D - Almost Identity Permutations /* Codeforces Round 888 D */

Posted 2020-10-14 不可爱的小可爱

D. Almost Identity Permutations

 

Description

    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 p_i?=?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.

Examples

Input

4 1

Output

1

Input

4 2

Output

7

Input

5 3

Output

31

Input

5 4

Output

76

题意：
    就是有一个集合里面是1~n个元素，但是这些元素的排列是随机的。问最多有多少个这样的排列，使得至少有n-k个 pi=i。
代码：（？？？？）

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5 ////先交一发别人的代码回头研究  = = 这个规律很膨胀啊 ！！！
 6     long long n,k,ar=1;
 7     cin>>n>>k;
 8     if(k>=2) ar+=n*(n-1)/2;
 9     if(k>=3) ar+=n*(n-1)*(n-2)/3;
10     if(k>=4) ar+=n*(n-1)*(n-2)*(n-3)*3/8;
11     cout<<ar;
12     return 0;
13 }

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