HDU 5461 Largest Point

最新推荐文章于 2020-05-25 15:05:06 发布

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本文介绍了一个算法问题，即从给定序列中选择两个不同的元素，通过特定公式计算最大值。文章提供了一段AC代码实现，该算法首先对元素进行排序，并巧妙地处理了选取元素时的特殊情况。

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

注意题目中i！=j 的条件

AC代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <string.h>
using namespace std;
struct number
{
    long long shu,xulie;
}x[500005],y[500005];
int cmp(number k,number l)
{
    return k.shu<l.shu;
}
int main()
{
    long long c,t,n,a,b,i,temp;
    scanf("%lld",&t);
    c=0;
    while(t--)
    {
        scanf("%lld%lld%lld",&n,&a,&b);
        for(i=0;i<n;i++)
        {
            scanf("%lld",&temp);
            x[i].shu=a*temp*temp;
            x[i].xulie=i;
            y[i].shu=b*temp;
            y[i].xulie=i;
        }
        sort(x,x+n,cmp);
        sort(y,y+n,cmp);
        printf("Case #%lld: %lld\n",++c,(x[n-1].xulie==y[n-1].xulie)?(max(x[n-1].shu+y[n-2].shu,x[n-2].shu+y[n-1].shu)):(x[n-1].shu+y[n-1].shu));
    }
    return 0;
}