Problem p03676 (C++)

Buggy submission (status: Time Limit Exceeded):
#include <bits/stdc++.h>
#define rep(a, n) for (ll a = 0; a < (n); ++a)
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<ll, P> PP;
typedef vector<vector<ll>> Graph;
template <class T> inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const ll INF = 1e9;

// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const int mod = 1000000007;
// const int mod = 998244353;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint &operator+=(const mint a) {
    if ((x += a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator-=(const mint a) {
    if ((x += mod - a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const { return mint(*this) += a; }
  mint operator-(const mint a) const { return mint(*this) -= a; }
  mint operator*(const mint a) const { return mint(*this) *= a; }
  mint pow(ll t) const {
    if (!t)
      return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1)
      a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint &operator/=(const mint a) { return *this *= a.inv(); }
  mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream &operator>>(istream &is, const mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }

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;
}

// combination mod prime
// mintが必要
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
  vector<mint> fact, ifact;
  combination(int n) : fact(n + 1), ifact(n + 1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i)
      fact[i] = fact[i - 1] * i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i)
      ifact[i - 1] = ifact[i] * i;
  }
  mint c(int n, int k) {
    if (k < 0 || k > n)
      return 0;
    return fact[n] * ifact[k] * ifact[n - k];
  }
  mint p(int n, int k) {
    if (k < 0 || k > n)
      return 0;
    return fact[n] * ifact[n - k];
  }
  mint f(int n) { return fact[n]; }
};

int main() {
  ll n;
  cin >> n;
  vector<ll> a(n + 1);
  rep(i, n + 1) cin >> a[i];
  vector<ll> cnt(n + 1, 0);
  rep(i, n + 1) { cnt[a[i]]++; }
  ll j;
  rep(i, n + 1) {
    if (cnt[i] == 2)
      j = i;
  }
  ll x, z;
  rep(i, n + 1) {
    if (a[i] == j) {
      x = i;
      break;
    }
  }
  for (int i = n; i >= 0; i--) {
    if (a[i] == j) {
      z = n - i;
      break;
    }
  }
  combination c(100010);
  for (int i = 1; i <= n + 1; i++) {
    mint ans = c.c(n + 1, i);
    for (int l = 0; l <= i - 1; l++) {
      mint tmp = c.c(x, l);
      tmp *= c.c(z, i - 1 - l);
      ans -= tmp;
    }
    cout << ans << endl;
  }

  return 0;
}


Fixed submission:
#include <bits/stdc++.h>
#define rep(a, n) for (ll a = 0; a < (n); ++a)
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<ll, P> PP;
typedef vector<vector<ll>> Graph;
template <class T> inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const ll INF = 1e9;

// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
const int mod = 1000000007;
// const int mod = 998244353;
struct mint {
  ll x; // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint &operator+=(const mint a) {
    if ((x += a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator-=(const mint a) {
    if ((x += mod - a.x) >= mod)
      x -= mod;
    return *this;
  }
  mint &operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const { return mint(*this) += a; }
  mint operator-(const mint a) const { return mint(*this) -= a; }
  mint operator*(const mint a) const { return mint(*this) *= a; }
  mint pow(ll t) const {
    if (!t)
      return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1)
      a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint &operator/=(const mint a) { return *this *= a.inv(); }
  mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream &operator>>(istream &is, const mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }

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;
}

// combination mod prime
// mintが必要
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
  vector<mint> fact, ifact;
  combination(int n) : fact(n + 1), ifact(n + 1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i)
      fact[i] = fact[i - 1] * i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i)
      ifact[i - 1] = ifact[i] * i;
  }
  mint c(int n, int k) {
    if (k < 0 || k > n)
      return 0;
    return fact[n] * ifact[k] * ifact[n - k];
  }
  mint p(int n, int k) {
    if (k < 0 || k > n)
      return 0;
    return fact[n] * ifact[n - k];
  }
  mint f(int n) { return fact[n]; }
};

int main() {
  ll n;
  cin >> n;
  vector<ll> a(n + 1);
  rep(i, n + 1) cin >> a[i];
  vector<ll> cnt(n + 1, 0);
  rep(i, n + 1) { cnt[a[i]]++; }
  ll j;
  rep(i, n + 1) {
    if (cnt[i] == 2)
      j = i;
  }
  ll x, z;
  rep(i, n + 1) {
    if (a[i] == j) {
      x = i;
      break;
    }
  }
  for (int i = n; i >= 0; i--) {
    if (a[i] == j) {
      z = n - i;
      break;
    }
  }
  combination c(100010);
  for (int i = 1; i <= n + 1; i++) {
    mint ans = c.c(n + 1, i);
    ans -= c.c(x + z, i - 1);
    cout << ans << endl;
  }

  return 0;
}