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 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
解题思路：
n个面的骰子 求每个面至少扔到一次的期望值
设dp[i]为已经扔了i个不同面的期望值 dp[n] = 0 求dp[0]
因为dp[i]为还需要扔i个不同的面 每次可能扔中已经扔过的面或者没有扔到过的面2中情况
所以dp[i] = (i/n)*dp[i] + (n-i)/n*dp[i+1] +1 等号2边都有dp[i]
移项得dp[i] = dp[i+1]+n/(n-i)
程序代码：
#include <cstdio> using namespace std; const int L=100010; double d[L]; int n; int main() { int t,Case=0; scanf("%d",&t); while(t--) { scanf("%d",&n); d[0]=0; for(int i=0;i<n;i++) d[i+1]=d[i]+n*1.0/(n-i); printf("Case %d: %.10f ",++Case,d[n]); } return 0; }