HDOJ 5461 Largest Point(瀋陽ネット試合)

Largest Point
Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 548    Accepted Submission(s): 234
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


    
    
    
         ，      ，      ，   

ac  ：

    
    
    
    
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define INF 0x7fffffffff;
using namespace std;
int num[5000100];
int v[5000100];
int main()
{
	int t,k;
	long long sum;
	int i,j,a,b,n;
	long long mi,M;
	int cas=0;
	scanf("%d",&t);
	while(t--)
	{
		sum=0;
		scanf("%d%d%d",&n,&a,&b);
		for(i=0;i<n;i++)
		scanf("%d",&num[i]);
		memset(v,0,sizeof(v));
		if(a<0)
		{
			mi=INF;
			for(i=0;i<n;i++)
			{
				if(fabs(num[i])<mi)
				{
					k=i;
					mi=fabs(num[i]);
				}
			}
			sum+=(a*mi*mi);
			v[k]=1;
		}
		else
		{
			if(a>0)
			{
				M=-INF;
				for(i=0;i<n;i++)
				{
					if(fabs(num[i])>M)
					{
						M=fabs(num[i]);
						k=i;
					}
				}
				sum+=(a*M*M);
				v[k]=1;
			}
		}
		if(b<0)
		{
			mi=INF;
			for(i=0;i<n;i++)
			{
				if(num[i]<mi&&!v[i])
				{
					mi=num[i];
				}
			}
			sum+=(b*mi);
		}
		else
		{
			if(b>0)
			{
				M=-INF;
				for(i=0;i<n;i++)
				{
					if(num[i]>M&&!v[i])
					{
						M=num[i];
					}
				}
				sum+=(b*M);
			}
		}
		printf("Case #%d: %lld
",++cas,sum);
	}
	return 0;
}