BOJ 12844 : XOR

• 문제 링크 : boj.kr/12844
• 난이도 : P3
• 태그 : 느리게 갱신되는 세그먼트 트리

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코드

#include <bits/stdc++.h>

#define all(x) (x).begin(), (x).end()

#define INF 0x7FFFFFFF

using namespace std;

using ll = long long;
using ld = long double;
using pii = pair<int,int>;
using pll = pair<ll, ll>;

vector<ll> lazy, tree, arr;

void build(int node, int s, int e) {
    if(s == e) {
        tree[node] = arr[s];
        return;
    }

    build(node*2, s, (s+e)/2);
    build(node*2+1, (s+e)/2+1, e);
    tree[node] = tree[node*2] ^ tree[node*2+1];
}

void push(int node, int s, int e, ll value) {
    if(value) {
        tree[node] ^= ((e-s+1)%2) * value;
        if(s != e) {
            lazy[node*2] ^= value;
            lazy[node*2+1] ^= value;
        }
        lazy[node] = 0;
    }
}

void update(int node, int s, int e, int l, int r, ll value) {
    push(node, s, e, lazy[node]);
    if(e < l || r < s) return;
    if(l <= s && e <= r) {
        push(node, s, e, value);
        return;
    }
    update(node*2, s, (s+e)/2, l, r, value);
    update(node*2+1, (s+e)/2+1, e, l, r, value);
    tree[node] = tree[node*2] ^ tree[node*2+1];
}

ll query(int node, int s, int e, int l, int r) {
    push(node, s, e, lazy[node]);
    if(e < l || r < s) return 0;
    if(l <= s && e <= r) return tree[node];

    ll lq = query(node*2, s, (s+e)/2, l, r);
    ll rq = query(node*2+1, (s+e)/2+1, e, l, r);
    return lq ^ rq;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);

    int n;
    cin >> n;
    arr.resize(n+1);
    lazy.resize(n*4, 0);
    tree.resize(n*4, 0);

    for(int i = 1; i <= n; i++) {
        cin >> arr[i];
    }
    build(1, 1, n);

    int m;
    cin >> m;
    for(int i = 0; i < m; i++) {
        int q;
        cin >> q;
        if(q == 1) {
            int a, b, c;
            cin >> a >> b >> c;
            update(1, 1, n, a+1, b+1, c);
        }
        if(q == 2) {
            int a, b;
            cin >> a >> b;
            cout << query(1, 1, n, a+1, b+1) << "\n";
        }
    }

    return 0;
}

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풀이

A^A=0을 이용한다.