[BZOJ 1420 && BZOJ 1319]

原根与指标QAQ。
我们要求的是$x^k = a (modp)$，给定$p，k，a，$求解$x$
具体做法是给两边同时取$log$
底数为原根$g$
则有$k * log_g x = log_g a(mod p)$
然后可以BSGS求出$log_g a$
Exgcd就可以解出$log_g x$的所有解。然后快速幂一下即可
注意$log_ga$一定要整除gcd(k, p-1)，否则无解
特判a=0的情况。
$x^A≡B (mod C)$
$Alnx≡lnB (mod φ(C))$

判定一个数是否为原根。
对于模数$p，p-1 = \prod{p_i^{a_i}}$
对于任意的$p_i$都有$g^{\frac{p-1}{p_i}}modp$不等于1

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <map>
#define maxn 1000010
using namespace std;
typedef long long ll;

ll p, k, a, g;
//g原根, g^x = a

ll P[maxn], cnt;

void pre_prime(){
    ll now = p - 1;
    for(ll i = 2; i * i <= now; i ++){
        while(now % i == 0){
            now /= i;
            if(P[cnt] != i)
                P[++ cnt] = i;
        }
    }
    if(now != 1) P[++ cnt] = now;
    sort(P + 1, P + 1 + cnt);
    cnt = unique(P + 1, P + 1 + cnt) - P - 1;
}

ll power_mod(ll a, ll b, ll md){
    ll ret = 1;
    while(b > 0){
        if(b & 1) ret = ret * a % md;
        b >>= 1;
        a = a * a % md;
    }return ret;
}

bool check(){
    for(int i = 1; i <= cnt; i ++)
        if(power_mod(g, (p - 1) / P[i], p) == 1)
            return false;
    return true;
}

void find_g(){
    for(g = 2; ; g ++)
        if(check())return;
}

map<ll, int> M;

ll BSGS(){
    if(a == 1)return 0;
    ll q = sqrt(p) + 1, now = 1;
    M[1] = 1;
    for(int i = 1; i <= q; i ++){
        now = now * g % p;
        if(now == a)return i;
        M[now] = i + 1;
    }
    ll inv = power_mod(now, p - 2, p), attempt = a;
    for(int i = 1; i <= q; i ++){
        attempt = attempt * inv % p;
        if(M.count(attempt))
            return M[attempt] - 1 + i * q;
    }return -1;
}

void Exgcd(ll a, ll b, ll& d, ll& x, ll& y){
    if(!b){d = a, x = 1, y = 0; return;}
    Exgcd(b, a%b, d, y, x); y -= x * (a / b);
}

ll ans[maxn];

int main(){
    scanf("%lld%lld%lld", &p, &k, &a);

    if(a == 0)return puts("1\n0"), 0;

    pre_prime();
    find_g();
    a = BSGS();//find ind_a
    ll gcd, x, y;
    Exgcd(k, p - 1, gcd, x, y);
    if(a % gcd)return puts("0"), 0;

    ll b = (p - 1) / gcd;
    while(x < 0) x += b;
    x = x * a / gcd % b;

    int amt = 0;
    for(; x < p - 1; x += b)
        ans[++ amt] = power_mod(g, x, p);
    sort(ans+1, ans+1+amt);

    printf("%d\n", amt);
    for(int i = 1; i <= amt; i ++)
        printf("%lld\n", ans[i]);
    return 0;
}