hdu 5461 Largest Point

最大化序列值算法

原创于 2017-06-22 20:34:11 发布 · 252 阅读

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本文介绍了一种算法，用于从整数序列中选择两个元素，并通过特定公式计算最大化的值。该算法根据不同系数的正负性分四类处理，确保了在不同情况下都能找到最优解。

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

2

3 2 1
1 2 3

5 -1 0
-3 -3 0 3 3

Sample Output

Case #1: 20
Case #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;
}