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<<b>n<=50000). The next line will contain the ninput 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：给出n个数字, 要求你求出它们的supermean, supermean的定义是: n个数先两两相邻求平均值, 那么得到n-1个数, 已知循环做这件事, 直到剩下的数字只有1个, 那么这数就是supermean.

思路：该题在刘汝佳的紫书上《入门经典第二版》上有涉及（具体320页），没有书的也可查看我博客中的uva 1635题，和该题类似，比如假设5个数a1，a2，a3，a4，a5，最后退出的结果是a1+4a2+6a3+4a4+a5，聪明的小伙伴或许已经想到杨辉三角，就是杨辉三角，二项式系数c40到c44，然后去求即可，打不了二维表，对于每个n，打一个一维表，最后一步关键的高精度处理，这里采用取对数用double保存;

code:

#include <iostream>#include <cstdio>#include <vector>#include <cmath>using namespace std;

const int N=50011;double f[N];

void my_way(int n)  //对于每个n，打一个一维表{    //f[0]=1;    f[0]=f[n]=0;    for (int i=1;i<=n/2;i++)        f[n-i]=f[i]=f[i-1]+log10(n-i+1)-log10(i);    //高精度处理1}int main(){    int n,T,ca=1;    double ans,a;    scanf("%d",&T);    while (T--)    {        scanf("%d",&n);        ans=0;        my_way(n-1);        for (int i=0;i<n;i++)        {            scanf("%lf",&a);            ans+=1.0*pow(10,f[i]-(n-1)*log10(2))*a;  //高精度处理2        }        printf("Case #%d: %.3lf\n",ca++,ans);    }}