Problem p02572 (C++) — buggy submission, verdict: Time Limit Exceeded

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
using namespace std;
using ll = long long;
using d = double;

const int mod = 1000000007;
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, mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }

int main() {
  int n;
  cin >> n;
  vector<mint> a(n);
  mint sum = 0;
  rep(i, n) { cin >> a[i]; }
  vector<mint> b(n);
  rep(i, n) sum += a[i];
  for (int i = 0; i < n; ++i) {
    b[i] = sum;
    sum -= a[i];
  }

  mint ans = 0;
  b.pop_back();
  rep(i, n) {
    b[i] -= a[i];
    ans += a[i] * b[i];
  }
  cout << ans << endl;
  return 0;
}

Error: TLE

Fixed submission:

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
using namespace std;
using ll = long long;
using d = double;

const int mod = 1000000007;
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, mint &a) { return is >> a.x; }
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }

int main() {
  int n;
  cin >> n;
  vector<ll> a(n);
  rep(i, n) scanf("%lld", &a[i]);

  mint ans = 0;
  rep(i, n - 1) {
    for (int j = i + 1; j < n; ++j) {
      ans += a[i] * a[j];
    }
  }
  printf("%lld\\n", ans);
  return 0;
}