HDU 4652 Dice （概率DP）

版权声明：欢迎关注我的博客，本文为博主【炒饭君】原创文章，未经博主同意不得转载 https://blog.csdn.net/a1061747415/article/details/36685493

Dice

Problem Description

You have a dice with m faces, each face contains a distinct number. We assume when we tossing the dice, each face will occur randomly and uniformly. Now you have T query to answer, each query has one of the following form:
0 m n: ask for the expected number of tosses until the last n times results are all same.
1 m n: ask for the expected number of tosses until the last n consecutive results are pairwise different.

 

Input

The first line contains a number T.(1≤T≤100) The next T line each line contains a query as we mentioned above. (1≤m,n≤10⁶) For second kind query, we guarantee n≤m. And in order to avoid potential precision issue, we guarantee the result for our query will not exceeding 10⁹ in this problem.

 

Output

For each query, output the corresponding result. The answer will be considered correct if the absolute or relative error doesn't exceed 10^-6.

 

Sample Input

6 0 6 1 0 6 3 0 6 5 1 6 2 1 6 4 1 6 6 10 1 4534 25 1 1232 24 1 3213 15 1 4343 24 1 4343 9 1 65467 123 1 43434 100 1 34344 9 1 10001 15 1 1000000 2000

 

Sample Output

1.000000000 43.000000000 1555.000000000 2.200000000 7.600000000 83.200000000 25.586315824 26.015990037 15.176341160 24.541045769 9.027721917 127.908330426 103.975455253 9.003495515 15.056204472 4731.706620396

 

Source

2013 Multi-University Training Contest 5

 

题目大意：

m边形的骰子，问你出现连续同样（不同）n次须要掷的次数的数学期望。

解题思路：

利用递归方式的DP的思想推公式

（1）若询问为0，则：


dp[i] 记录的是已经连续i个同样，到n个同样同须要的次数的数学期望
dp[0]= 1+dp[1]
dp[1]= 1+( 1/m*dp[2]+(m-1)/m*dp[1])=1+(dp[2]+(m-1)*dp[1])/m;
dp[2]= 1+(dp[3]+(m-1)*dp[2])/m;
....................
dp[n]= 0


推出：

dp[i]   = 1 + ( (m-1)*dp[1] + dp[i+1] ) / m
dp[i+1] = 1 + ( (m-1)*dp[1] + dp[i+2] ) / m

因此。m*(dp[i+1]-dp[i])=(dp[i+2]-dp[i+1])

我们发现是等比数列

dp[0]-dp[1]=1;
dp[1]-dp[2]=m;
..........
dp[n-1]-dp[n]=m^(n-1)

累加，得：dp[0]-dp[n]=1+m+m^2+..........m^(n-1)=(1-m^n)/(1-m)

所以：dp[0]=(1-m^n)/(1-m);

（2）若询问为1，则：

 dp[0] = 1 + dp[1]
 dp[1] = 1 + (dp[1] + (m-1) dp[2]) / m
 dp[2] = 1 + (dp[1] + dp[2] + (m-2) dp[3]) / m
 dp[i] = 1 + (dp[1] + dp[2] + ... dp[i] + (m-i)*dp[i+1]) / m
dp[i+1]= 1 + (dp[1] + dp[2] + ... dp[i] + dp[i+1] + (m-i-1)*dp[i+1]) / m
 ...
 dp[n] = 0;

选出 dp[i] 和 dp[i+1] 这两行相减 得

dp[i] - dp[i+1] = (m-i-1)/m * (dp[i+1] - dp[i+2]);

因此  dp[i+1] - dp[i+2] = m/(m-i-1)*(dp[i]-dp[i+1]);

所以：
dp[0]-dp[1]=1;
dp[1]-dp[2]=1*m/(m-1);
dp[2]-dp[3]=1*m/(m-1)*m/(m-2);
..........

dp[n-1]-dp[n]=1*m/(m-1)*m/(m-2)*.......*m/(m-n+1);

累加得到答案

解题代码：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

inline double solve(){
    int op,m,n;
    scanf("%d%d%d",&op,&m,&n);
    double ans=0;
    if(op==0){
        for(int i=0;i<=n-1;i++){
            ans+=pow(1.0*m,i);
        }
    }else{
        double tmp=1.0;
        for(int i=1;i<=n;i++){
            ans+=tmp;
            tmp*=m*1.0/(m-i);
        }
    }
    return ans;
}

int main(){
    int t;
    while(scanf("%d",&t)!=EOF){
        while(t-- >0){
            printf( "%.9lf
",solve() );
        }
    }
    return 0;
}