Yuki 550 夏休みの思い出（１）

Problem Description:
Solve f(x) = x * x * x + A * x * x + B * x + C, given A, B, and C. Output the solutions in increasing order.

Constraints:
-1e18 < A, B, C < 1e18
-1e9 < min(sol) < max(sol) < 1e9
It is guaranteed that there are exactly 3 distinct solutions

Solution:
pekempey showed this clever solution:
We want to make use of the knowledge that the solution is bounded by integer range.
Consider deciding a mod = sqrt(1e9).
We can first complete search on this modulo equivalence relation:
g(x) = f(x) % mod = 0
It takes only O(mod = sqrt(1e9)) to solve for solutions under mod equivalence.
We will have a constant number (at least 1, less than 3) solutions, suppose one of them is x, then we know that we should search for values that are of form x + mod * i, for i in -mod..mod. Again this takes only O(mod) time.

Code:

#include <bits/stdc++.h>

template<typename T>
std::ostream& operator<<(std::ostream &os, const std::vector<T> &vec) {
  for (size_t i = 0; i < vec.size(); ++i) {
    os << vec[i] << " \n"[i + 1 == vec.size()];
  }
  return os;
}

std::vector<long long> solve_mod(long long a, long long b, long long c) {
  constexpr int maxs = 1e9;
  constexpr int mod = std::sqrt(maxs) + 10;
  std::vector<int> sol_mod;
  for (int i = 0; i < mod; ++i) {
    int aa = a % mod;
    int bb = b % mod;
    int cc = c % mod;
    int s = i * i % mod * i % mod;
    (s += aa * i % mod * i % mod) %= mod;
    (s += bb * i % mod) %= mod;
    (s += cc) %= mod;
    if (s == 0) sol_mod.emplace_back(i);
  }
  std::vector<long long> sol;
  for (int x : sol_mod) {
    for (int i = -mod; i < mod; ++i) {
      __int128 xx = x + static_cast<__int128>(i) * mod;
      __int128 s = xx * xx * xx + xx * xx * a + xx * b + c;
      if (s == 0) sol.emplace_back(xx);
    }
  }
  std::sort(sol.begin(), sol.end());
  return sol;
}

signed main() {
  std::ios::sync_with_stdio(false);
  long long A, B, C;
  std::cin >> A >> B >> C;
  std::cout << solve_mod(A, B, C) << std::endl;
  return 0;
}