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 ;
 }