HDU 4652 Dice(概率dp)

题目的大意是：dp[i]表示当前在已经投掷出i个不相同/相同这个状态时期望还需要投掷多少次。

求得到连续的相同的期望：是1->m^n的前n项和。这里的解释可以看这里：http://www.matrix67.com/blog/archives/3638

另一种情况：dp[i]=1+(dp[1]+……+dp[i]+(m-i)*dp[i+1])/m

可以迭代系数也可以系数做差：

dp[n]=0
令d[i]=dp[i]-dp[i+1]
则有d[i]=m*dp[i-1]/(m-i),d[0]=1

Dice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 382    Accepted Submission(s): 248
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

60 6 10 6 30 6 51 6 21 6 41 6 6101 4534 251 1232 241 3213 151 4343 241 4343 91 65467 1231 43434 1001 34344 91 10001 151 1000000 2000

 

Sample Output

1.00000000043.0000000001555.0000000002.2000000007.60000000083.20000000025.58631582426.01599003715.17634116024.5410457699.027721917127.908330426103.9754552539.00349551515.0562044724731.706620396

 

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define eps 1e-9///#define M 1000100#define LL __int64///#define LL long long#define INF 0x7ffffff#define PI 3.1415926535898using namespace std;const int maxn = 1010000;double dp[maxn];int main(){    int T;    while(cin >>T)    {        while(T--)        {            int u, n, m;            cin >>u>>m>>n;            if(!u)            {                printf("%.9lf\n", (pow(m,n)-1.0)/(m*1.0-1.0));                continue;            }            double sum = 1.0;            double d = 1.0;            for(int i = 1; i < n; i++)            {                d = 1.0*m/(m-i)*d;                sum += d;            }            printf("%.9lf\n",sum);        }    }    return 0;}