Problem p02602 (C++) — buggy submission, verdict: Runtime Error

/*~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=
*$* WRITER:kakitamasziru/OxOmisosiru *$*
~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=*/
#ifdef LOCAL_JUDGE
#define _GLIBCXX_DEBUG
#endif
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <cstdint>   // int64_t, int*_t
#include <iomanip>
#include <iostream> // cout, endl, cin
#include <limits>   //setprecision
#include <string>   // string, to_string, stoi
#include <tuple>    // tuple, make_tuple
#include <utility>  // pair, make_pair
#include <vector>   // vector
// #include <cstdio> // printf
#include <map>   // map
#include <queue> // queue, priority_queue
#include <set>   // set
// #include <unordered_set> //unordered_set
#include <bitset> // bitset
#include <cmath>  //abs,,,
#include <deque>  // deque
#include <math.h> //pow,,,
#include <stack>  // stack
#define endl "\\n";
using namespace std;

const long long INF = 100100100100100;
const int MOD = 1000000007;
const int inf = 1001001001;
typedef pair<int, int> P;

// Solve N^M. This, mod_pow use Iterative Square Method.
long long mod_pow(long long N, long long M) {
  if (M == 0)
    return 1;
  long long res = mod_pow((N * N) % MOD, M / 2);
  // 最下位ビット(*N)が1の時は単独でNをかける
  if (M & 1)
    res = (res * N) % MOD;

  return res %= MOD;
}

int gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }

int main() {
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);

  int N, K;
  cin >> N >> K;
  vector<long long> retu(N);
  for (int i = 0; i < N; i++)
    cin >> retu.at(i);
  queue<long long> Q;
  for (int i = 0; i < K; i++) {
    Q.push(retu.at(i));
  }
  for (int i = K; i < N; i++) {
    long long from = Q.front();
    long long to = retu.at(i);
    Q.push(to);
    Q.pop();
    if (from >= to) {
      cout << "No" << endl;
    } else {
      cout << "Yes" << endl;
    }
  }
}

Error: 0

Fixed submission:

/*~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=
*$* WRITER:kakitamasziru/OxOmisosiru *$*
~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=*/
#ifdef LOCAL_JUDGE
#define _GLIBCXX_DEBUG
#endif
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <cstdint>   // int64_t, int*_t
#include <iomanip>
#include <iostream> // cout, endl, cin
#include <limits>   //setprecision
#include <string>   // string, to_string, stoi
#include <tuple>    // tuple, make_tuple
#include <utility>  // pair, make_pair
#include <vector>   // vector
// #include <cstdio> // printf
#include <map>   // map
#include <queue> // queue, priority_queue
#include <set>   // set
// #include <unordered_set> //unordered_set
#include <bitset> // bitset
#include <cmath>  //abs,,,
#include <deque>  // deque
#include <math.h> //pow,,,
#include <stack>  // stack
#define endl "\\n";
using namespace std;

const long long INF = 100100100100100;
const int MOD = 1000000007;
const int inf = 1001001001;
typedef pair<int, int> P;

// Solve N^M. This, mod_pow use Iterative Square Method.
long long mod_pow(long long N, long long M) {
  if (M == 0)
    return 1;
  long long res = mod_pow((N * N) % MOD, M / 2);
  // 最下位ビット(*N)が1の時は単独でNをかける
  if (M & 1)
    res = (res * N) % MOD;

  return res %= MOD;
}

int gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }

int main() {
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);

  int N, K;
  cin >> N >> K;
  vector<long long> retu(N);
  for (int i = 0; i < N; i++)
    cin >> retu.at(i);
  queue<long long> Q;
  for (int i = 0; i < K; i++) {
    Q.push(retu.at(i));
  }
  for (int i = K; i < N; i++) {
    long long from = Q.front();
    long long to = retu.at(i);
    Q.pop();
    if (from >= to) {
      cout << "No" << endl;
    } else {
      cout << "Yes" << endl;
    }
  }
}