permutation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 167 Accepted Submission(s): 96
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.
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
题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=2583
1 #include<stdio.h> 2 int main() 3 { 4 int n,k; 5 int dp[102][102]; 6 for(int i=0;i<=101;i++) 7 { 8 dp[i][0]=1; dp[0][i]=0; 9 } 10 for(int i=1;i<=101;i++) 11 for(int j=1;j<=101;j++) 12 { 13 if(i-j==1) dp[i][j]=1; 14 else if(i-j>1) dp[i][j]=( dp[i-1][j]*(j+1) + dp[i-1][j-1]*(i-j) ) % 2009; 15 else dp[i][j]=0; 16 } 17 while(scanf("%d%d",&n,&k)!=EOF) 18 { 19 printf("%d ",dp[n][k]); 20 } 21 } 22 /*状态转移是: 23 DP（n，k）=dp（n-1，k）*（k+1）+dp（n-1）（k-1）*（n-k） 24 n代表考虑n个数字的状态，k代表小于号的个数。 25 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. 26 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， 27 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。 28 */