【代码】 四法求逆元

本文应该真的差不多就只有代码
不要脸地说一句其实要学的话只看代码也能知道方法了，还是写得比较结构化的

#include <cstdio>typedef long long ll;ll Qpow(ll x, ll n, ll p){    ll ans = 1;    while (n){    if (n & 1) ans = ans * x % p;    x = x * x % p; n >>= 1;    }return ans;}void Exgcd(ll a, ll p, ll &x, ll &y){    if (!p){x = 1, y = 0;}    else {Exgcd(p, a % p, y, x); y -= x * (a / p);}}ll GetExgcd(ll a, ll p){    ll res, tmp;    Exgcd(a, p, res, tmp);    return (res % p + p) % p;}struct Euler{    static const int maxn = 1e5;    bool book[maxn];    int prime[maxn], tot;    int phi[maxn];    void Work(int p){    tot = 0; phi[1] = 1;    for (int i = 2; i <= p; ++i){        if (!book[i]){        prime[++tot] = i;        phi[i] = i - 1;        }        for (int j = 1, nxt; j <= tot; ++i){        if ((nxt = i * prime[j]) > p) break;        book[nxt] = 1;        if (i % prime[j] == 0){            phi[nxt] = phi[i] * prime[j]; break;        }        phi[nxt] = phi[i] * (prime[j] - 1);        }    }    }    ll Get(ll a, ll p){    Work(p);    return Qpow(a, phi[p] - 1, p);    }};ll LinearRecurrence(ll a, ll p){    if (a == 1) return 1;    return LinearRecurrence(p % a, p) * (p - p / a);}ll Femat(ll a, ll p){    return Qpow(a, p - 2, p);}int main (){    ll a, p; scanf ("%lld%lld", &a, &p);    Euler res;    printf ("Femat: %lld\nExgcd: %lld\nEuler: %lld\n", Femat(a, p), GetExgcd(a, p), res.Get(a, p));    if (a <= p) printf ("LinearRecurrence: %lld", LinearRecurrence(a, p));      return 0;}