乘法逆元总结

参考：非常好的博客！

#include <bits/stdc++.h>
using namespace std;
long long n,p;
long long inv[3000001],fac[3000001],finv[3000001];
typedef long long ll;

void xianxing()//线性
{
	inv[1] = 1;
	cout << 1 << endl;
	for (int i = 2; i <= n; i++)
	{
		inv[i] = (p - p / i) * inv[p % i] % p;
		printf("%lld\n",inv[i]);
	}
}

void gcd(ll a,ll b,ll &x,ll &y)//扩展欧几里得
{
	if(!b)x = 1,y = 0;
	else gcd(b,a % b,y,x),y -= (a / b) * x;
}

void exou()
{
	for (int i = 1; i <= n; i++)
	{
		ll x = 0,y = 0;
		gcd(i,p,x,y);
		x = ( x % p + p) % p;
		printf("%lld\n",x);
	}
}

void fm()
{
	for (int i = 1; i <= n; i++)
	{
		ll ans = 1,base = i,pp = p - 2;
		while(pp)
		{
			if(pp & 1)
			{
				ans = (ans * base) % p;
			}
			pp >>= 1;
			base = (base * base) % p;
		}
		printf("%lld\n",ans);
	}
}

void factorial()
{
	fac[0] = 1;
	for (int i = 1; i <= n; i++)
	{
		fac[i] = (fac[i - 1] * i ) % p;
	}
	ll x = 0,y = 0;
	gcd(fac[n],p,x,y);
	x = (x % p + p) % p;
	finv[n] = x;
	for (int i = n - 1; i > 0; i--)
	{
		finv[i] = finv[i + 1] * (i + 1) % p;
	}
	for (int i = 1; i <= n; i++)
	{
		printf("%lld\n",finv[i] * fac[i - 1] % p);
	}
}

int main()
{
	cin >> n >> p;
	factorial();
	return 0;
}

没什么好说的，明天就csps了，再见