O2版

普通版

代码完全一样

#include<iostream>
#include<cmath>
#include<algorithm>
#include<stdio.h>
using namespace std;
const int N = 4 * (200000);
int n,m;
struct tree{
	int sum,tag;
}a[N];
/*
>>k /2^k
<<k *2^k
*/
int data[N];
void pushdown(int now,int l,int r) {
	if(a[now].tag) {
		int mid = (l+r)>>1,len = r-l+1;
		a[now<<1|1].tag ^= 1;
		a[now<<1].tag ^= 1;
		a[now<<1].sum = (len - (len>>1)) - a[now<<1].sum;
		a[now<<1|1].sum = (len>>1) - a[now<<1|1].sum;
		a[now].tag = 0;
	}
}
void build(int l,int r,int now) {
	if(l == r) {
		a[now].sum = data[l];
		return ;
	}
	int mid = (l+r)>>1;
	build(l,mid,now<<1);
	build(mid+1,r,now<<1|1);
	a[now].sum = a[now<<1].sum + a[(now<<1)|1].sum;
}
void change(int l,int r,int x,int y,int now) {//from x to y
	pushdown(now,l,r);
	if(x <= l && r <= y) {
		a[now].tag ^= 1;
		a[now].sum = r-l+1 - a[now].sum;
		return;
	}
	int mid = (l+r)>>1;
	if(x <= mid)change(l,mid,x,y,now<<1);
	if(y >= mid+1 )change(mid+1,r,x,y,now<<1|1);
	a[now].sum = a[now<<1].sum + a[now<<1|1].sum;
}
int query(int l,int r,int x,int y,int now) {
	pushdown(now,l,r);
	if(x <= l && r <= y)return a[now].sum;
	int mid = (l+r)>>1;
	int ans = 0;
	if(x <= mid)ans += query(l,mid,x,y,now<<1);
	if(y >= mid+1)ans += query(mid+1,r,x,y,now<<1|1);
	return ans;
}
int main() {
	scanf("%d%d",&n,&m);
	char t;
	for(int i = 1; i <= n; i++) {
		cin >> t;
		data[i] = int(t-'0') ; 
	}
	build(1,n,1);
	int op,l,r;
	for(int i = 1; i <= m; i++) {
		scanf("%d%d%d",&op,&l,&r);
		if(op == 0) change(1,n,l,r,1);
		else cout << query(1,n,l,r,1) << "\n";
	}
	return 0;
} 

玄学啊