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
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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])); } }
