本文介绍了一个算法问题，即如何从给定序列中选取两个元素，通过特定公式计算得出最大值。该问题通过将序列分为两部分并分别进行排序来解决，并提供了一段C++代码实现。
思路：直接分成两部分考虑，排个序就可以了
#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn = 5*1e6+7;
LL a[maxn];
struct Node
{
LL w;
int id;
}a1[maxn],b1[maxn];
bool cmp(Node a,Node b){return a.w>b.w;}
int main()
{
int T,cas=1;
scanf("%d",&T);
while(T--)
{
int n;
LL aa,bb;
scanf("%d%lld%lld",&n,&aa,&bb);
for(int i = 1;i<=n;i++)
{
scanf("%lld",&a[i]);
a1[i].w = aa*a[i]*a[i];
a1[i].id = i;
b1[i].w = bb*a[i];
b1[i].id = i;
}
sort(a1+1,a1+1+n,cmp);
sort(b1+1,b1+1+n,cmp);
if(a1[1].id!=b1[1].id)
{
printf("Case #%d: %lld\n",cas++,a1[1].w+b1[1].w);
continue;
}
else
printf("Case #%d: %lld\n",cas++,max(a1[1].w+b1[2].w,a1[2].w+b1[1].w));
}
}
Description
Given the sequence 
with 
integers 
.
Given the integral coefficients 
and 
.
The fact that select two elements 
and 
of and 
to
maximize the value of 
,
becomes the largest point.
with integers .
Given the integral coefficients and .
The fact that select two elements and of and to
maximize the value of ,
becomes the largest point.Input
An positive integer 
,
indicating there are test
cases.
For each test case, the first line contains three integers corresponding to
and .
The second line contains integers where for .
The sum of for
all cases would not be larger than .
,
indicating there are test
cases. For each test case, the first line contains three integers corresponding to
and .
The second line contains integers where for . The sum of
for
all cases would not be larger than .Output
The output contains exactly lines.
For each test case, you should output the maximum value of.
lines. For each test case, you should output the maximum value of
.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
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