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 2007 on a line.
Sample Input
5 2
Sample Output
66
题意大概就是求n的全排列中，用大于小于号连接，其中小于号有k个的情况有多少种。注意，ac代码是%2009，不是2007
显然n的全排列有n！个，考虑下dp算法
状态转移是DP（n，k）=dp（n-1，k）*（k+1）+dp（n-1）（k-1）*（n-k）
n代表考虑n个数字的状态，k代表小于号的个数。
先作特殊情况，序列1 2 3 4 5 6，一共5个小于号，如果我要加入数字7，那么我有7种加法，并且只有一种加法会使小于号+1.
序列a1 a2 a3 a4 a5 ……a(n-1)，一共有n-1个数字，有k个小于号，那么我加入an有n种加法，其中会使小于号+1的有n-k种（就是加在大于号的位置上），其他的k+1种不会改变小于号的个数。
#include<stdio.h>
int main()
{
int n,k,i,j,s;
int dp[111][111];
for(i=0;i<=101;i++)
{
dp[i][0]=1;
dp[0][i]=0;
}
while(scanf("%d%d",&n,&k)!=EOF)
{
for(i=1;i<=101;i++)
for(j=1;j<=101;j++)
{
dp[i][j]=(dp[i-1][j]*(j+1)+dp[i-1][j-1]*(i-j))%2009;
}
printf("%d\n",dp[n][k]);
}
}
hdu 2583 permutation 动态规划