Dice (III) （light oj 1248 概率dp）_given a dice with n sides, you have to find the…-优快云博客

本文链接：https://blog.youkuaiyun.com/Baiyi_destroyer/article/details/81258318

本文探讨了对于一个n面公平骰子，为了看到所有面至少一次所需的平均掷骰次数问题。通过数学推导，给出了计算期望掷骰次数的公式，并提供了一个C++实现的示例代码。

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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

题解：假设已经有k个点数已经出现过, 则掷下一次可能得到未出现过的点数, 期望为1, 也可能得到已出现过的点数, 概率为k / n, 若得到已出现的点数, 则继续掷色子, 所以期望

E = 1 + k / n (1 + k / n (1 + n / k * (...)))

= 1 + k / n + (k / n) ^ 2 + (k / n) ^ 3 + ...//等比数列求和，循环看做无限大（1- (k / n) ^ max）就可以看做1

= 1 + k / (n - k)

代码如下：

#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <ctime>
#define maxn 1007
#define N 100005
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define eps 0.000000001
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define Debug(x) cout<<x<<" "<<endl
using namespace std;
typedef long long ll;


int main()
{
    int t,cnt=1;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        double ans=0;
        for(int i=0;i<n;i++)
            ans+=1+1.0*i/(n-i);
        printf("Case %d: %lf\n",cnt++,ans);
    }
    return 0;
}