solution
{ i ∗ a i % p = b ， i < p − 1 , ( t − k ) ∗ a t % p = b ， i ≥ p − 1 ， i = ( p − 1 ) k + t . \begin{cases} i*a^i\%p=b，i<p-1, \\(t-k)*a^{t}\%p=b，i≥p-1，i=(p-1)k+t. \end{cases} { i∗ai%p=b，i<p−1,(t−k)∗at%p=b，i≥p−1，i=(p−1)k+t.
分别枚举 i ， t . i，t. i，t.
code
/*Siberian Squirrel*/
/*Cute JinFish*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const double PI = acos(-1), eps = 1e-8;
/*const int MOD = 998244353, r = 119, k = 23, g = 3;
const int MOD = 1004535809, r = 479, k = 21, g = 3;*/
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int M = 1e7 + 10, N= 2e6 + 10;
int sgn(double x) {
if(fabs(x) < eps) return 0;
return x < 0? -1: 1;
}
//inline int rnd(){static int seed=2333;return seed=(((seed*666666ll+20050818)%998244353)^1000000007)%1004535809;}
//double Rand() {return (double)rand() / RAND_MAX;}
ll a, b, p, x;
int inv[N];
void solve(ll res = 0) {
ll temp = a;
for(int i = 1; i < min(p - 1, x + 1); ++ i) {
if(1ll * i * temp % p == b) res ++;
temp = temp * a % p;
}
temp = 1;
for(int i = 0; i <= p - 2; ++ i) {
ll t = ((b * inv[temp] - i + p) % p) * inv[p-1] % p;
res += ((x - i) / (p - 1) - t) / p;
if((t && t * (p - 1) + i <= x)) res ++;
temp = temp * a % p;
}
cout << res << endl;
}
int main() {
ios::sync_with_stdio(false);cin.tie(0);cout.tie(nullptr);
// srand(time(0));
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
#endif
int o = 1;
// cin >> o;
while(o --) {
cin >> a >> b >> p >> x;
inv[1] = 1;
for(int i = 2; i < N; ++ i)
inv[i] = 1ll * (p - p / i) * inv[p % i] % p;
solve();
}
return 0;
}