Problem p02596 (C++) — buggy submission, verdict: Time Limit Exceeded

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctype.h>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <math.h>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
#include <utility>
#include <vector>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define REP(i, n) FOR(i, 0, n - 1)
#define NREP(i, n) FOR(i, 1, n)
using ll = long long;
using pii = pair<int, int>;
using PII = pair<ll, ll>;
using piii = pair<pii, pii>;
using Pi = pair<int, pii>;
using Graph = vector<vector<int>>;
const int dx[4] = {0, -1, 1, 0};
const int dy[4] = {-1, 0, 0, 1};
bool check(int x, int y) {
  if (0 <= x && x < 55 && 0 <= y && y < 55)
    return true;
  else
    return false;
}
const ll INF = 1e9 + 7;
int gcd(int x, int y) {
  if (x < y)
    swap(x, y);
  if (y == 0)
    return x;
  return gcd(y, x % y);
}
void mul(ll a, ll b) { a = a * b % INF; }
using Graph = vector<vector<int>>;

// iの逆元mod inf
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;
}
const double PI = 3.14159265358979323846;
const int MAX = 510000;
ll fac[MAX], finv[MAX], inv[MAX];
// テーブルをつくる前処理
void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1; // mod pにおける1,2,,,nの逆元
  for (int i = 2; i < MAX; ++i) {
    fac[i] = fac[i - 1] * i % INF;
    inv[i] = INF - inv[INF % i] * (INF / i) % INF;
    finv[i] = finv[i - 1] * inv[i] % INF;
  }
}

ll 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] % INF) % INF;
}
double Euclidean_distance(double x1, double y1, double x2, double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}

// 素因数分解
map<int, int> prime_factor(int n) {
  map<int, int> res;
  for (int i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      ++res[i];
      n /= i;
    }
  }
  if (n != 1)
    res[n] = 1;
  return res;
}

ll powmod(ll a, ll k, ll mod) {
  ll ap = a, ans = 1;
  while (k) {
    if (k & 1) {
      ans *= ap;
      ans %= mod;
    }
    ap = ap * ap;
    ap %= mod;
    k >>= 1;
  }
  return ans;
}
ll invi(ll a, ll mod) { return powmod(a, mod - 2, mod); }
#define PI 3.14159265358979323846 // 円周率
// 逆元Aを足したときのmodで割った余りは
//+=invi(A)
// Yで割ることはY^(10^9+7)-2を掛けることは同値
///////////////////////////////////////

int main() {
  ll k;
  cin >> k;
  ll cnt = 1;
  ll now = 7;
  while (true) {
    if (now % k == 0) {
      cout << cnt << endl;
      return 0;
    }
    now *= 10;
    now += 7;
    cnt++;
    if (now % k == 0) {
      cout << cnt << endl;
      return 0;
    }
    if (now > k)
      break;
  }
  ll key;
  key = now % k;
  while (true) {
    ll a = key;
    key *= 10;
    key += 7;
    key %= k;
    cnt++;
    if (key == 0)
      break;
    if (a == key || cnt >= k) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << cnt << endl;
  return 0;
}


Error: TLE

Fixed submission:

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <cerrno>
#include <cfloat>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctype.h>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <math.h>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
#include <utility>
#include <vector>
using namespace std;
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define REP(i, n) FOR(i, 0, n - 1)
#define NREP(i, n) FOR(i, 1, n)
using ll = long long;
using pii = pair<int, int>;
using PII = pair<ll, ll>;
using piii = pair<pii, pii>;
using Pi = pair<int, pii>;
using Graph = vector<vector<int>>;
const int dx[4] = {0, -1, 1, 0};
const int dy[4] = {-1, 0, 0, 1};
bool check(int x, int y) {
  if (0 <= x && x < 55 && 0 <= y && y < 55)
    return true;
  else
    return false;
}
const ll INF = 1e9 + 7;
int gcd(int x, int y) {
  if (x < y)
    swap(x, y);
  if (y == 0)
    return x;
  return gcd(y, x % y);
}
void mul(ll a, ll b) { a = a * b % INF; }
using Graph = vector<vector<int>>;

// iの逆元mod inf
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;
}
const double PI = 3.14159265358979323846;
const int MAX = 510000;
ll fac[MAX], finv[MAX], inv[MAX];
// テーブルをつくる前処理
void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1; // mod pにおける1,2,,,nの逆元
  for (int i = 2; i < MAX; ++i) {
    fac[i] = fac[i - 1] * i % INF;
    inv[i] = INF - inv[INF % i] * (INF / i) % INF;
    finv[i] = finv[i - 1] * inv[i] % INF;
  }
}

ll 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] % INF) % INF;
}
double Euclidean_distance(double x1, double y1, double x2, double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}

// 素因数分解
map<int, int> prime_factor(int n) {
  map<int, int> res;
  for (int i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      ++res[i];
      n /= i;
    }
  }
  if (n != 1)
    res[n] = 1;
  return res;
}

ll powmod(ll a, ll k, ll mod) {
  ll ap = a, ans = 1;
  while (k) {
    if (k & 1) {
      ans *= ap;
      ans %= mod;
    }
    ap = ap * ap;
    ap %= mod;
    k >>= 1;
  }
  return ans;
}
ll invi(ll a, ll mod) { return powmod(a, mod - 2, mod); }
#define PI 3.14159265358979323846 // 円周率
// 逆元Aを足したときのmodで割った余りは
//+=invi(A)
// Yで割ることはY^(10^9+7)-2を掛けることは同値
///////////////////////////////////////

int main() {
  ll k;
  cin >> k;
  ll cnt = 1;
  ll now = 7;
  while (true) {
    if (now % k == 0) {
      cout << cnt << endl;
      return 0;
    }
    now *= 10;
    now += 7;
    cnt++;
    if (now % k == 0) {
      cout << cnt << endl;
      return 0;
    }
    if (now > k)
      break;
  }
  ll key;
  key = now % k;
  while (true) {
    ll a = key;
    key *= 10;
    key += 7;
    key %= k;
    cnt++;
    if (key == 0)
      break;
    if (a == key) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << cnt << endl;
  return 0;
}