Problem p02726 (C++)

Buggy submission (status: Time Limit Exceeded):
#include "bits/stdc++.h"
using namespace std;
const int MAX = 700000;
const int MOD = 1000000007;

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;
}
/*第二引数で第一引数を割ったときの切り上げの計算*/
long long int maxtime(long long int x, long long int y) {
  return (x + y - 1) / y;
}
/*最大公約数*/
long long int lcm(long long int number1, long long int number2) {
  long long int m = number1;
  long long int n = number2;

  if (number2 > number1) {
    m = number2;
    n = number1;
  }
  long long int s = -1;
  while (s != 0) {
    s = m % n;
    m = n;
    n = s;
  }
  return m;
}
/*最大公倍数*/
long long int gcd(long long int number1, long long int number2) {
  long long int m = number1;
  long long int n = number2;
  return m / lcm(m, n) * n;
}
/*逆元計算*/
long long int modinv(long long a, long long m) {
  long long int b = m, u = 1, v = 0;
  while (b) {
    long long int t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0)
    u += m;
  return u;
}
// index が条件を満たすかどうか
vector<long long int> meguru;
bool isOK(int index, int key) {
  if (meguru[index] >= key)
    return true;
  else
    return false;
}
// 汎用的な二分探索のテンプレ
int binary_search(int key) {
  int left = -1; // 「index = 0」が条件を満たすこともあるので、初期値は -1
  int right = (int)meguru.size(); // 「index =
                                  // a.size()-1」が条件を満たさないこともあるので、初期値は
                                  // a.size()
  /* どんな二分探索でもここの書き方を変えずにできる！ */
  while (right - left > 1) {
    int mid = left + (right - left) / 2;

    if (isOK(mid, key))
      right = mid;
    else
      left = mid;
  }
  /* left は条件を満たさない最大の値、right は条件を満たす最小の値になっている
   */
  return right;
}
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;
}
int dp[3000][3000] = {};
int main() {
  long long int n, x, y;
  cin >> n >> x >> y;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == x && j == y) {
        dp[i][j] = 1;
      } else if (i == y && j == x) {
        dp[j][i] = 1;
      } else {
        dp[i][j] = abs(i - j);
      }
    }
  }
  for (int k = 1; k <= n; k++) {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
      }
    }
  }
  int ans[3000] = {};
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans[dp[i][j]]++;
    }
  }
  for (int i = 1; i <= n - 1; i++) {
    cout << ans[i] << endl;
  }
}


Fixed submission:
#include "bits/stdc++.h"
using namespace std;
const int MAX = 700000;
const int MOD = 1000000007;

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;
}
/*第二引数で第一引数を割ったときの切り上げの計算*/
long long int maxtime(long long int x, long long int y) {
  return (x + y - 1) / y;
}
/*最大公約数*/
long long int lcm(long long int number1, long long int number2) {
  long long int m = number1;
  long long int n = number2;

  if (number2 > number1) {
    m = number2;
    n = number1;
  }
  long long int s = -1;
  while (s != 0) {
    s = m % n;
    m = n;
    n = s;
  }
  return m;
}
/*最大公倍数*/
long long int gcd(long long int number1, long long int number2) {
  long long int m = number1;
  long long int n = number2;
  return m / lcm(m, n) * n;
}
/*逆元計算*/
long long int modinv(long long a, long long m) {
  long long int b = m, u = 1, v = 0;
  while (b) {
    long long int t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0)
    u += m;
  return u;
}
// index が条件を満たすかどうか
vector<long long int> meguru;
bool isOK(int index, int key) {
  if (meguru[index] >= key)
    return true;
  else
    return false;
}
// 汎用的な二分探索のテンプレ
int binary_search(int key) {
  int left = -1; // 「index = 0」が条件を満たすこともあるので、初期値は -1
  int right = (int)meguru.size(); // 「index =
                                  // a.size()-1」が条件を満たさないこともあるので、初期値は
                                  // a.size()
  /* どんな二分探索でもここの書き方を変えずにできる！ */
  while (right - left > 1) {
    int mid = left + (right - left) / 2;

    if (isOK(mid, key))
      right = mid;
    else
      left = mid;
  }
  /* left は条件を満たさない最大の値、right は条件を満たす最小の値になっている
   */
  return right;
}
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;
}
int dp[3000][3000] = {};
int main() {
  long long int n, x, y;
  cin >> n >> x >> y;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == x && j == y) {
        dp[i][j] = 1;
      } else if (i == y && j == x) {
        dp[j][i] = 1;
      } else {
        dp[i][j] = abs(i - j);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      int t = abs(i - x) + abs(j - y) + 1;
      dp[i][j] = min(t, abs(i - j));
    }
  }
  int ans[3000] = {};
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      ans[dp[i][j]]++;
    }
  }
  for (int i = 1; i <= n - 1; i++) {
    cout << ans[i] << endl;
  }
}