这道题我优化到最简了，可还是只能对3个样例先发出来吧，等我找到过的方法再修改

#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include <cstdio>
#define ll long long

inline ll read() {
    ll x = 0; char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
    return x;
}

inline void write(ll x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}

inline ll mul(ll a, ll b, ll mod) {
    return (__int128)a * b % mod;
}

inline ll pow_mod(ll a, ll b, ll mod) {
    ll res = 1;
    for (; b; b >>= 1, a = mul(a, a, mod))
        if (b & 1) res = mul(res, a, mod);
    return res;
}

int main() {
    ll p = read(), b = read(), n = read();
    
    ll phi = p - 1, c2 = 0, c3 = 0, t = phi;
    while (!(t & 1)) t >>= 1, c2++;
    while (t % 3 == 0) t /= 3, c3++;
    
    ll p2 = 1, p3 = 1;
    for (int i = 0; i < c2; i++) p2 <<= 1;
    for (int i = 0; i < c3; i++) p3 *= 3;
    
    auto solve_log = [&](ll g, ll h, ll mod, ll order) {
        ll res = 1;
        for (ll i = 0; i < order; i++) {
            if (res == h) return i;
            res = mul(res, g, mod);
        }
        return -1LL;
    };
    
    ll x1 = 0, x2 = 0;
    if (p2 > 1) {
        ll g1 = pow_mod(b, phi / p2, p);
        ll h1 = pow_mod(n, phi / p2, p);
        x1 = solve_log(g1, h1, p, p2);
        if (x1 == -1) {puts("-1"); return 0;}
    }
    if (p3 > 1) {
        ll g2 = pow_mod(b, phi / p3, p);
        ll h2 = pow_mod(n, phi / p3, p);
        x2 = solve_log(g2, h2, p, p3);
        if (x2 == -1) {puts("-1"); return 0;}
    }
    
    if (p2 == 1) write(x2);
    else if (p3 == 1) write(x1);
    else {
        for (ll i = x1; i < p2 * p3; i += p2) {
            if (i % p3 == x2) {
                write(i);
                return 0;
            }
        }
        puts("-1");
    }
    
    return 0;
}

#include <iostream>
using namespace std;

// 快速幂取模函数，计算 (base^exponent) % mod
int fastPower(int base, int exponent, int mod) {
    int result = 1;
    base %= mod;
    while (exponent > 0) {
        if (exponent & 1) {
            result = (result * base) % mod;
        }
        base = (base * base) % mod;
        exponent >>= 1;
    }
    return result;
}

int main() {
    int p, b, n;
    cin >> p >> b >> n;
    for (int l = 0; l < p; l++) {
        if (fastPower(b, l, p) == n) {
            cout << l << endl;
            return 0;
        }
    }
    cout << -1 << endl;
    return 0;
}

#include <iostream>
using namespace std;

// 计算最大公约数，用于判断两个数是否互质
long long gcd(long long a, long long b) {
    return b == 0? a : gcd(b, a % b);
}

// 快速幂取模函数，用于高效计算 (a^n) % m
long long powMod(long long a, long long n, long long m) {
    long long res = 1;
    a %= m;
    while (n > 0) {
        if (n & 1) {
            // 如果n的二进制最低位为1，累乘当前的a
            res = res * a % m;
        }
        a = a * a % m;
        n >>= 1;  // 右移一位，相当于n /= 2
    }
    return res;
}

int main() {
    long long p, b, n;
    cin >> p >> b >> n;
    // 判断b与p是否互质，如果不互质直接输出 -1
    if (gcd(b, p)!= 1) {
        cout << -1 << endl;
        return 0;
    }

    bool found = false;
    for (long long l = 1; l < p; l++) {
        if (powMod(b, l, p) == n) {
            cout << l << endl;
            found = true;
            break;
        }
    }
    if (!found) {
        cout << -1 << endl;
    }
    return 0;
}