핵심 아이디어

<aside> 💡

$[L,R]$구간에 켜져있는 스위치의 수를 $c$라고 하면

$[L,R]$구간의 스위치를 반전하면 켜져있는 스위치의 수는 $R-L+1-c$이다.

</aside>

코드

#include <bits/stdc++.h>
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
using namespace std;
typedef long long ll;

class Seg
{
    private:
        int n;
        vector<int> tree;
        vector<int> lazy;

        void prop(int node, int start, int end)
        {
            if(lazy[node] == 0) return;

            tree[node] = (end-start+1) - tree[node];

            if(start != end){
                int mid = start+end>>1;
                lazy[2*node] ^= lazy[node];
                lazy[2*node+1] ^= lazy[node];
            }

            lazy[node] = 0;
        }

        void upd(int node, int start, int end, int left, int right, int val)
        {
            prop(node, start, end);
            if(right<start || end<left) return;
            if(left<=start && end<=right){
                lazy[node] ^= val;
                prop(node, start, end);
                return;
            }

            int mid = start+end>>1;
            upd(2*node, start, mid, left, right, val);
            upd(2*node+1, mid+1, end, left, right, val);

            tree[node] = tree[2*node]+tree[2*node+1];
        }

        int qry(int node, int start, int end, int left, int right)
        {
            prop(node, start, end);
            if(right<start || end<left) return 0;
            if(left<=start && end<=right) return tree[node];

            int mid = start+end>>1;
            int l = qry(2*node, start, mid, left, right);
            int r = qry(2*node+1, mid+1, end, left, right);

            return l+r;
        }

    public:
        Seg(int n): n(n)
        {
            tree.resize(4*n);
            lazy.resize(4*n);
        }

        void upd(int left, int right, int val) {upd(1, 1, n, left, right, val);}
        int qry(int left, int right) {return qry(1, 1, n, left, right);}
};

signed main()
{
    FASTIO;
    int n, m;
    cin >> n >> m;
    Seg seg(n);

    for(int i = 0; i<m; i++){
        int o, a, b;
        cin >> o >> a>> b;
        if(o == 0){
            seg.upd(a, b, 1);
        }
        else{
            cout << seg.qry(a, b) <<"\\n";
        }
    }
    
    return 0;
}