【集训DAY11】Calc【数学】

思路：

n!k肯定满足$2^{a_1}\ast 3^{a_2}\dots \dots \ast n^{a_n}$
我们枚举k来找a的规律
发现是前缀和
求k-1次前缀和，计算出每个a的值，最后再分解质因数来求

$code$

#include<iostream>
#include<cstdio>

#define ll long long

using namespace std;

const ll mod = 1e9 + 9;

ll n, k;
ll a[1010], s[1010];

int main() {
    
	scanf("%lld%lld", &n, &k);
	for(ll i = 1; i <= n; i ++) a[i] = 1;
	for(ll i = 2; i <= k; i ++)
		for(ll j = 1; j <= n; j ++) a[j] += a[j - 1], a[j] %= mod;
	for(ll i = 1, j = n; i < n; i ++, j --) {
    
		ll p = j;
		for(ll g = 2; g <= p; g ++) {
    
			ll tot = 0;
			while(p % g == 0) p /= g, tot ++, tot %= mod;
			s[g] += (tot * a[i]) % mod;
			s[g] %= mod;
		}
	}
	ll ans = 1;
	for(int i = 2; i <= n; i ++) {
    
		if(s[i] == 0) continue;
		ans = ans * (s[i] + 1) % mod;
	}
	printf("%lld", ans);
	return 0;
}