Submission #1086377

#	Time	Username	Problem	Language	Result	Execution time	Memory
1086377		Sergio_2357	Monkey and Apple-trees (IZhO12_apple)	C++17	0 / 100	500 ms	262144 KiB

apple

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define all(a) a.begin(), a.end()

#ifdef IS_LOCAL

#define dbg(x) {cout << #x << " = " << x << endl;}

template <class T>
ostream& operator<<(ostream& os, vector<T> v) {
	os << "[";
	for (int i = 0; i < (int)v.size(); i++) {
		os << v[i];
		if (i != (int)v.size()-1)
			os << ", ";
	}
	os << "]";
	return os;
}

#else

#define dbg(x) 0

#endif

struct SegTree {
	struct Node {
		Node* L = NULL;
		Node* R = NULL;
		int v = 0;
		int lz = 0;
	};
	typedef Node* pnode;
	pnode R = NULL;
	int n;
	SegTree(int n_) {
		n = n_;
	}
	void destroy(pnode& r) {
		if (!r) return;
		destroy(r->L);
		destroy(r->R);
		delete r;
	}
	~SegTree() {
		destroy(R);
	}
	void push_lz(pnode& n, int l, int r) {
		if (!n) {
			n = new Node;
		}
		if (!n->L) {
			n->L = new Node;
		}
		if (!n->R) {
			n->R = new Node;
		}
		int m = (l+r)/2;
		if (n->lz) {
			n->lz = 0;
			n->L->lz = n->R->lz = 1;
			n->L->v = (m-l);
			n->R->v = (r-m);
		}
	}
	void update(pnode& R, int l, int r, int ql, int qr) {
		push_lz(R, l, r);
		if (ql <= l && r <= qr) {
			R->v = (r-l);
			R->lz = 1;
			return;
		}
		if (r <= ql || qr <= l || r == l) {
			return;
		}
		int m = (r+l)/2;
		update(R->L, l, m, ql, qr);
		update(R->R, m, r, ql, qr);
		R->v = R->L->v + R->R->v;
	}
	int query(pnode& R, int l, int r, int ql, int qr) {
		push_lz(R, l, r);
		if (ql <= l && r <= qr) {
			return R->v;
		}
		if (r <= ql || qr <= l || r == l) {
			return 0;
		}
		int m = (r+l)/2;
		return query(R->L, l, m, ql, qr)+
				query(R->R, m, r, ql, qr);
	}
	void set_1(int l, int r) {
		update(R, 0, n, l, r);
	}
	int get_sum(int l, int r) {
		return query(R, 0, n, l, r);
	}
};

signed main() {
	int M;
	cin >> M;
	int C = 0;
	SegTree st(1e9+1000);
	while (M--) {
		int d, x, y;
		cin >> d >> x >> y;
		if (d == 1) {
			C = st.get_sum(x+C-1, y+C);
			cout << C << endl;
		} else {
			st.set_1(x+C-1, y+C);
		}
	}
}

• Subtask 1: 0 / 100