【Number Theory】Inverse element

use

Since division cannot be modulo, there is such a thing as a multiplicative inverse
(a/b)%p is equivalent to finding a∗(b’s inverse)%p

Fermat's little theorem

Since the modulus p is always a prime number in algorithmic competitions, Fermat's little theorem can be used:

b^(p−1)%p=1

It can be directly obtained that b^(p-2) is the inverse element of b in the sense of mod p

ll pow(ll a, ll n, ll p)    //快速幂 a^n % p
{
    
    
    ll ans = 1;
    while(n)
    {
    
    
        if(n & 1) ans = ans * a % p;
        a = a * a % p;
        n >>= 1;
    }
    return ans;
}

ll niyuan(ll a, ll p)   //费马小定理求逆元
{
    
    
    return pow(a, p - 2, p);
}

Extended Euclid

For using the extended Euclidean algorithm to find the inverse element, obviously, if bx%p=1, then bx+py=1 (p=0, solve x), directly use exgcd(b, p, x, y), Then (x%p+p)%p is the inverse of b ((x%p+p)%p is the smallest positive integer solution of x).

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

ll niyuan(ll a, ll b)   //求a对b取模的逆元
{
    
    
    ll x, y;
    exgcd(a, b, x, y);
    return (x + b) % b;
}

example

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define P 998244353
ll powmod(ll a,ll n, ll p)
{
    
    
	ll ans=1;
	while(n)
	{
    
    
		if(n&1)ans=ans*a%p;
		a=a*a%p;
		n>>=1;
	}
	return ans;
}
ll c[1000010];
ll f[1000010];
int main() {
    
    
	ios::sync_with_stdio(0);
	int n;
	cin >> n;
	for(int i=0;i<n;i++)
	{
    
    
		int k;
		cin>>k;
		for(int j=0;j<k;j++)
		{
    
    
			int e;
			cin>>e;
			c[e]++;
			f[e]=(f[e]+powmod(n,P-2,P)%P*powmod(k,P-2,P)%P)%P;
		}

	}
	ll ans=0;
	for(int i=1;i<=1e6;i++)
	{
    
    
		ans=(ans+c[i]*f[i]%P*powmod(n,P-2,P)%P)%P;
	}
	cout<<ans<<endl;
	return 0;
}

references

https://www.cnblogs.com/-citywall123/p/10673212.html