UVA

Description

Problem F
Supermean
Time Limit: 2 second

"I have not failed. I've just found 10,000 ways that won't work."

Thomas Edison

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	Sample Output
4 1 10.4 2 1.0 2.2 3 1 2 3 5 1 2 3 4 5	Case #1: 10.400 Case #2: 1.600 Case #3: 2.000 Case #4: 3.000

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Problemsetter: Igor Naverniouk

题意：给出n个数，每相邻的两个数求平均数。将得到n-1个，然后再两两求平均数，依次类推直到最后一个。求这个数是多少

思路：系数的话非常easy想到是杨辉三角的系数，可是由于n大太，所以为了防止溢出我们用log来储存。每一项的通式是：∑i=0n−1C[n−1][i]∗num[i]2n−1

然后就是在推组合数的同一时候对数处理

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 50005;

double C[maxn], num[maxn];

int main() {
	int t, n, cas = 1;
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		for (int i = 0; i < n; i++)
			scanf("%lf", &num[i]);

		double ans = 0.0, tmp = log10(1);
		for (int i = 0; i < n; i++) { 
			if (i) 
				tmp = tmp + log10(n-i) - log10(i);
			
			if (num[i] < 0)
				ans -= pow(10, tmp + log10(-num[i]) - (n-1)*log10(2));
			else ans += pow(10, tmp + log10(num[i]) - (n-1)*log10(2));
		}

		printf("Case #%d: %.3lf
", cas++, ans);
	}
	return 0;
}