Almost Identity Permutations CodeForces

A permutation p of size n is an array such that every integer from 1 to n occurs exactly once in this array.

Let’s call a permutation an almost identity permutation iff there exist at least n - k indices i (1 ≤ i ≤ n) such that pi = i.

Your task is to count the number of almost identity permutations for given numbers n and k.

Input
The first line contains two integers n and k (4 ≤ n ≤ 1000, 1 ≤ k ≤ 4).

Output
Print the number of almost identity permutations for given n and k.

Example
Input
4 1
Output
1
Input
4 2
Output
7
Input
5 3
Output
31
Input
5 4
Output
76

这题说有index-k个数要求和index一致，那么其他和自己的index不一致，当然是错排公式了，但是这个错排最多只算到4个。。。
其他枚举组合们就是组合数了。。。用java了，组合数挺大的。

import java.math.BigInteger;import java.util.Scanner;public class Main{    static BigInteger fac[]=new BigInteger[1005];    static BigInteger w[]=new BigInteger[6];    static BigInteger get(int n,int m)    {        return fac[n].divide(fac[m]).divide(fac[n-m]);    }    public static void main(String[]args)    {        Scanner sc=new Scanner(System.in);        int n=sc.nextInt();        int k=sc.nextInt();        fac[0]=BigInteger.ONE;        for(int i=1;i<=n;i++)            fac[i]=fac[i-1].multiply(BigInteger.valueOf(i));        w[0]=BigInteger.ONE;        w[1]=BigInteger.ZERO;        w[2]=BigInteger.ONE;        for(int i=3;i<=5;i++)            w[i]=BigInteger.valueOf(i-1).multiply(w[i-1].add(w[i-2]));        BigInteger ans=BigInteger.ZERO;        for(int i=n-k;i<=n;i++)        {            ans=ans.add(get(n,i).multiply(w[n-i]));        }        System.out.println(ans);    }}