Codeforces 919E

求有多少个$n \in [1,x]$，满足

$$na^n \equiv b \pmod p$$

$$2 \leq p \leq 10^6 + 3,1 \leq a,b <p,1 \leq n \leq 10^{12}$$

保证$p$为质数。

将$b$移到等式左边，再将$n \in [1,p]$时的$b^{-1}n$存到哈希表中，枚举$a^1$到$a^{p-1}$，在哈希表中查有没有值是它的逆元，查到了后用CRT计算答案。

const int MAXN = 1000000 + 10;

int MOD;

IL int add(int a, int b) {
  a += b;
  return a >= MOD ? a - MOD : a;
}

IL int mul(int a, int b) {
  return 1LL * a * b % MOD;
}

IL int FastPow(int a, int p) {
  int result = 1;
  for (int base = a; p; p >>= 1, base = mul(base, base))
    if (p & 1)
      result = mul(result, base);
  return result;
}

int a[MAXN];
std::unordered_multimap < int, int > mp;

int main() {
  int a, b;
  LL x;
  std::cin >> a >> b >> MOD >> x;
  int base = FastPow(b, MOD - 2);
  For(i, 1, mymin(x, 1LL * (MOD - 1)))
    mp.insert(std::make_pair(mul(base, i), i));
  int cur = 1, inv = FastPow(MOD - 1, MOD - 2);
  LL answer = 0;
  For(i, 1, mymin(1LL * (MOD - 1), x)) {
    cur = mul(cur, a);
    auto cnt = mp.equal_range(FastPow(cur, MOD - 2));
    for (auto it = cnt.F; it != cnt.S; ++ it) {
      LL tot= (1LL * it->S * inv * (MOD - 1) + 1LL * (i == MOD - 1 ? 0 : i) * MOD) % (1LL * MOD * (MOD - 1));
      answer += x >= tot ? (x - tot) / (1LL * MOD * (MOD - 1)) + 1 : 0;
    }
  }
  std::cout << answer << '\n';
  return 0;
}