HDOJ 4652 Dice

 

期望DP +数学推导

Dice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 337    Accepted Submission(s): 223
Special Judge

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

题意：一个m个面的筛子。两种询问：（1）平均抛多少次后使得最后n次的面完全一样；（2）平均抛多少次后使得最后n次的面完全不同？

思路：设dp[i]表示i次完全相同、不同时还需要抛的次数期望。

（1）下面首先讨论完全相同的情况。

（2）完全不同的情况：

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>

 using namespace std;

 int main()
 {
     int t,m,n,s;
     while(scanf("%d",&t)!=EOF)
     {
         while(t--)
         {
             scanf("%d%d%d",&s,&m,&n);
             if(s==)
             {
                 printf("%.9lf\n",(.-pow(m,n))/(-m));
             }
             else if(s==)
             {
                 double sum=.,tmp=.;
                 for(int i=;i<n;i++)
                 {
                     tmp=tmp*m/(m-i);
                     sum+=tmp;
                 }
                 printf("%.9lf\n",sum);
             }
         }
     }
     return ;
 }