Supermean

　　　　　　　　　　　　　　Supermean

Do you know how to compute the mean (or average) of n numbers? Well, that’s not good enough
for me. I want the supermean! “What’s a supermean,” you ask? I’ll tell you. List the n given numbers
in non-decreasing order. Now compute the average of each pair of adjacent numbers. This will give
you n − 1 numbers listed in non-decreasing order. Repeat this process on the new list of numbers until
you are left with just one number - the supermean. I tried writing a program to do this, but it’s too
slow. :-( Can you help me?
Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line
containing n (0 < n ≤ 50000). The next line will contain the n input numbers, each one between −1000
and 1000, in non-decreasing order.
Output
For each test case, output one line containing ‘Case #x:’ followed by the supermean, rounded to 3
fractional digits.

Sample Input
4
1
10.4
2
1.0 2.2
3
1 2 3
5
1 2 3 4 5

Sample Output
Case #1: 10.400
Case #2: 1.600
Case #3: 2.000
Case #4: 3.000

题意：这个题，，，读题以为就是求平均数了，而且测试数据还是那样的尽人意。。。

　　实际上这个题是对每两个连续的数求平均数，，，直到求到只剩一个数字，，，他就是答案

tip：半天没想出来，搜题解了（不搜估计我都不知道自己题意都错了）；注意有三点

　　1.每个数字的使用次数就是C（n-1，k）（k从0到n-1），把他们全加起来就是sum，当然要除以2^n-1;即C(n-1, k)*A[k]/(2^n-1)

　　2.测试数据会超出范围，咱来取对数处理；即e^(ln C(n-1, k)*A[k]/(2^n-1) )==> e^( ln C(n-1,k) + ln A[k] - (n-1)*ln2 );

　　3.C组合数，你该咋算，模拟？我用了，超时，有这么个规律 C(n, k) = C(n, k-1)*(n-k+1)/k;但是咋用就看你了，比如下面是

　　这样子用的：：：：lnc+=log(n-i+1)-log(i)；

好了，看代码吧。。。。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>

using namespace std;

int n;
double value;

int main()
{
    int caseNum,num=1;
    scanf("%d",&caseNum);
    while(caseNum--)
    {
        scanf("%d",&n);
        n=n-1;
        double ln2=n*log(2),lnc=0;
        double ans=0.0;
        scanf("%lf", &value);

        if(value>0)///C（n，0）的
            ans+=exp(log(value)+lnc-ln2);
        else
            ans-=exp(log(-value)+lnc-ln2);

        for(int i=1;i<=n;++i)
        {
            scanf("%lf",&value);
            lnc+=log(n-i+1)-log(i);
            if(value>0)
                ans+=exp(log(value)+lnc-ln2);
            else
                ans-=exp(log(-value)+lnc-ln2);
        }

        printf("Case #%d: %.3lf
",num++,ans);
    }

    return 0;
}