light oj 1248 第六周E题（期望）

E - 期望（经典问题）

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

 

Description

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 than10^-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

题解：n个面的骰子 求每个面至少扔到一次的期望值

比如给你6个面的 他的期望就是6*1+6*（1/2）+6*（1/3）+6*（1/4）+6*（1/5）+6*（1/6）

注意输出格式，小数点后十位。

#include<iostream>
#include<cstdio>
using namespace std;
int n,t,k=1;
double dp[100005];
int main()
{
    cin>>t;
    while(t--)
    {
       cin>>n;
       dp[n]=0;
       for(int i=n-1;i>=0;i--)
        dp[i]=dp[i+1]+(double)n/(i+1);
        printf("Case %d: %.10lf
",k++,dp[0]);
    }
    return 0;
}