Educational Codeforces Round 3 D
D. Almost Identity Permutations
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.
Examples
Input
4 1
Output
1
Input
4 2
Output
7
Input
5 3
Output
31
Input
5 4
Output
76
由于k只有4个数 所以在纸上算就不难算出
k == 1时 答案为1
k == 2时 答案为1+C(2,n)*1
k == 3时 答案为1+C(2,n)*1+C(3,n)*2
k == 4时 答案为1+C(2,n)*1+C(3,n)*2+C(4,n)*9
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 int main() { 5 ll n, k; 6 cin >> n >> k; 7 if(k == 1) printf("1\n"); 8 else if(k == 2) printf("%lld\n",1+n*(n-1)/2); 9 else if(k == 3) printf("%lld\n",1+n*(n-1)/2+n*(n-1)*(n-2)/3); 10 else if(k == 4) printf("%lld\n",1+n*(n-1)/2+n*(n-1)*(n-2)/3+n*(n-1)*(n-2)*(n-3)*3/8); 11 return 0; 12 }
