daily 2

CF 1644E

#define int ll
void solve() {
  int n;
  cin >> n;
  string s;
  cin >> s;
  s = '0' + s;
  bool flag = false;
  auto calc = [&] (string &s, char c) {
    int pre = 0;
    // first we calculate how many characters left in this column
    for (char& cc: s) {
      pre += (cc == c);
    }
    bool pre_have = false;
    ll ans = 0;
    for (int i = 0 ; i <= s.size() - 1; ++i) {
      if (s[i] == c) {
        pre_have = true;
        --pre;
      }
      if (i == s.size() - 1 || s[i + 1] == (c == 'D' ? 'R' : 'D')) {
        ans += pre_have ? pre : n - 1;
      }
    }
    return ans;
  };
  int numr = 0, numd = 0;
  for (int i = 0 ; i < s.size(); ++i) {
    if (s[i] == 'R') numr++;
    if (s[i] == 'D') numd++;
  }
  if (!numr || !numd) {
    cout << n << '\n';
    return;
  }
  cout << n * n - calc(s, 'R') - calc(s, 'D') << '\n';
  return ;
}

CF 1673D

int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
void solve() {
  ll bi, bd, bl, ci, cd, cl;
  cin >> bi >> bd >> bl >> ci >> cd >> cl;
  ll endb = bi + (bl - 1) * bd;
  ll endc = ci + (cl - 1) * cd;
  // cases when no answer could show off
  if (cd % bd != 0    // case 1: the difference could not divided entirely
      || endc > endb  // case 2: no intersection
      || ci < bi      // case 2: no intersection
      || abs(bi - ci) % bd != 0  // case 3: similar to case 2
  ) {
    cout << 0 << '\n';
    return;
  }
  // cases when there are infinite answer
  if (ci - cd <
          bi ||  // case 1: when leftesth - 1 element is less than the initial
                 // element of b, then a could starts in any position
      endc + cd > endb  // case 2: same as case1
  ) {
    cout << -1 << '\n';
    return;
  }
  // cases when there are finite answer
  vector<int> fac;
  int i;
  for (i = 2; i * i < cd; ++i) {
    if (cd % i == 0) {  // all factors
      fac.push_back(i);
      fac.push_back(cd / i);
    }
  }
  if (i * i == cd) {
    fac.push_back(i);
  }
  ll ans = 1;
  for (auto& f : fac) {
    int g = gcd(f, bd);
    int lcm = f * bd / g;
    if (lcm != cd) {  // we have to let the lcm equal to cd otherwise it is not
                      // possible to intersect
      continue;
    }
    ans = (ans + 1ll * cd / f * cd / f) % MOD;
  }
  if (cd == bd) {
    ans = (ans + 1ll * bd * bd) % MOD;
  }
  cout << ans << '\n';
  return;
}