【代碼】 四法求逆元

本文應該真的差不多就只有代碼
不要臉地說一句其實要學的話只看代碼也能知道方法了，還是寫得比較結構化的

#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;
}