伯努利试验性质--概率期望

Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see all its faces at least once. Assume that the dice is fair, that means when you throw the dice, the probability of occurring any face is equal.

For example, for a fair two sided coin, the result is 3. Because when you first throw the coin, you will definitely see a new face. If you throw the coin again, the chance of getting the opposite side is 0.5, and the chance of getting the same side is 0.5. So, the result is

1 + (1 + 0.5 * (1 + 0.5 * ...))

= 2 + 0.5 + 0.52 + 0.53 + ...

= 2 + 1 = 3

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 105).

Output

For each case, print the case number and the expected number of times you have to throw the dice to see all its faces at least once. Errors less than 10-6 will be ignored.

Sample Input

5

1

2

3

6

100

Sample Output

Case 1: 1

Case 2: 3

Case 3: 5.5

Case 4: 14.7

Case 5: 518.7377517640

题意就是掷骰子直到所有面都出现的期望。

在伯努利试验中，成功的概率为p，若ξ表示出现首次成功时的试验次数，则ξ是离散型随机变量，它只取正整数，且有P(ξ=k)=(1-p)的(k-1)次方乘以p (k=1，2，…，0<p<1)，此时称随机变量ξ服从几何分布。它的期望为1/p，方差为(1-p)/(p的平方)

那么第一个面出现概率1，第二个新的一面概率是(n-1)/n，接着是(n-1)/n，他们的倒数是1、n/(n-1)、n/(n-2).......那么最后的期望就是n*(1+1/2+1/3+...）

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
    int t;
    cin>>t;
    for(int k=1;k<=t;k++)
    {
        int n;
        cin>>n;
        //int sum=(1+n)*n/2;
        double ans=0;
        for(int i=1;i<=n;i++)
        {
            ans+=1.0/i;
        }
        printf("Case %d: %.7lf\n",k,n*ans);
    }
}