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;
}