AC 的：

#include <iostream>
#define lc (o << 1)
#define rc (o << 1 | 1)
using namespace std;
const int S = 2e5 + 5;
int n, q;
int a[S], sum[S * 4];
int tag[S * 4];
void pushup(int o) {
	sum[o] = sum[lc] + sum[rc];
	return;
} 
void pushdown(int o, int l, int r) {
	if (tag[o] > 0) {
		tag[lc] = (tag[lc] + tag[o]) % 2;
		tag[rc] = (tag[rc] + tag[o]) % 2;
		tag[o] = 0;
		int mid = l + r >> 1;
		sum[lc] = (mid - l + 1) - sum[lc];
		sum[rc] = (r - mid) - sum[rc];
	}
	return;
}
void build(int o, int l, int r) {
	if (l == r) {
		sum[o] = a[l];
		return;
	}
	int mid = l + r >> 1;
	build(lc, l, mid);
	build(rc, mid + 1, r);
	pushup(o); 
}  
void update(int o, int l, int r, int ql, int qr) {
	if (ql <= l && r <= qr) {
		tag[o] ^= 1;
		sum[o] = (r - l + 1) - sum[o];
		return;
	}
	pushdown(o, l, r);
	int mid = l + r >> 1;
	if (ql <= mid) {
		update(lc, l, mid, ql, qr);
	}
	if (mid < qr) {
		update(rc, mid + 1, r, ql, qr);
	}
	pushup(o);
}
int query(int o, int l, int r, int ql, int qr) {
	if (ql <= l && r <= qr) {
		return sum[o];
	}
	int mid = l + r >> 1;
	pushdown(o, l, r);
	int ans = 0;
	if (ql <= mid) {
		ans += query(lc, l, mid, ql, qr);
	}
	if (mid < qr) {
		ans += query(rc, mid + 1, r, ql, qr);
	}
	return ans;
}   
int main() {
	cin >> n >> q;
	char c;
	for (int i = 1; i <= n; i++) {
		cin >> c;
		a[i] = c - '0';
	}
	build (1, 1, n); 
	int op, l, r;
	while (q--) {
		cin >> op >> l >> r;
		if (!op) {
			update(1, 1, n, l, r);
		} else {
			cout << query(1, 1, n, l, r) << endl;
		}
	}
	return 0;
}

WA 的：

#include <iostream>
#define lc (o << 1)
#define rc (o << 1 | 1)
using namespace std;
const int S = 2e5 + 5;
int n, q;
int a[S], sum[S * 4];
int tag[S * 4];
void pushup(int o) {
	sum[o] = sum[lc] + sum[rc];
	return;
} 
void pushdown(int o, int l, int r) {
	tag[lc] = (tag[lc] + tag[o]) % 2;
	tag[rc] = (tag[rc] + tag[o]) % 2;
	tag[o] = 0;
	int mid = l + r >> 1;
	if (tag[lc]) {
		sum[lc] = (mid - l + 1) - sum[lc];
	}
	if (tag[rc]) {
		sum[rc] = (r - mid) - sum[rc];
	} 
	return;
}
void build(int o, int l, int r) {
	if (l == r) {
		sum[o] = a[l];
		return;
	}
	int mid = l + r >> 1;
	build(lc, l, mid);
	build(rc, mid + 1, r);
	pushup(o); 
}  
void update(int o, int l, int r, int ql, int qr) {
	if (ql <= l && r <= qr) {
		tag[o] ^= 1;
		if (tag[o]) {
			sum[o] = (r - l + 1) - sum[o];
		}
		return;
	}
	pushdown(o, l, r);
	int mid = l + r >> 1;
	if (ql <= mid) {
		update(lc, l, mid, ql, qr);
	}
	if (mid < qr) {
		update(rc, mid + 1, r, ql, qr);
	}
	pushup(o);
}
int query(int o, int l, int r, int ql, int qr) {
	if (ql <= l && r <= qr) {
		return sum[o];
	}
	int mid = l + r >> 1;
	pushdown(o, l, r);
	int ans = 0;
	if (ql <= mid) {
		ans += query(lc, l, mid, ql, qr);
	}
	if (mid < qr) {
		ans += query(rc, mid + 1, r, ql, qr);
	}
	return ans;
}   
int main() {
	cin >> n >> q;
	char c;
	for (int i = 1; i <= n; i++) {
		cin >> c;
		a[i] = c - '0';
	}
	build (1, 1, n); 
	int op, l, r;
	while (q--) {
		cin >> op >> l >> r;
		if (!op) {
			update(1, 1, n, l, r);
		} else {
			cout << query(1, 1, n, l, r) << endl;
		}
	}
	return 0;
}

两段代码的差异仅仅是在修改 sum 数组时判断 tag[o] 是否等于 0

我觉得 tag[o] 等于 0 的话不就是改了偶数次等于没改吗，为什么不对啊

求大佬指教 Orz