2014
02-10

permutation

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 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.

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.

5 2

66

/*the number of n-permutations with k"<"symbol is F[n][k]
find the turning
F[i][0]=1,F[i][i-1]=1,i>j,i(n),j(k) 0<i<=100,0<=k<=100
以1 3 2 5 4为例，如果我们要插入6,如何办，分<和>分析可知
F[n+1][k] = F[n][k]*(k+1)+F[n][k-1]*(n-k+1)
F[1][0]=F[2][0]=F[3][0]=F[4][0]=.....=F[n][0]=1;
F[1][0]=F[2][1]=F[3][2]...............F[n][n-1]=1;
转移式:
F[n][k] = F[n-1][k]*(k+1)+F[n-1][k-1]*(n-k)*/

#include<iostream>
using namespace std;

int F[105][105];
int n,k;

void DP()
{
int i,j;
for(i=1; i<=100; ++i)
F[i][0] = F[i][i-1] = 1;
for(i=1; i<=100; ++i)
{
for(j=0; j<i-1; ++j)
{
F[i][j] = (F[i-1][j]*(j+1)+F[i-1][j-1]*(i-j)) % 2009;
}
}
}
int main()
{
DP();
while(cin >> n >> k)
{
cout << F[n][k] << endl;
}
return 0;
}