Uva 10883 - Supermean 解题报告（对数）

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 InputSample 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

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    解题报告： 本人自己WA了好几次，然后看解题报告的。实在很难想到用对数。

    公式很容易推出来，每一项为C(n-1, i)*a[i]/2^(n-1)。难点在于n很大，以至于C(n-1, i)和2^(n-1)都很大，无法直接计算。

    方法呢也很简单，先用对数处理，再做幂运算还原。对数可以递推。

    代码可以改进，可以不开数组，也不用每次都计算log10(2^(n-1))，懒得改了，代码如下：

#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>using namespace std;const int maxn = 55555;double c[maxn], a[maxn];int cas=1;void work(){    int n;    scanf("%d",&n);    for(int i=0;i<n;i++)        scanf("%lf",a+i);    c[0]=0;    for(int i=1;i<n;i++)        c[i]=c[i-1]+log10(n-i)-log10(i);    double ans=0;    for(int i=0;i<n;i++)        ans+=pow(10.0, c[i]-(n-1)*log10(2))*a[i];    printf("Case #%d: %.3lf\n", cas++, ans);}int main(){    int T;    scanf("%d",&T);    while(T--)        work();}