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
题意：有一个n面的骰子，每次投掷每个面都是等概率出现，求所有面都出现的期望投掷次数。
思路：题目转化为假设现在有k面被投掷出来了，要使被投掷出来的面达到k+1个平均需要投掷多少次。求概率的话就是（从n-k中选1个）/（从n个中选一个）即n-k/n 期望为概率的倒数即n/(n-k) 故答案为n（1/1+1/2+1/3+1/4+……+1/n）
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 using namespace std;
6 int a[110];
7 double dp[110];
8 int n;
9 int casen,ca=1;
10 int main()
11 {
12 scanf("%d",&casen);
13 while(casen--)
14 {
15 scanf("%d",&n);
16 double sum=0;
17 for(int i=1;i<=n;i++)
18 {
19 sum+=(double)n/i;
20 }
21 printf("Case %d: %.8f
",ca++,sum);
22 }
23 }