HDU 2583 permutation

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HDU 2583 permutation

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
题意为告诉我们n个排列求其中含有k个'<'的排列数目mod2009是多少

思路：开始2了，都想到边上了给成功绕过去了
设dp[i][j]表示的是i个排列有k个<
设前N-1个的情况已经全部知道
则对于N，k个<来说就是有
1：前N-1个的k个将N放到最左边以及a<b和a<n>b，所以前dp[i-1][j]的情况乘以可以k的个数加1
2：对于前N-1个的k-1个有a>b
插入N:a<N>b增加了一个<故有
方程dp[i][j]=dp[i-1][j]*(j+1)+dp[i-1][j-1]*(i-j);

代码如下：#include<cstdio>#include<cstring>#include<iostream>using namespace std;int main(){ int dp[101][101]; int m,n; int i,j; memset(dp,0,sizeof(dp)); for(i=0;i<=100;i++) dp[i][0]=1; for(i=0;i<=100;i++) for(j=0;j<=100;j++) { if(j>=i) dp[i][j]=0; else if(j==i-1) dp[i][j]=1; else dp[i][j]=dp[i-1][j]*(j+1)+dp[i-1][j-1]*(i-j); if(dp[i][j]>2009) dp[i][j]%=2009; } while(scanf("%d%d",&m,&n)!=EOF) { printf("%d\n",dp[m][n]); } return 0;}</iostream></cstring></cstdio>