結果

問題 No.833 かっこいい電車
ユーザー gazelle
提出日時 2019-05-24 21:50:52
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 565 ms / 2,000 ms
コード長 4,703 bytes
コンパイル時間 1,796 ms
コンパイル使用メモリ 133,756 KB
最終ジャッジ日時 2025-01-07 03:58:14
ジャッジサーバーID
（参考情報） 		judge3 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 30
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 #
プレゼンテーションモードにする

#include<iostream>
#include<iomanip>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<unordered_map>
#include<stack>
#include<queue>
#include<math.h>
#include<functional>
#include<bitset>
#include<cassert>
#include<random>
using namespace std;
using lint = long long;
using ld = long double;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
#define INF 1000000000LL
#define EPS 1e-12
#define FOR(i,n,m) for(lint i=n;i<(int)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v)  sort(ALL(v));v.erase(unique(ALL(v)),v.end());
#define pb push_back
#include <cstdint>
using i64 = std::int_fast64_t;
const static i64 mod = 1000000007;
struct mint {
    i64 n;
public:
    mint(const i64 n = 0) : n((n % mod + mod) % mod) {}
    mint pow(int m) const {
        i64 a = n, r = 1;
        while(m > 0) {
            if(m & 1) { r *= a; r %= mod; }
            a = (a * a) % mod; m /= 2;
        }
        return mint(r);
    }
    mint &operator++() { *this += 1; return *this; }
    mint &operator--() { *this -= 1; return *this; }
    mint operator++(int) { mint ret = *this; *this += 1; return ret; }
    mint operator--(int) { mint ret = *this; *this -= 1; return ret; }
    mint operator~() const { return (this -> pow(mod - 2)); } // inverse
    friend bool operator==(const mint& lhs, const mint& rhs) {
        return lhs.n == rhs.n;
    }
    friend bool operator<(const mint& lhs, const mint& rhs) {
        return lhs.n < rhs.n;
    }
    friend bool operator>(const mint& lhs, const mint& rhs) {
        return lhs.n > rhs.n;
    }
    friend mint &operator+=(mint& lhs, const mint& rhs) {
        lhs.n += rhs.n;
        if (lhs.n >= mod) lhs.n -= mod;
        return lhs;
    }
    friend mint &operator-=(mint& lhs, const mint& rhs) {
        lhs.n -= rhs.n;
        if (lhs.n < 0) lhs.n += mod;
        return lhs;
    }
    friend mint &operator*=(mint& lhs, const mint& rhs) {
        lhs.n = (lhs.n * rhs.n) % mod;
        return lhs;
    }
    friend mint &operator/=(mint& lhs, const mint& rhs) {
        lhs.n = (lhs.n * (~rhs).n) % mod;
        return lhs;
    }
    friend mint operator+(const mint& lhs, const mint& rhs) {
        return mint(lhs.n + rhs.n);
    }
    friend mint operator-(const mint& lhs, const mint& rhs) {
        return mint(lhs.n - rhs.n);
    }
    friend mint operator*(const mint& lhs, const mint& rhs) {
        return mint(lhs.n * rhs.n);
    }
    friend mint operator/(const mint& lhs, const mint& rhs) {
        return mint(lhs.n * (~rhs).n);
    }
};
istream& operator>>(istream& is, mint& m) { is >> m.n; return is; }
ostream& operator<<(ostream& os, mint& m) { os << m.n; return os; }
template <typename T>
class SegTree {
private:
    int n;
    const function<T(T, T)> op; // 演算
    const T ie; // 演算の単位元
    vector<T> seq;
public:
    /// op: 演算, ie: 演算の単位元
    SegTree(int _n, function<T(T, T)> op, const T ie) : op(op), ie(ie) {
        n = 1;
        while(n < _n) n *= 2;
        seq.resize(2 * n - 1);
        for(int i = 0; i < 2 * n - 1; i++) seq[i] = ie;
    }
    /// k 番目(0-indexed)の要素を e で更新
    void update(int k, const T e) {
        k += n - 1;
        seq[k] = e;
        while(k > 0) {
            k = (k - 1) / 2;
            seq[k] = op(seq[k * 2 + 1], seq[k * 2 + 2]);
        }
    }
    // k 番目(0-indexed)の要素を取得
    T get(int k) {
        k += n - 1;
        return seq[k];
    }
    /// [a, b) 番目(0-indexed)の要素全体の演算結果を返す
    T query(int a, int b, int k = 0, int l = 0, int r = -1) {
        if(r == -1) r = n;
        if(r <= a || b <= l) return ie;
        if(a <= l && r <= b) return seq[k];
        T vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
        T vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
        return op(vl, vr);
    }
};
// [Range Sum Query] rsq(n, [] (ll a, ll b) { return a + b; }, 0)
// [Range Max Query] rMq(n, [] (ll a, ll b) { return max(a, b); }, -1e18)
// [Range Min Query] rmq(n, [] (ll a, ll b) { return min(a, b); }, 1e18)
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, q;
    cin >> n >> q;
    SegTree<lint> rsq1(n, [] (lint a, lint b) { return a + b; }, 0);
    SegTree<lint> rsq2(n - 1, [] (lint a, lint b) { return a + b; }, 0);
    REP(i, n) {
        lint a;
        cin >> a;
        rsq1.update(i, a);
    }
    REP(i, q) {
        int qq;
        lint x;
        cin >> qq >> x;
        x--;
        if(qq == 1) {
            rsq2.update(x, 1);
        }
        else if(qq == 2) {
            rsq2.update(x, 0);
        }
        else if(qq == 3) {
            rsq1.update(x, rsq1.get(x) + 1);
        } else {
            int ng = -1, ok = x;
            while(abs(ok - ng) > 1) {
                int m = (ng + ok) / 2;
                if(rsq2.query(m, x) == x - m) ok = m;
                else ng = m;
            }
            int l = ok;
            ok = x; ng = n;
            while(abs(ok - ng) > 1) {
                int m = (ng + ok) / 2;
                if(rsq2.query(x, m) == m - x) ok = m;
                else ng = m;
            }
            int r = ok;
            cout << rsq1.query(l, r + 1) << endl;
        }
    }
    return 0;
}
/* --------------------------------------- */
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