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 795    Accepted Submission(s): 327

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



模拟太难想了，种类繁多。要快速解决这问题，直接暴力，只要考虑a,b,的值就行了。。。








#include<iostream>
#include<algorithm>
#include<cmath> 
using namespace std;
long long a[5000001];
long long ans1,ans2,s1,s2,ans3,ans4;
int main()
{
	int n,t,ca=1,o;
	long long m,k;
    scanf("%d",&t);
	while(t--)
	{	
		ans1=ans2=ans3=ans4=0;
		scanf("%d%I64d%I64d",&n,&m,&k);
		for(int i=1;i<=n;i++)
		{
			scanf("%I64d",&a[i]);	
		}
		//sort(a+1,a+1+n);
		s1=LLONG_MIN;
		for(int i=1;i<=n;i++)
		{
			s2=m*a[i]*a[i];
			if(s1<s2)
			{
				s1=s2;
				o=i;
			 } 
		}
		ans1=s1;
		s1=LLONG_MIN; 
		for(int i=1;i<=n;i++)
		{
			if(i!=o)
			{
			s2=a[i]*k;
			s1=max(s1,s2);
			}
		}
		ans2=s1; 
	   s1=LLONG_MIN;
		for(int i=1;i<=n;i++)
		{
			s2=a[i]*k;
			if(s1<s2)
			{
				s1=s2;
				o=i;
			 } 
		}
		ans3=s1;
		s1=LLONG_MIN; 
		for(int i=1;i<=n;i++)
		{
			if(i!=o)
			{
			s2=m*a[i]*a[i];
			s1=max(s1,s2);
			}
		}
		ans4=s1;
		ans1=ans1+ans2;
		ans3=ans3+ans4;
		ans4=max(ans1,ans3);
		printf("Case #%d: %I64d\n",ca++,ans4);
	}
	return 0;
 }