結果

問題 No.259 セグメントフィッシング＋
コンテスト
ユーザー fumiphys
提出日時 2019-10-18 02:13:02
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 167 ms / 2,000 ms
コード長 5,890 bytes
コンパイル時間 2,204 ms
コンパイル使用メモリ 183,572 KB
実行使用メモリ 21,632 KB
最終ジャッジ日時 2024-06-25 13:59:11
合計ジャッジ時間 6,208 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 3
other AC * 23
権限があれば一括ダウンロードができます

ソースコード

diff # 

// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define bit(n) (1LL<<(n))
// functions
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(0); cout << fixed << setprecision(20);}
} fast_io_;

ll cntr[5010], cntl[5010];
ll tmpr[5010], tmpl[5010];

template<typename T, typename E>
struct SegmentTree_ {
  function<T(T, T)> f; // aggregate function
  function<T(T, E)> g; // update function
  int n;
  T def;
  vector<T> vec;
  SegmentTree_(){}
  SegmentTree_(int n_, function<T(T, T)> f, function<T(T, E)> g, T def, vector<T> v=vector<T>()): f(f), g(g), def(def){

    // initialize vector
    n = 1;
    while(n < n_){
      n *= 2;
    }
    vec = vector<T>(2*n -1, def);

    // initialize segment tree
    for(int i = 0; i < v.size(); i++){
      vec[i + n - 1] = v[i];
    }
    for(int i = n - 2; i >= 0; i--){
      vec[i] = f(vec[2*i+1], vec[2*i+2]);
    }
  }
  void update(int k, const E &val){
    k = k + n - 1;
    vec[k] = g(vec[k], val);
    while(k > 0){
      k = (k - 1) / 2;
      vec[k] = f(vec[2*k+1], vec[2*k+2]);
    }
  }
  // [l, r) -> [a, b) (at k)
  T query(int a, int b, int k, int l, int r){
    if(r <= a || b <= l)return def;
    if(a <= l && r <= b)return vec[k];
    T ld = query(a, b, 2*k+1, l, (l+r)/2);
    T rd = query(a, b, 2*k+2, (l+r)/2, r);
    return f(ld, rd);
  }
  T query(int a, int b){
    return query(a, b, 0, 0, n);
  }
};

template<typename T, typename E>
using SegmentTree = struct SegmentTree_<T, E>;
using SegmentTreeI = SegmentTree<int, int>;
using SegmentTreeL = SegmentTree<long long, long long>;


int main(int argc, char const* argv[])
{
  int n, q; cin >> n >> q;
  vector<char> c(q);
  vector<ll> t(q), L(q), R(q); rep(i, q)cin >> c[i] >> t[i] >> L[i] >> R[i];
  SegmentTreeL segl = SegmentTreeL(2 * n, [](ll a, ll b){return a + b;},
                                   [](ll a,ll b){return a + b;}, 0, vector<ll>());
  SegmentTreeL segr = SegmentTreeL(2 * n, [](ll a, ll b){return a + b;},
                                   [](ll a,ll b){return a + b;}, 0, vector<ll>());
  rep(i, q){
    if(c[i] == 'L'){
      int idx = (L[i] + t[i]) % (2 * n);
      segl.update(idx, R[i]);
    }else if(c[i] == 'R'){
      int idx = (L[i] - t[i]) % (2 * n);
      if(idx < 0)idx += 2 * n;
      segr.update(idx, R[i]);
    }else{
      R[i]--;
      ll res = 0;
      int l = (L[i] + t[i]) % (2 * n), r = (R[i] + t[i]) % (2 * n);
      if(l <= r)res += segl.query(l, r + 1);
      else res += segl.query(l, 2 * n) + segl.query(0, r + 1);
      l = (2 * n - 1 - R[i] + t[i]) % (2 * n), r = (2 * n - 1 - L[i] + t[i]) % (2 * n);
      if(l <= r)res += segl.query(l, r + 1);
      else res += segl.query(l, 2 * n) + segl.query(0, r + 1);
      l = (L[i] - t[i]) % (2 * n), r = (R[i] - t[i]) % (2 * n);
      if(l < 0)l += 2 * n;
      if(r < 0)r += 2 * n;
      if(l <= r)res += segr.query(l, r + 1);
      else res += segr.query(l, 2 * n) + segr.query(0, r + 1);
      l = (2 * n - 1 - R[i] - t[i]) % (2 * n), r = (2 * n - 1 - L[i] - t[i]) % (2 * n);
      if(l < 0)l += 2 * n;
      if(r < 0)r += 2 * n;
      if(l <= r)res += segr.query(l, r + 1);
      else res += segr.query(l, 2 * n) + segr.query(0, r + 1);
      cout << res << endl;
    }
  }
  return 0;
}
0