Problem p02660 (C++)

Buggy submission (status: Time Limit Exceeded):
#include <algorithm>
#include <cassert>
#include <cmath>
#include <deque>
#include <functional>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <vector>
using namespace std;
typedef long long ll;
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
const int INF = 1 << 29;
#define LNF __LONG_LONG_MAX__
const ll MOD = 1000000007;
const int MAX = 200000;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}

// 二項係数計算
long long COM(int n, int k) {
  if (n < k)
    return 0;
  if (n < 0 || k < 0)
    return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
// 法mでのaの逆元を計算
long long modinv(long long a, long long m) {
  long long b = m, u = 1, v = 0;
  while (b) {
    long long t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0)
    u += m;
  return u;
}
// a^n mod を計算する
long long modpow(long long a, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1)
      res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
// 最大公約数
int gcd(int x, int y) { return (x % y) ? gcd(y, x % y) : y; }
// 最小公倍数
int lcm(int x, int y) { return x / gcd(x, y) * y; }
// nが大きく,kが10^7程度の時の二項係数
long long com[200005];
void combination(long long n, long long k) {
  com[0] = 1;
  for (int i = 1; i <= k; i++) {
    com[i] = (com[i - 1] * (n - i + 1) % MOD) * modinv(i, MOD) % MOD;
  }
}

int main() {
  ll N;
  cin >> N;
  ll out = 0;
  for (int i = 2; i * i <= N; i++) {
    if (N % i != 0)
      continue;
    ll c = 0;
    ll o = 1;
    while (N % i == 0) {
      N /= i;
      c++;
      if (o == c) {
        o++;
        c = 0;
      }
    }
    // cerr << o << "\t" << c << endl;
    o--;
    out += o;
  }
  if (N != 1) {
    out++;
  }
  cout << out << endl;
  return 0;
}


Fixed submission:
#include <algorithm>
#include <cassert>
#include <cmath>
#include <deque>
#include <functional>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <vector>
using namespace std;
typedef long long ll;
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
const int INF = 1 << 29;
#define LNF __LONG_LONG_MAX__
const ll MOD = 1000000007;
const int MAX = 200000;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}

// 二項係数計算
long long COM(int n, int k) {
  if (n < k)
    return 0;
  if (n < 0 || k < 0)
    return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
// 法mでのaの逆元を計算
long long modinv(long long a, long long m) {
  long long b = m, u = 1, v = 0;
  while (b) {
    long long t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0)
    u += m;
  return u;
}
// a^n mod を計算する
long long modpow(long long a, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1)
      res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}
// 最大公約数
int gcd(int x, int y) { return (x % y) ? gcd(y, x % y) : y; }
// 最小公倍数
int lcm(int x, int y) { return x / gcd(x, y) * y; }
// nが大きく,kが10^7程度の時の二項係数
long long com[200005];
void combination(long long n, long long k) {
  com[0] = 1;
  for (int i = 1; i <= k; i++) {
    com[i] = (com[i - 1] * (n - i + 1) % MOD) * modinv(i, MOD) % MOD;
  }
}

int main() {
  ll N;
  cin >> N;
  ll out = 0;
  for (ll i = 2; i * i <= N; i++) {
    if (N % i != 0)
      continue;
    ll c = 0;
    ll o = 1;
    while (N % i == 0) {
      N /= i;
      c++;
      if (o == c) {
        o++;
        c = 0;
      }
    }
    // cerr << o << "\t" << c << endl;
    o--;
    out += o;
  }
  if (N != 1) {
    out++;
  }
  cout << out << endl;
  return 0;
}