Submission #735093

#	Time	Username	Problem	Language	Result	Execution time	Memory
735093		makrav	XORanges (eJOI19_xoranges)	C++14	100 / 100	150 ms	18148 KiB

xoranges

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<int> vei;
typedef vector<vei> vevei;

#define coa               \
            for (auto i : a) {    \
                cout << i << ' '; \
            }                     \
            cout << '\n';
#define cia             \
            for (auto& i : a) { \
                cin >> a;       \
            }
#define cna                       \
            int n;                        \
            cin >> n;                     \
            vector<int> a(n);             \
            for (int i = 0; i < n; i++) { \
                cin >> a[i];              \
            }
#define cnka                      \
            int n, k;                     \
            cin >> n >> k;                \
            vector<int> a(n);             \
            for (int i = 0; i < n; i++) { \
                cin >> a[i];              \
            }
#define cnab                      \
            int n;                        \
            cin >> n;                     \
            vector<int> a(n);             \
            for (int i = 0; i < n; i++) { \
                cin >> a[i];              \
            }                             \
            vector<int> b(n);             \
            for (int i = 0; i < n; i++) { \
                cin >> b[i];              \
            }
#define all(a) (a).begin(), (a).end()
#define sz(a) (int) a.size()
#define con cout << "NO\n"
#define coe cout << "YES\n";
#define str string
#define pb push_back
#define ff first
#define sc second
#define pii pair<int, int>
#define mxe max_element
#define mne min_element
#define stf shrink_to_fit
#define f(i, l, r) for (int i = (l); i < (r); i++)
#define double ld
#define int long long

struct segtree {
    int n;
    vector<int> a;
    vector<int> t;

    segtree() = default;
    segtree(int n_, vector<int> a_) {
        n = n_;
        a = a_;
        t.assign(4 * n, 0);
        build(1, 0, n);
    }

    void build(int v, int tl, int tr) {
        if (tl + 1 == tr) {
            t[v] = a[tl];
            return;
        }
        int tm = (tl + tr) / 2;
        build(v * 2, tl, tm);
        build(v * 2 + 1, tm, tr);
        t[v] = (t[v * 2] ^ t[v * 2 + 1]);
    }

    void upd(int v, int tl, int tr, int pos, int val) {
        if (tl + 1 == tr) {
            t[v] = val;
            return;
        }
        int tm = (tl + tr) / 2;
        if (pos < tm) upd(v * 2, tl, tm, pos, val);
        else upd(v * 2 + 1, tm, tr, pos, val);
        t[v] = (t[v * 2] ^ t[v * 2 + 1]);
    }

    int req(int v, int tl, int tr, int l, int r) {
        if (l <= tl && tr <= r) return t[v];
        if (tr <= l || tl >= r) return 0;
        int tm = (tl + tr) / 2;
        return (req(v * 2, tl, tm, l, r) ^ req(v * 2 + 1, tm, tr, l, r));
    }
};

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int n, q; cin >> n >> q;
    vector<int> a(n);
    for (auto& u : a) cin >> u;

    vector<int> ch, nch;
    f(i, 0, n) {
        if (i % 2 == 0) ch.pb(a[i]);
        else nch.pb(a[i]);
    }

    segtree sg0(sz(ch), ch), sg1(sz(nch), nch);
    while (q--) {
        int t; cin >> t;
        if (t == 1) {
            int p, v; cin >> p >> v;
            p--;
            if (p % 2 == 0) {
                sg0.upd(1, 0, sz(ch), p / 2, v);
            }
            else {
                sg1.upd(1, 0, sz(nch), p / 2, v);
            }
        }
        else {
            int l, r; cin >> l >> r;
            l--; r--;
            if ((r - l + 1) % 2 == 0) {
                cout << "0\n";
            }
            else {
                if (l % 2 == 0) {
                    cout << sg0.req(1, 0, sz(ch), l / 2, r / 2 + 1) << '\n';
                }
                else {
                    cout << sg1.req(1, 0, sz(nch), l / 2, r / 2 + 1) << '\n';
                }
            }
        }
    }

    return 0;
}

• Subtask 1: 12 / 12
• Subtask 2: 18 / 18
• Subtask 3: 25 / 25
• Subtask 4: 20 / 20
• Subtask 5: 25 / 25