Problem p02601 (C++)

Buggy submission (status: Time Limit Exceeded):
#include <bits/stdc++.h>
#include <string.h>

typedef long long int ll;
#define all(x) (x).begin(), (x).end()

using namespace std;

int nxt() {
  int x;
  cin >> x;
  return x;
}

ll nxtl() {
  ll x;
  cin >> x;
  return x;
}

void SieveOfEratosthenes(int n, vector<int> &primes) {
  // Create a boolean array "prime[0..n]" and initialize
  // all entries it as true. A value in prime[i] will
  // finally be false if i is Not a prime, else true.
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));

  for (int p = 2; p * p <= n; p++) {
    // If prime[p] is not changed, then it is a prime
    if (prime[p] == true) {
      // Update all multiples of p greater than or
      // equal to the square of it
      // numbers which are multiple of p and are
      // less than p^2 are already been marked.
      for (int i = p * p; i <= n; i += p)
        prime[i] = false;
    }
  }

  // Print all prime numbers
  for (int p = 2; p <= n; p++)
    if (prime[p])
      primes.push_back(p);
}

ll max(ll a, ll b) {
  if (a > b)
    return a;
  return b;
}

ll power(ll x, ll y, ll mod) {
  ll temp;
  if (y == 0)
    return 1;
  temp = power(x, y / 2, mod);
  if (y % 2 == 0)
    return (temp * temp) % mod;
  else
    return (((x * temp) % mod) * temp) % mod;
}

ll modPow(ll a, ll x, ll p) {
  // calculates a^x mod p in logarithmic time.
  ll res = 1;
  while (x > 0) {
    if (x % 2 != 0) {
      res = (res * a) % p;
    }
    a = (a * a) % p;
    x /= 2;
  }
  return res;
}

ll modInverse(ll a, ll p) {
  // calculates the modular multiplicative of a mod m.
  //(assuming p is prime).
  return modPow(a, p - 2, p);
}
ll modBinomial(ll n, ll k, ll p) {
  // calculates C(n,k) mod p (assuming p is prime).

  ll numerator = 1; // n * (n-1) * ... * (n-k+1)
  for (int i = 0; i < k; i++) {
    numerator = (numerator * (n - i)) % p;
  }

  ll denominator = 1; // k!
  for (int i = 1; i <= k; i++) {
    denominator = (denominator * i) % p;
  }

  // numerator / denominator mod p.
  return (numerator * modInverse(denominator, p)) % p;
}

ll f[200001];

ll pow(ll a, ll b, ll MOD) {
  ll x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD)
        x %= MOD;
    }
    y = (y * y);
    if (y > MOD)
      y %= MOD;
    b /= 2;
  }
  return x;
}

/* Modular Multiplicative Inverse
 Using Euler's Theorem
 a^(phi(m)) = 1 (mod m)
 a^(-1) = a^(m-2) (mod m) */
ll InverseEuler(ll n, ll MOD) { return pow(n, MOD - 2, MOD); }

ll C(ll n, ll r, ll MOD) {

  return (f[n] *
          ((InverseEuler(f[r], MOD) * InverseEuler(f[n - r], MOD)) % MOD)) %
         MOD;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int a = nxt(), b = nxt(), c = nxt();
  int k = nxt();
  while (k) {
    if (b <= a) {
      k--;
      b *= 2;
    } else if (c <= b) {
      k--;
      c *= 2;
    }
  }
  if (b > a && c > b)
    cout << "Yes\n";
  else
    cout << "No\n";

  return 0;
}


Fixed submission:
#include <bits/stdc++.h>
#include <string.h>

typedef long long int ll;
#define all(x) (x).begin(), (x).end()

using namespace std;

int nxt() {
  int x;
  cin >> x;
  return x;
}

ll nxtl() {
  ll x;
  cin >> x;
  return x;
}

void SieveOfEratosthenes(int n, vector<int> &primes) {
  // Create a boolean array "prime[0..n]" and initialize
  // all entries it as true. A value in prime[i] will
  // finally be false if i is Not a prime, else true.
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));

  for (int p = 2; p * p <= n; p++) {
    // If prime[p] is not changed, then it is a prime
    if (prime[p] == true) {
      // Update all multiples of p greater than or
      // equal to the square of it
      // numbers which are multiple of p and are
      // less than p^2 are already been marked.
      for (int i = p * p; i <= n; i += p)
        prime[i] = false;
    }
  }

  // Print all prime numbers
  for (int p = 2; p <= n; p++)
    if (prime[p])
      primes.push_back(p);
}

ll max(ll a, ll b) {
  if (a > b)
    return a;
  return b;
}

ll power(ll x, ll y, ll mod) {
  ll temp;
  if (y == 0)
    return 1;
  temp = power(x, y / 2, mod);
  if (y % 2 == 0)
    return (temp * temp) % mod;
  else
    return (((x * temp) % mod) * temp) % mod;
}

ll modPow(ll a, ll x, ll p) {
  // calculates a^x mod p in logarithmic time.
  ll res = 1;
  while (x > 0) {
    if (x % 2 != 0) {
      res = (res * a) % p;
    }
    a = (a * a) % p;
    x /= 2;
  }
  return res;
}

ll modInverse(ll a, ll p) {
  // calculates the modular multiplicative of a mod m.
  //(assuming p is prime).
  return modPow(a, p - 2, p);
}
ll modBinomial(ll n, ll k, ll p) {
  // calculates C(n,k) mod p (assuming p is prime).

  ll numerator = 1; // n * (n-1) * ... * (n-k+1)
  for (int i = 0; i < k; i++) {
    numerator = (numerator * (n - i)) % p;
  }

  ll denominator = 1; // k!
  for (int i = 1; i <= k; i++) {
    denominator = (denominator * i) % p;
  }

  // numerator / denominator mod p.
  return (numerator * modInverse(denominator, p)) % p;
}

ll f[200001];

ll pow(ll a, ll b, ll MOD) {
  ll x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD)
        x %= MOD;
    }
    y = (y * y);
    if (y > MOD)
      y %= MOD;
    b /= 2;
  }
  return x;
}

/* Modular Multiplicative Inverse
 Using Euler's Theorem
 a^(phi(m)) = 1 (mod m)
 a^(-1) = a^(m-2) (mod m) */
ll InverseEuler(ll n, ll MOD) { return pow(n, MOD - 2, MOD); }

ll C(ll n, ll r, ll MOD) {

  return (f[n] *
          ((InverseEuler(f[r], MOD) * InverseEuler(f[n - r], MOD)) % MOD)) %
         MOD;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int a = nxt(), b = nxt(), c = nxt();
  int k = nxt();
  while (k) {
    if (b <= a) {
      k--;
      b *= 2;
    } else {
      k--;
      c *= 2;
    }
  }
  if (b > a && c > b)
    cout << "Yes\n";
  else
    cout << "No\n";

  return 0;
}