HDU - 3664 Permutation Counting 排列规律dp

Permutation Counting

Given a permutation a1, a2, … aN of {1, 2, …, N}, we define its E-value as the amount of elements where ai > i. For example, the E-value of permutation {1, 3, 2, 4} is 1, while the E-value of {4, 3, 2, 1} is 2. You are requested to find how many permutations of {1, 2, …, N} whose E-value is exactly k.

InputThere are several test cases, and one line for each case, which contains two integers, N and k. (1 <= N <= 1000, 0 <= k <= N). 
OutputOutput one line for each case. For the answer may be quite huge, you need to output the answer module 1,000,000,007.Sample Input

3 0
3 1

Sample Output

1
4

Hint

There is only one permutation with E-value 0: {1,2,3}, and there are four permutations with E-value 1: {1,3,2}, {2,1,3}, {3,1,2}, {3,2,1}
 
数列1-n，可以随意排列组合，求恰有k个a[i]>i的排列个数。

一道找规律的dp。看数据范围就知道不能暴力求解，但可以用暴力找出n较小的几种小数列排列数，发现规律。类似杨辉三角，就像两数和靠拢，于是可以发现状态转移方程f[i][j]=(((i+1)*f[i][j-1])%MOD+((j-i)*f[i-1][j-1])%MOD)%MOD

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<math.h>
#include<queue>
#include<set>
#include<stack>
#include<algorithm>
#include<vector>
#include<iterator>
#define MAX 1005
#define INF 0x3f3f3f3f
#define MOD 1000000007
using namespace std;
 
typedef long long ll;
 
ll f[MAX][MAX];
 
int main()
{
    int n,m,i,j;
    for(i=;i<=;i++){
        f[][i]=;
    }
    for(i=;i<=;i++){
        for(j=i+;j<=;j++){
            f[i][j]=(((i+)*f[i][j-])%MOD+((j-i)*f[i-][j-])%MOD)%MOD;
        }
    }
    while(~scanf("%d%d",&n,&m)){
        if(n==m){
            printf("0\n");
            continue;
        }
        printf("%lld\n",f[m][n]);
    }
    return ;
}