HDU 4652 Dice (概率DP)

B - Dice

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4652

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≤10 ⁶) 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

 

先附個大牛的網址:http://blog.csdn.net/ten_three/article/details/18924459

 

第一個圖片中最後四，五行中的 1/m 應爲 m .... 圖片裏的不對

 

 

#include <cstdio>  
#include <iostream>  
#include <cstring>  
#include <algorithm>  
#include <cmath>  
using namespace std;  
double solve0(int m,int n)  
{  
    double ans=0;  
    for(int i=0;i<=n-1;i++)  
        ans+=pow(1.0*m,i);  
    return ans;  
}  
double solve1(int m,int n)  
{  
    double ans=0;  
    double tmp=1;  
    for(int i=1;i<=n;i++)  
    {  
        ans+=tmp;  
        tmp*=(m+0.0)/(m-i);  
    }  
    return ans;  
}  
int main()  
{  
    int t;  
    while(scanf("%d",&t)!=EOF)  
    {  
        while(t--)  
        {  
            int n,m,op;  
            scanf("%d%d%d",&op,&m,&n);  
            if(op==0)  
                printf("%.9lf\n",solve0(m,n));  
            else  
                printf("%.9lf\n",solve1(m,n));  
        }  
    }  
    return 0;  
}