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≤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 10 9 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：

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 
 6 using namespace std;
 7 
 8 int main()
 9 {
10     int t,m,n,s;
11     while(scanf("%d",&t)!=EOF)
12     {
13         while(t--)
14         {
15             scanf("%d%d%d",&s,&m,&n);
16             if(s==0)
17             {
18                 printf("%.9lf\n",(1.-pow(m,n))/(1-m));
19             }
20             else if(s==1)
21             {
22                 double sum=1.,tmp=1.;
23                 for(int i=1;i<n;i++)
24                 {
25                     tmp=tmp*m/(m-i);
26                     sum+=tmp;
27                 }
28                 printf("%.9lf\n",sum);
29             }
30         }
31     }
32     return 0;
33 }