Favorite Dice（期望dp）

最新推荐文章于 2021-03-30 14:32:48 发布

原创最新推荐文章于 2021-03-30 14:32:48 发布 · 663 阅读

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期望dp和期望概率问题

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本文探讨了一个神奇的N面骰子，通过概率论和动态规划的方法，计算了要使每个面至少出现一次所需的平均投掷次数。文章提供了一种逆向DP的解题思路，并附带了C++代码实现。

BuggyD loves to carry his favorite die around. Perhaps you wonder why it's his favorite? Well, his die is magical and can be transformed into an N-sided unbiased die with the push of a button. Now BuggyD wants to learn more about his die, so he raises a question:

What is the expected number of throws of his die while it has N sides so that each number is rolled at least once?

Input

The first line of the input contains an integer t, the number of test cases. t test cases follow.

Each test case consists of a single line containing a single integer N (1 <= N <= 1000) - the number of sides on BuggyD's die.

Output

For each test case, print one line containing the expected number of times BuggyD needs to throw his N-sided die so that each number appears at least once. The expected number must be accurate to 2 decimal digits.

Example

Input:
2
1
12

Output:
1.00
37.24

题目大概:

给出一个n个面的筛子，问要使得每个面都朝上一次，需要投掷多少次。

思路：

期望dp一般逆推。

一般都是dp[i]表示已经出了i个面，距离n个面还差的期望。

第i次投掷可以落在前i个面，概率为i/n，也可以落在另外的（n-i）个面，概率是（n-i）/n。

每次投掷都百分百的增加投掷次数1.

dp[i]=i/n*dp[i]+（n-i）/n*dp[i+1]+1.

化简的dp[i]=dp[i+1]+n/(n-i).

代码：

#include <bits/stdc++.h>

using namespace std;

double dp[1100];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        double n;
        scanf("%lf",&n);
        dp[(int)n]=0;
        for(int i=n-1;i>=0;i--)
        {
            dp[i]=dp[i+1]+n/(n-(double)i);
        }
        printf("%lf\n",dp[0]);
    }
    return 0;
}