Dice (III) LightOJ - 1248

Posted 啦啦啦

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Dice (III) LightOJ - 1248相关的知识,希望对你有一定的参考价值。

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

http://blog.csdn.net/u014552756/article/details/51462135

突然发现我的概率之菜

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int maxn=1e5+5;
 5 
 6 int kase,n;
 7 double dp[maxn];
 8 
 9 int main()
10 {   cin>>kase;
11     for(int t=1;t<=kase;t++){
12         cin>>n;
13         dp[1]=1.0;
14         for(int i=2;i<=n;i++) dp[i]=dp[i-1]+double(n)/(double(i)-1.0);
15         printf("Case %d: %lf\n",t,dp[n]);
16     }
17     return 0;
18 }

 

以上是关于Dice (III) LightOJ - 1248的主要内容,如果未能解决你的问题,请参考以下文章

F - Dice (III) LightOJ - 1248

LightOJ 1248 Dice (III)

lightoj 1248-G - Dice (III) (概率dp)

非原创LightOj 1248 - Dice (III)几何分布+期望

LightOJ 1248 - Dice (III) 给一个质地均匀的n的骰子, 求投掷出所有点数至少一次的期望次数。(概率)

LightOJ 1248 - Dice (III) 给一个质地均匀的n的骰子, 求投掷出所有点数至少一次的期望次数。(概率)