HDU 5461 沈阳网络赛

Largest Point

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 235    Accepted Submission(s): 112

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
   
   
   
   

 

Source

2015 ACM/ICPC Asia Regional Shenyang Online
题意,求公式的最大值, at2i+

 

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
#include<cmath>
#define INF 0x3f3f3f3f
using namespace std;

struct node
{
    long long data;
    int num;
}arr[5000100],arr1[5000100];
int cmp(node p1,node p2)
{
    return p1.data<p2.data;
}
int main()
{
    int t,a,b,x,y;
    int n,m,i,j,k,flag,flag2,nnum=0;
    scanf("%d",&t);
    while(t--)
    {
        nnum++;
        scanf("%d%d%d",&n,&a,&b);
        long long ans=0;
        for(i=0; i<n; i++)
        {
            scanf("%lld",&arr[i].data);
            arr1[i].data=arr[i].data*arr[i].data;
            arr[i].num=arr1[i].num=i;
        }
        sort(arr,arr+n,cmp);
        sort(arr1,arr1+n,cmp);
        if(a>=0&&b>=0)
        {
           if(arr1[n-1].num==arr[n-1].num)
           {
               ans=max(a*arr1[n-1].data+b*arr[n-2].data,a*arr1[n-2].data+b*arr[n-1].data);
           }
           else
            ans=a*arr1[n-1].data+b*arr[n-1].data;
        }
        else if(a>=0&&b<=0)
        {
           if(arr1[n-1].num==arr[0].num)
           {
               ans=max(a*arr1[n-1].data+b*arr[1].data,a*arr1[n-2].data+b*arr[0].data);
           }
           else
            ans=a*arr1[n-1].data+b*arr[0].data;
        }
        else if(a<=0&&b>=0)
        {
            if(arr1[0].num==arr[n-1].num)
           {
               ans=max(a*arr1[0].data+b*arr[n-2].data,a*arr1[1].data+b*arr[n-1].data);
           }
           else
            ans=a*arr1[0].data+b*arr[n-1].data;
        }
        else if(a<=0&&b<=0)
        {
              if(arr1[0].num==arr[0].num)
           {
               ans=max(a*arr1[0].data+b*arr[1].data,a*arr1[1].data+b*arr[0].data);
           }
           else
            ans=a*arr1[0].data+b*arr[0].data;
        }
        printf("Case #%d: %lld\n",nnum,ans);
    }
    return 0;
}