Dice (III) LightOJ - 1248

最新推荐文章于 2020-03-25 09:13:15 发布

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本文探讨了对于一个n面公平骰子，计算掷出所有面至少一次所需平均次数的问题。通过数学解析，我们了解到，对于一个两面的公平硬币，期望次数为3次。文章提供了详细的计算过程，并附带了一个C++实现的示例程序，用于计算不同面数的骰子掷出所有面至少一次的期望次数。

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

100

Sample Output

Case 1: 1

Case 2: 3

Case 3: 5.5

Case 4: 14.7

Case 5: 518.7377517640

#include<bits/stdc++.h>
using namespace std;
int main()
{
  int n,oo=0;
  int t;
  scanf("%d",&t);
  while(t--)
  {
      scanf("%d",&n);
      double k1=0;
      for(int i=1;i<=n;i++)
      {
          k1+=(n*1.0)/(i*1.0);
      }
      printf("Case %d: %.8lf\n",++oo,k1);
  }

}