結果

問題 No.2359 A in S ?
ユーザー sgfc
提出日時 2023-06-23 22:39:59
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 524 ms / 2,000 ms
コード長 3,716 bytes
コンパイル時間 4,182 ms
コンパイル使用メモリ 241,084 KB
実行使用メモリ 11,904 KB
最終ジャッジ日時 2024-07-01 02:24:22
合計ジャッジ時間 15,322 ms
ジャッジサーバーID
（参考情報） 		judge5 / 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 * 18
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 "bits/stdc++.h"
#include <stdio.h>
#include <math.h>
#pragma warning(push)
#pragma warning(disable:26498 26450 26451 26812)
#include "atcoder/all"
#pragma warning(pop)
//#include <boost/multiprecision/cpp_int.hpp>
//using namespace boost::multiprecision;
// std::fixed << std::setprecision(10) <<
using namespace std;
using namespace atcoder;
using ll = long long;
using ull = unsigned long long;
using mint = modint998244353;
using mint2 = modint1000000007;
const double eps = 1e-9;
#define REP(i, n) for (ll i = 0; i < ll(n); ++i)
#define REPI(i, init, n) for (ll i = init; i < ll(n); ++i)
#define REPD(i, init) for (ll i = init; i >=0; --i)
#define REPDI(i, init, n) for (ll i = init; i >=n; --i)
#define REPE(i, c) for (const auto& i : c)
using vl = vector<ll>;
#define VL(a, n) vector<ll> a; a.assign(n, 0);
#define VLI(a, n) vector<ll> a; a.assign(n, 0); for(auto& x : a) cin >> x;
#define VSI(a, n) vector<string> a; a.assign(n, ""); for(auto& x : a) cin >> x;
using vvl = vector<vector<ll>>;
#define VVLI(a, n, m, init) vector<vector<ll>> a(n); for(auto& x : a) x.assign(m, init);
struct uv { ll u; ll v; ll c; };
#define VUVI(a, n) vector<uv> a; a.assign(n, {0, 0, 1}); for(auto& x : a) {cin >> x.u >> x.v; x.u--; x.v--;}
#define VUVCI(a, n) vector<uv> a; a.assign(n, {0, 0, 0}); for(auto& x : a) {cin >> x.u >> x.v >> x.c; x.u--; x.v--;}
vvl to_edge(const int n, const vector<uv>& v) { vvl ret(n); for (auto& x : v) ret[x.u].push_back(x.v); return ret; }
vvl to_edge_d(const int n, const vector<uv>& v) { vvl ret(n); for (auto& x : v) { ret[x.u].push_back(x.v); ret[x.v].push_back(x.u); } return ret; }
template <class T = long long> using pql = priority_queue<T>;
template <class T = long long> using pqg = priority_queue<T, vector<T>, greater<T>>;
const char* YES = "Yes";
const char* NO = "No";
void yn(bool f) { std::cout << (f ? YES : NO) << endl; };
template<class T> void ov(const T& v) { for (auto it = v.begin(); it != v.end(); it++) { if (it != v.begin()) cout << " "; cout << *it; } };
const ll llhuge = 1LL << 60; //十分でかいが多少足しても溢れない数値
template<class T, class U> bool chmin(T& a, const U& b) { a = min(a, b); return b == a; }
template<class T, class U> bool chmax(T& a, const U& b) { a = max(a, b); return b == a; }
template<class T> void mysort(T& v) { std::sort(begin(v), end(v)); };
template<class T, class U> void mysort(T& v, U pr) { std::sort(begin(v), end(v), pr); };
template<class T> void myrev(T& v) { std::reverse(begin(v), end(v)); };
struct qe {
    ll i;
    ll x;
    ll y;
};
int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    ll n, m;
    cin >> n >> m;
    vector<qe> ql;
    vector<qe> qr;
    REP(i, n) {
        ll l, r, x, y;
        cin >> l >> r >> x >> y;
        ql.push_back({ l, x, y });
        qr.push_back({ r, x, y });
    }
    mysort(ql, [](const qe& a, const qe& b) {return a.i < b.i; });
    mysort(qr, [](const qe& a, const qe& b) {return a.i < b.i; });
    VLI(a, m);
    ll amax = 100000;
    ll b = ll(sqrt(amax));
    vl hugeans(amax + 1, 0);
    vvl smallans(b, vl(b, 0));
    ll li = 0;
    ll ri = 0;
    vl lastans(amax + 1, 0);
    REPI(i, 1, amax+1) {
        while (li < n && ql[li].i <= i) {
            ll x = ql[li].x;
            ll y = ql[li].y;
            if (x < b) {
                smallans[x][y]++;
            }
            else {
                for (ll j = y; j <= amax; j += x) {
                    hugeans[j]++;
                }
            }
            li++;
        }
        ll ans = 0;
        REPI(j, 1, b) {
            ans += smallans[j][i % j];
        }
        lastans[i] = ans + hugeans[i];
        while (ri < n && qr[ri].i <= i) {
            ll x = qr[ri].x;
            ll y = qr[ri].y;
            if (x < b) {
                smallans[x][y]--;
            }
            else {
                for (ll j = y; j <= amax; j += x) {
                    hugeans[j]--;
                }
            }
            ri++;
        }
    }
    REP(i, m) {
        cout << lastans[a[i]] << endl;
    }
    
    return 0;
}
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