結果

問題 No.728 ギブ and テイク
ユーザー ats5515
提出日時 2018-08-24 22:08:46
言語 C++11
(gcc 13.3.0)
結果
AC  
実行時間 407 ms / 3,000 ms
コード長 3,027 bytes
コンパイル時間 740 ms
コンパイル使用メモリ 87,156 KB
実行使用メモリ 26,112 KB
最終ジャッジ日時 2024-06-23 07:35:17
合計ジャッジ時間 5,891 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge1
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 * 3
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 <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
#include <cstdint>
#include <cstring>
using namespace std;
#define repi(i,a,b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(i,n) repi(i,0,n)
template<int N> class FID {
    static const int bucket = 512, block = 16;
    static char popcount[];
    int n, B[N / bucket + 10];
    unsigned short bs[N / block + 10], b[N / block + 10];
public:
    FID() {}
    FID(int n, bool s[]) : n(n) {
        if (!popcount[1]) for (int i = 0; i < (1 << block); i++) popcount[i] = __builtin_popcount(i);
        bs[0] = B[0] = b[0] = 0;
        for (int i = 0; i < n; i++) {
            if (i%block == 0) {
                bs[i / block + 1] = 0;
                if (i%bucket == 0) {
                    B[i / bucket + 1] = B[i / bucket];
                    b[i / block + 1] = b[i / block] = 0;
                }
                else b[i / block + 1] = b[i / block];
            }
            bs[i / block] |= short(s[i]) << (i%block);
            b[i / block + 1] += s[i];
            B[i / bucket + 1] += s[i];
        }
        if (n%bucket == 0) b[n / block] = 0;
    }
    int count(bool val, int r) { return val ? B[r / bucket] + b[r / block] + popcount[bs[r / block] & ((1 << (r%block)) - 1)] : r - count(1, r); }
    int count(bool val, int l, int r) { return count(val, r) - count(val, l); }
};
template<int N> char FID<N>::popcount[1 << FID<N>::block];
template<class T, int N, int D> class wavelet {
    int n, zs[D];
    FID<N> dat[D];
    int freq_dfs(int d, int l, int r, T val, T a, T b) {
        if (l >= r or val >= b) return 0;
        if (d == D) return a <= val ? r - l : 0;
        T nv = 1LL << (D - d - 1) | val, nnv = ((1ULL << (D - d - 1)) - 1) | nv;
        if (nnv < a) return 0;
        if (a <= val and nnv < b) return r - l;
        int lc = dat[d].count(1, l), rc = dat[d].count(1, r);
        return freq_dfs(d + 1, l - lc, r - rc, val, a, b) + freq_dfs(d + 1, lc + zs[d], rc + zs[d], nv, a, b);
    }
public:
    wavelet() {}
    wavelet(int n, T seq[]) : n(n) {
        T f[N], l[N], r[N];
        bool b[N];
        memcpy(f, seq, sizeof(T)*n);
        for (int d = 0; d < D; d++) {
            int lh = 0, rh = 0;
            for (int i = 0; i < n; i++) {
                bool k = (f[i] >> (D - d - 1)) & 1;
                if (k) r[rh++] = f[i];
                else l[lh++] = f[i];
                b[i] = k;
            }
            dat[d] = FID<N>(n, b);
            zs[d] = lh;
            swap(l, f);
            memcpy(f + lh, r, rh * sizeof(T));
        }
    }
    int freq(int l, int r, T a, T b) { return freq_dfs(0, l, r, 0, a, b); }
};
const int _N = 300100, D = 40;
wavelet<long long, _N, D> a;
#define int long long
int ri[300100];
signed main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cerr << (int)3e9 << endl;
    int N;
    cin >> N;
    vector<int> A(N);
    vector<int> L(N);
    vector<int> R(N);
    int res = 0;
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }
    for (int i = 0; i < N; i++) {
        cin >> L[i] >> R[i];
    }
    for (int i = 0; i < N; i++) {
        ri[i] = A[i] + R[i];
    }
    a = wavelet<long long, _N, D>(N, ri);
    for (int i = 0; i < N; i++) {
        int le = lower_bound(A.begin(), A.end(), A[i] - L[i]) - A.begin();
        res += a.freq(le, i, A[i], (int)3e9 + 5);
    }
    cout << res << endl;
}
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