UVA - 10883 Supermean

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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 then given numbers in non-decreasing order. Now compute the average of each pair of adjacent numbers. This will give youn - 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 containingn (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
4110.421.0 2.231 2 351 2 3 4 5	Case #1: 10.400Case #2: 1.600Case #3: 2.000Case #4: 3.000

Problemsetter: Igor Naverniouk

題意：給出n個數，每相鄰的兩個數求平均數，將得到n-1個，然後再兩兩求平均數，依次類推直到最後一個，求這個數是多少

思路：係數的話很容易想到是楊輝三角的係數，但是因為n大太，所以為了防止溢出我們用log來儲存，每一項的通式是：∑i=0n?1C[n?1][i]?num[i]2n?1

然後就是在推組合數的同時對數處理

#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\n", cas++, ans);}return 0;}