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;
}