Codeforces #737Div2A. A. Ezzat and Two Subsequences（模拟）

Ezzat has an array of n integers (maybe negative). He wants to split it into two non-empty subsequences a and b, such that every element from the array belongs to exactly one subsequence, and the value of f(a)+f(b) is the maximum possible value, where f(x) is the average of the subsequence x.

A sequence x is a subsequence of a sequence y if x can be obtained from y by deletion of several (possibly, zero or all) elements.

The average of a subsequence is the sum of the numbers of this subsequence divided by the size of the subsequence.

For example, the average of [1,5,6] is (1+5+6)/3=12/3=4, so f([1,5,6])=4.

Input

The first line contains a single integer t (1≤t≤103)— the number of test cases. Each test case consists of two lines.

The first line contains a single integer n (2≤n≤105).

The second line contains n integers a1,a2,…,an (−109≤ai≤109).

It is guaranteed that the sum of n over all test cases does not exceed 3⋅105.

Output

For each test case, print a single value — the maximum value that Ezzat can achieve.

Your answer is considered correct if its absolute or relative error does not exceed 10−6.

Formally, let your answer be a, and the jury’s answer be b. Your answer is accepted if and only if |a−b|max(1,|b|)≤10−6.

样例输入

4

3

3 1 2

3

-7 -6 -6

3

2 2 2

4

17 3 5 -3

样例输出

4.500000000

-12.500000000

4.000000000

18.666666667

题意

将一个给定的数组分成两个数组，将每个数组的和除以这个数组的个数，将两个数相加。

找一个最合理的分配方式使相加后的和最大，输出这个最大的和;

这个简单，就是把最大的拿出来，加上前面的平均数；

如果有测试点2，超限的，提醒一下不要用double，在数组定义里面

听懂了记得给个赞鼓励一下，码字不易，用爱发电。

上ac代码。

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学习算法

#include<bits/stdc++.h>
using namespace std;
int main()
{
  int t;
  cin>>t;
  while(t--)
  {
  long long sum=0;int m;
    long long  c[100000];//这里如果用double会提示测试点2不能通过，甚至可能出现错误
  scanf("%d",&m);
    for(int i=1;i<=m;i++)
   scanf("%lld",&c[i]);
    sort(c+1,c+m+1);    
    for(int i=1;i<=m-1;i++)
    sum+=c[i];
printf("%.9f\n",(1.0*sum/(m-1)+c[m]));
  }
 }