[LightOJ 1248] Dice (III)

标签：ace   一个   close   没有   scan   because   color   cstring   --   

Dice (III)

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

题目的意思是,给你一个均匀的N面体,问你期望投掷多少次,使每一面至少出现过1次.

我们设E[i]表示还有i面没有出现,则:

E[i]=未成功掷出新的一面的期望+成功掷出新的一面的期望=E[i]*i/n+(E[i+1]*(n-i)/n+1)

整理得,E[i]=E[i+1]+n/(n-i).那么边界E[n]=0,我们求出E[0]就行了.

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<cmath>
 5 using namespace std;
 6 int n;
 7 int main(){
 8     int T; scanf("%d",&T);
 9     for (int Ts=1; Ts<=T; Ts++){
10         scanf("%d",&n); double ans=0;
11         for (int i=n-1; i>=0; i--) ans+=(double)n/(double)(n-i);
12         printf("Case %d: %.10f\n",Ts,ans);
13     }
14     return 0;
15 }

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[LightOJ 1248] Dice (III)

标签：ace   一个   close   没有   scan   because   color   cstring   --   

原文地址：http://www.cnblogs.com/whc200305/p/7204834.html