Problem p02690 (C++) — buggy submission, verdict: Time Limit Exceeded

#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>

// increment
#define rep_n(_1, _2, _3, NAME, ...) NAME
#define rep_2(i, n) for (int(i) = 0; (i) < (int)(n); ++(i))
#define rep_3(i, initial, n)                                                   \\
  for (int(i) = (int)(initial); (i) < (int)(n); ++(i))
#define rep(...) rep_n(__VA_ARGS__, rep_3, rep_2, rep_1)(__VA_ARGS__)
// decrement
#define rep_r_n(_1, _2, _3, NAME, ...) NAME
#define rep_r_2(i, n) for (int(i) = (int)(n); (i) > 0; (i)--)
#define rep_r_3(i, initial, n)                                                 \\
  for (int(i) = (int)(initial); (i) > (int)(n); (i)--)
#define rep_r(...) rep_r_n(__VA_ARGS__, rep_r_3, rep_r_2)(__VA_ARGS__)
// define type definition
using namespace std;
typedef long long ll;
using mpint = boost::multiprecision::cpp_int; // 多倍長整数（任意精度）
typedef pair<int, int> P;

// define construct variable
const double EPS = numeric_limits<double>::epsilon();
// 10^9 + 7
const ll MOD = ll(1e9) + 7;
const ll INF = ll(1e10);
const string ALPHABET = "abcdefghijklmnopqrstuvwxyz";

// define methods
// 最大公約数
ll gcd(const ll a, const ll b) { return b ? gcd(b, a % b) : a; }
// 最小公倍数
ll lcm(const ll a, const ll b) { return a / gcd(a, b) * b; }
// char -> int
int ctoi(char c) { return (int)(c - '0'); }
// 素因数分解
vector<pair<ll, ll>> prime_factorize(ll n) {
  vector<pair<ll, ll>> p;
  for (ll i = 2; i * i <= n; i++) {
    if ((n % i) != 0)
      continue;
    p.push_back(make_pair(i, 0));
    while ((n % i) == 0) {
      n /= i;
      p[i].second += 1;
    }
    p.push_back(make_pair(n, 0));
  }

  if (n != 1)
    p.push_back(make_pair(n, 1));
  return p;
}
// 組み合わせの整合性を保つ（後ほど解析する）
ll modinv(ll a) {
  ll b = MOD, u = 1, v = 0;
  while (b) {
    ll t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= MOD;
  if (u < 0) {
    u += MOD;
  }
  return u;
}
// 順列
ll permutation(const ll n, const ll m) {
  ll result = 1;
  for (ll i = n; i > n - m; i--) {
    result = (result * i) % MOD;
  }
  return result;
}

// 繰り返し n 乗法
ll repeat_square(const ll n, const ll x) {
  if (x == 0)
    return 1;
  ll res = repeat_square(n, x / 2);
  res = res * res % MOD;
  if ((x % 2) == 1)
    res = res * n % MOD;
  return res;
}
// 組み合わせ
ll combination(const ll n, const ll a) {
  ll x = 1, y = 1;
  for (ll i = 0; i < a; i++) {
    x = x * (n - i) % MOD;
    y = y * (i + 1) % MOD;
  }
  return x * repeat_square(y, MOD - 2) % MOD;
}

int main() {
  ll X;
  cin >> X;

  // for(int i = 0; i < 1001; i++) {
  // 	for(int j = -1000; j < i; j++) {
  // 		ll x1 = i * i * i * i * i;
  // 		ll x2 = j * j * j * j * j;
  // 		if ((x1 - x2) == X) {
  // 			cout << i << " " << j << endl;
  // 			return 0;
  // 		}
  // 	}
  // }
  for (int i = 0; i < 1001; i++) {
    for (int j = -1000; j < i; j++) {
      if (pow(i, 5) - pow(j, 5) == X) {
        cout << i << " " << j << endl;
        return 0;
      }
    }
  }
  return 0;
}


Error: TLE

Fixed submission:

#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>

// increment
#define rep_n(_1, _2, _3, NAME, ...) NAME
#define rep_2(i, n) for (int(i) = 0; (i) < (int)(n); ++(i))
#define rep_3(i, initial, n)                                                   \\
  for (int(i) = (int)(initial); (i) < (int)(n); ++(i))
#define rep(...) rep_n(__VA_ARGS__, rep_3, rep_2, rep_1)(__VA_ARGS__)
// decrement
#define rep_r_n(_1, _2, _3, NAME, ...) NAME
#define rep_r_2(i, n) for (int(i) = (int)(n); (i) > 0; (i)--)
#define rep_r_3(i, initial, n)                                                 \\
  for (int(i) = (int)(initial); (i) > (int)(n); (i)--)
#define rep_r(...) rep_r_n(__VA_ARGS__, rep_r_3, rep_r_2)(__VA_ARGS__)
// define type definition
using namespace std;
typedef long long ll;
using mpint = boost::multiprecision::cpp_int; // 多倍長整数（任意精度）
typedef pair<int, int> P;

// define construct variable
const double EPS = numeric_limits<double>::epsilon();
// 10^9 + 7
const ll MOD = ll(1e9) + 7;
const ll INF = ll(1e10);
const string ALPHABET = "abcdefghijklmnopqrstuvwxyz";

// define methods
// 最大公約数
ll gcd(const ll a, const ll b) { return b ? gcd(b, a % b) : a; }
// 最小公倍数
ll lcm(const ll a, const ll b) { return a / gcd(a, b) * b; }
// char -> int
int ctoi(char c) { return (int)(c - '0'); }
// 素因数分解
vector<pair<ll, ll>> prime_factorize(ll n) {
  vector<pair<ll, ll>> p;
  for (ll i = 2; i * i <= n; i++) {
    if ((n % i) != 0)
      continue;
    p.push_back(make_pair(i, 0));
    while ((n % i) == 0) {
      n /= i;
      p[i].second += 1;
    }
    p.push_back(make_pair(n, 0));
  }

  if (n != 1)
    p.push_back(make_pair(n, 1));
  return p;
}
// 組み合わせの整合性を保つ（後ほど解析する）
ll modinv(ll a) {
  ll b = MOD, u = 1, v = 0;
  while (b) {
    ll t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= MOD;
  if (u < 0) {
    u += MOD;
  }
  return u;
}
// 順列
ll permutation(const ll n, const ll m) {
  ll result = 1;
  for (ll i = n; i > n - m; i--) {
    result = (result * i) % MOD;
  }
  return result;
}

// 繰り返し n 乗法
ll repeat_square(const ll n, const ll x) {
  if (x == 0)
    return 1;
  ll res = repeat_square(n, x / 2);
  res = res * res % MOD;
  if ((x % 2) == 1)
    res = res * n % MOD;
  return res;
}
// 組み合わせ
ll combination(const ll n, const ll a) {
  ll x = 1, y = 1;
  for (ll i = 0; i < a; i++) {
    x = x * (n - i) % MOD;
    y = y * (i + 1) % MOD;
  }
  return x * repeat_square(y, MOD - 2) % MOD;
}

void calc(vector<pair<ll, ll>> l, ll b) {
  int ans = 0;
  rep(i, l.size()) { ans = l[i].first; }
  cout << ans << " " << b << endl;
  return;
}

int pf(ll n) {
  int count, ans = 0;
  bool is_ok = false;
  for (ll i = 2; i * i <= n; i++) {
    if ((n % i) != 0)
      continue;
    while ((n % i) == 0) {
      n /= i;
      count += 1;
    }
    if (count == 5) {
      is_ok = true;
      ans = i;
      break;
    }
    count = 0;
  }

  return ans;
}

int main() {
  ll X;
  cin >> X;
  ll a = 0;

  while (true) {
    ll b = 0;
    while (true) {
      ll x1 = a * a * a * a * a;
      ll x2 = b * b * b * b * b;
      if (abs(x1 - x2) > X)
        break;

      if ((x1 - x2) == X) {
        cout << a << " " << b << endl;
        return 0;
      }

      x2 *= -1;
      if ((x1 - x2) == X) {
        cout << a << " " << (b * -1) << endl;
        return 0;
      }
      b += 1;
    }
    a += 1;
  }
  return 0;
}