luogu P2574 XOR的艺术 (线段树)

luogu P2574 XOR的艺术 (线段树)

算是比较简单的线段树.
当区间修改时.\(1 xor 1 = 0,0 xor 1 = 1\)所以就是区间元素个数减去以前的\(1\)的个数就是现在\(1\)的个数.

#include <iostream>
#include <cstdio>
#define lson now << 1
#define rson now << 1 | 1
const int maxN = 2e5 + 7;


struct Node {
	int l,r,Xor,sum;
}tree[maxN << 2];

void updata(int now) {
	tree[now].sum = tree[lson].sum + tree[rson].sum;
	return;
}

void build(int l,int r,int now) {
	tree[now].l = l;tree[now].r = r;
	if(l == r) {
		int a;
		scanf("%1d",&a);
		tree[now].sum = a ? 1 : 0;
		return;
	}
	int mid = (l + r) >> 1;
	build(l,mid,lson);
	build(mid + 1,r,rson);
	updata(now);
	return;
}

void work(int now) {
	tree[now].sum = tree[now].r - tree[now].l + 1 - tree[now].sum;
	tree[now].Xor ^= 1;
	return;
}

void pushdown(int now) {
	if(!tree[now].Xor) return;
	work(lson);
	work(rson);
	tree[now].Xor = 0;
	return;
}

void modify(int l,int r,int now) {
	if(tree[now].l >= l && tree[now].r <= r)  {
		work(now);
		return;
	}
	int mid = (tree[now].l + tree[now].r) >> 1;
	pushdown(now);
	if(mid >= l) modify(l,r,lson);
	if(mid < r) modify(l,r,rson);
	updata(now);
	return;
}

int query(int l,int r,int now) {
	if(tree[now].l >= l && tree[now].r <= r) 
		return tree[now].sum;
	int mid = (tree[now].l + tree[now].r) >> 1,sum = 0;
	pushdown(now);
	if(mid >= l) sum += query(l,r,lson); 
	if(mid < r) sum += query(l,r,rson);
	return sum;
}

int main() {
	int n,m;
	scanf("%d%d",&n,&m);
	build(1,n,1);
	int type,l,r;
	while(m --) {
		scanf("%d%d%d",&type,&l,&r);
		if(type) printf("%d\n", query(l,r,1));
		else modify(l,r,1);
	}
	return 0;
}