Problem p02615 (C++) — buggy submission, verdict: Runtime Error

#include <bits/stdc++.h>
#include <iostream>
using namespace std;
#define int long long
#define double long double
#define endl "\\n"
#define pb push_back
#define PI 3.1415926535897932384626433832795l
#define F first
#define S second
#define mp make_pair
#define f(i, n) for (int i = 0; i < n; i++)
#define fastio                                                                 \\
  ios::sync_with_stdio(0);                                                     \\
  cin.tie(0);                                                                  \\
  cout.tie(0);
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define gcd(a, b) __gcd((a), (b))
#define fill(a, value) memset(a, value, sizeof(a));
#define minn(v) *min_element(v.begin(), v.end());
#define maxx(v) *max_element(v.begin(), v.end());
#define print(x) cout << (x) << endl;
#define sum(v) +x accumulate(v.begin(), v.end(), x);
typedef pair<int, int> pii;
typedef vector<int> vi;
/*
max and min value in a set
max_element = *(my_set.rbegin());
min_element = *my_set.begin();
*/
// x.substr(start, length);
/*
Create a vector containing n
vectors of size m.
---vector<vector<int> > vec( n , vector<int> (m, 0));
*/

/*left shifting an integer “x” with an integer “y” (x<<y)
    is equivalent to multiplying x with 2^y (2 raise to power y).*/
/*Similarly right shifting (x>>y) is equivalent
    to dividing x with 2^y.
*/

struct type {
  int k, i, j;
};
// to check if a no is prime or not?
bool isPrime(int n) {
  if (n <= 1)
    return false;
  if (n <= 3)
    return true;
  if (n % 2 == 0 || n % 3 == 0)
    return false;

  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0)
      return false;

  return true;
}

// minimum subaaray sum of size k-sliding window approach
int findSubarray(int arr[], int n, int k) {
  int window_sum = 0;
  int min_window = INT_MAX;
  int last = 0;
  for (int i = 0; i < n; i++) {
    window_sum += arr[i];
    if (i + 1 >= k) {
      if (min_window > window_sum) {
        min_window = window_sum;
        last = i;
      }
      window_sum -= arr[i + 1 - k];
    }
  }
  int sum = 0;
  for (int i = last - k + 1; i <= last; i++) {
    sum += arr[i];
  }
  return sum;
}
// finds the next permutation to given sequence of nos
int nextpermutation(vi v) {
  vi t = v;
  sort(all(t));
  int x = 0;
  while (true) {
    x++;
    if (t == v) {
      return x;
    }
    next_permutation(all(t));
  }
}
vector<int> factors(int n) {
  // Note that this loop runs till square root
  vi ans;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n == i * i)
        ans.pb(i);
      else {
        ans.pb(i);
        if (i != n / i)
          ans.pb(n / i);
      }
    }
  }
  sort(all(ans));
  return ans;
}

// Recursive C++ program to check if a string is subsequence of another string
bool isSubSequence(char str1[], char str2[], int m, int n) {
  // Base Cases
  if (m == 0)
    return true;
  if (n == 0)
    return false;

  // If last characters of two strings are matching
  if (str1[m - 1] == str2[n - 1])
    return isSubSequence(str1, str2, m - 1, n - 1);

  // If last characters are not matching
  return isSubSequence(str1, str2, m, n - 1);
}
void findDivisors(int n) {

  int div[n + 1];
  memset(div, 0, sizeof div);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j * i <= n; j++)
      div[i * j]++;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    ans += i * div[i];
  cout << ans << endl;
}
int digits(int n) {
  int res = 0;
  while (n > 0) {
    res += n % 10;
    n /= 10;
  }
  return (res);
}
int power(int x, int y, int p) {
  int res = 1; // Initialize result

  x = x % p; // Update x if it is more than or
  // equal to p

  while (y > 0) {
    // If y is odd, multiply x with result
    if (y & 1)
      res = (res * x) % p;

    // y must be even now
    y = y >> 1; // y = y/2
    x = (x * x) % p;
  }
  return res;
}

// Returns n^(-1) mod p
int modInverse(int n, int p) { return power(n, p - 2, p); }

// Returns nCr % p using Fermat's little
// theorem.
int nCr(int n, int r, int p) {
  // Base case
  if (r == 0)
    return 1;
  int fac[n + 1];
  fac[0] = 1;
  for (int i = 1; i <= n; i++)
    fac[i] = (fac[i - 1] * i) % p;

  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
double dist(int x1, int y1, int x2, int y2) {
  double a = (x1 - x2) * (x1 - x2);
  double b = (y1 - y2) * (y1 - y2);

  return sqrt(a + b);
}
signed main() {
  fastio;
  cout << fixed << setprecision(12);
  int n;
  cin >> n;
  vi a(n);
  f(i, n) cin >> a[i];
  sort(rall(a));
  int ans = 0;
  ans += a[0];
  int x = 1;
  int y = 0;
  for (int i = 2; i < n; i++) {
    if (y == 2) {
      x += 1;
      y = 0;
    }
    ans += a[x];
    y += 1;
  }
  print(ans)

      return 0;
}

Error: 0

Fixed submission:

#include <bits/stdc++.h>
#include <iostream>
using namespace std;
#define int long long
#define double long double
#define endl "\\n"
#define pb push_back
#define PI 3.1415926535897932384626433832795l
#define F first
#define S second
#define mp make_pair
#define f(i, n) for (int i = 0; i < n; i++)
#define fastio                                                                 \\
  ios::sync_with_stdio(0);                                                     \\
  cin.tie(0);                                                                  \\
  cout.tie(0);
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define gcd(a, b) __gcd((a), (b))
#define fill(a, value) memset(a, value, sizeof(a));
#define minn(v) *min_element(v.begin(), v.end());
#define maxx(v) *max_element(v.begin(), v.end());
#define print(x) cout << (x) << endl;
#define sum(v) +x accumulate(v.begin(), v.end(), x);
typedef pair<int, int> pii;
typedef vector<int> vi;
/*
max and min value in a set
max_element = *(my_set.rbegin());
min_element = *my_set.begin();
*/
// x.substr(start, length);
/*
Create a vector containing n
vectors of size m.
---vector<vector<int> > vec( n , vector<int> (m, 0));
*/

/*left shifting an integer “x” with an integer “y” (x<<y)
    is equivalent to multiplying x with 2^y (2 raise to power y).*/
/*Similarly right shifting (x>>y) is equivalent
    to dividing x with 2^y.
*/

struct type {
  int k, i, j;
};
// to check if a no is prime or not?
bool isPrime(int n) {
  if (n <= 1)
    return false;
  if (n <= 3)
    return true;
  if (n % 2 == 0 || n % 3 == 0)
    return false;

  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0)
      return false;

  return true;
}

// minimum subaaray sum of size k-sliding window approach
int findSubarray(int arr[], int n, int k) {
  int window_sum = 0;
  int min_window = INT_MAX;
  int last = 0;
  for (int i = 0; i < n; i++) {
    window_sum += arr[i];
    if (i + 1 >= k) {
      if (min_window > window_sum) {
        min_window = window_sum;
        last = i;
      }
      window_sum -= arr[i + 1 - k];
    }
  }
  int sum = 0;
  for (int i = last - k + 1; i <= last; i++) {
    sum += arr[i];
  }
  return sum;
}
// finds the next permutation to given sequence of nos
int nextpermutation(vi v) {
  vi t = v;
  sort(all(t));
  int x = 0;
  while (true) {
    x++;
    if (t == v) {
      return x;
    }
    next_permutation(all(t));
  }
}
vector<int> factors(int n) {
  // Note that this loop runs till square root
  vi ans;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n == i * i)
        ans.pb(i);
      else {
        ans.pb(i);
        if (i != n / i)
          ans.pb(n / i);
      }
    }
  }
  sort(all(ans));
  return ans;
}

// Recursive C++ program to check if a string is subsequence of another string
bool isSubSequence(char str1[], char str2[], int m, int n) {
  // Base Cases
  if (m == 0)
    return true;
  if (n == 0)
    return false;

  // If last characters of two strings are matching
  if (str1[m - 1] == str2[n - 1])
    return isSubSequence(str1, str2, m - 1, n - 1);

  // If last characters are not matching
  return isSubSequence(str1, str2, m, n - 1);
}
void findDivisors(int n) {

  int div[n + 1];
  memset(div, 0, sizeof div);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j * i <= n; j++)
      div[i * j]++;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    ans += i * div[i];
  cout << ans << endl;
}
int digits(int n) {
  int res = 0;
  while (n > 0) {
    res += n % 10;
    n /= 10;
  }
  return (res);
}
int power(int x, int y, int p) {
  int res = 1; // Initialize result

  x = x % p; // Update x if it is more than or
  // equal to p

  while (y > 0) {
    // If y is odd, multiply x with result
    if (y & 1)
      res = (res * x) % p;

    // y must be even now
    y = y >> 1; // y = y/2
    x = (x * x) % p;
  }
  return res;
}

// Returns n^(-1) mod p
int modInverse(int n, int p) { return power(n, p - 2, p); }

// Returns nCr % p using Fermat's little
// theorem.
int nCr(int n, int r, int p) {
  // Base case
  if (r == 0)
    return 1;
  int fac[n + 1];
  fac[0] = 1;
  for (int i = 1; i <= n; i++)
    fac[i] = (fac[i - 1] * i) % p;

  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
double dist(int x1, int y1, int x2, int y2) {
  double a = (x1 - x2) * (x1 - x2);
  double b = (y1 - y2) * (y1 - y2);

  return sqrt(a + b);
}
signed main() {
  fastio;
  cout << fixed << setprecision(12);
  int n;
  cin >> n;
  vi a(n);
  f(i, n) cin >> a[i];
  sort(rall(a));
  int ans = 0;
  ans += a[0];
  ans += a[1];
  int x = n - 3;
  int i = 1;
  while (x--) {
    ans += a[i];
    i++;
  }
  print(ans)

      return 0;
}