hdu 2583 permutation

permutation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 141    Accepted Submission(s): 81

Problem Description

Permutation plays a very important role in Combinatorics. For example ,1 2 3 4 5 and 1 3 5 4 2 are both 5-permutations. As everyone's known, the number of n-permutations is n!. According to their magnitude relatives ,if we insert the sumbols "<" or ">"between every pairs of consecutive numbers of a permutations,we can get the permutations with symbols. For example,1 2 3 4 5 can be changed to 1<2<3<4<5, 1 3 5 4 2 can be changed to 1<3<5>4>2. Now it's yout task to calculate the number of n-permutations with k"<"symbol. Maybe you don't like large numbers ,so you should just geve the result mod 2009.

Input

Input may contai multiple test cases.
Each test case is a line contains two integers n and k .0<n<=100 and 0<=k<=100.
The input will terminated by EOF.

Output

The nonegative integer result mod 2009 on a line.

Sample Input

5 2

Sample Output

66

 

#include <iostream>



using namespace std;

const int m=2009;

int dp[110][110];



int main()

{

    int n,k,i,j;



    //预处理

    for(i=0;i<=100;i++)

    {

        dp[i][0]=1;

        dp[i][i]=0;

    }

    for(i=1;i<=100;i++)

        for(j=1;j<i;j++)

            if(j==i-1)//k==n-1时，结果为1

                dp[i][j]=1;

            else//状态转移方程dp[i][j]=(j+1)*dp[i-1][j]+(i-j)*dp[i-1][j-1];

                dp[i][j]=((j+1)*dp[i-1][j]%m+(i-j)*dp[i-1][j-1]%m)%m;



    while(cin>>n>>k)

        if(k>n)//n>=k时，结果必为0

            cout<<'0'<<endl;

        else

            cout<<dp[n][k]<<endl;



    return 0;

}