HDU 4652 Dice （概率DP）

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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

field=problem&key=2013%20Multi-University%20Training%20Contest%205&source=1&searchmode=source" rel="nofollow" style="color:rgb(26,92,200);text-decoration:none;">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\n",solve() );
        }
    }
    return 0;
}