【BZOJ5294】【BJOI2018】二进制

【题目链接】

• 点击打开链接
  

【思路要点】

• 考虑计算总区间数量-不合法区间的数量。
• 不合法区间分为两种：
  
• 1、区间和为1的区间。
• 2、区间和为大于1的奇数，且存在不足两个0。
• 分别用线段树统计上述区间个数即可。
• 时间复杂度\(O(QLogN)\)。

【代码】

#include
using namespace std;
const int MAXN = 100005;
const int MAXP = 200005;
template  void chkmax(T &x, T y) {x = max(x, y); }
template  void chkmin(T &x, T y) {x = min(x, y); } 
template  void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template  void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template  void writeln(T x) {
	write(x);
	puts("");
}
struct info {
	long long sum;
	int l, r, nl, nr;
	int len, zl, zr, znl, znr;
};
info operator + (info a, info b) {
	info ans;
	ans.sum = a.sum + b.sum;
	ans.len = a.len + b.len;
	ans.sum += 1ll * b.znl * a.zr;
	ans.sum += 1ll * b.zl * a.znr;
	int tmp[2], tnp[2];
	tmp[0] = a.r / 2;
	tmp[1] = a.r - tmp[0];
	tnp[0] = b.l / 2;
	tnp[1] = b.l - tnp[0];
	ans.sum += 1ll * tmp[0] * tnp[1];
	ans.sum += 1ll * tmp[1] * tnp[0];
	if (b.nl) {
		tmp[0] = a.r / 2;
		tmp[1] = a.r - tmp[0];
		tnp[0] = (b.nl + 1) / 2;
		tnp[1] = b.nl - tnp[0];
		if (b.l & 1) swap(tnp[0], tnp[1]);
		ans.sum += 1ll * tmp[0] * tnp[1];
		ans.sum += 1ll * tmp[1] * tnp[0];
		if (b.l == 0 && a.r != 0) ans.sum--;
	}
	if (a.nr) {
		tmp[0] = (a.nr + 1) / 2;
		tmp[1] = a.nr - tmp[0];
		tnp[0] = b.l / 2;
		tnp[1] = b.l - tnp[0];
		if (a.r & 1) swap(tmp[0], tmp[1]);
		ans.sum += 1ll * tmp[0] * tnp[1];
		ans.sum += 1ll * tmp[1] * tnp[0];
		if (a.r == 0 && b.l != 0) ans.sum--;
	}
	if (a.nl == 0) {
		ans.l = a.l + b.l;
		ans.nl = b.nl;
	} else {
		ans.l = a.l;
		if (a.l + a.nl == a.len) ans.nl = a.nl + b.l;
		else ans.nl = a.nl;
	}
	if (b.nr == 0) {
		ans.r = b.r + a.r;
		ans.nr = a.nr;
	} else {
		ans.r = b.r;
		if (b.r + b.nr == b.len) ans.nr = b.nr + a.r;
		else ans.nr = b.nr;
	}
	if (a.znl == 0) {
		ans.zl = a.zl + b.zl;
		ans.znl = b.znl;
	} else {
		ans.zl = a.zl;
		if (a.zl + a.znl == a.len) ans.znl = a.znl + b.zl;
		else ans.znl = a.znl;
	}
	if (b.znr == 0) {
		ans.zr = b.zr + a.zr;
		ans.znr = a.znr;
	} else {
		ans.zr = b.zr;
		if (b.zr + b.znr == b.len) ans.znr = b.znr + a.zr;
		else ans.znr = b.znr;
	}
	return ans;
}
struct SegmentTree {
	struct Node {
		int lc, rc;
		info ans;
	} a[MAXP];
	int root, size, n;
	void build(int &root, int l, int r) {
		root = ++size;
		if (l == r) {
			int x; read(x);
			a[root].ans.sum = x;
			a[root].ans.len = 1;
			a[root].ans.l = a[root].ans.r = x;
			a[root].ans.nl = a[root].ans.nr = !x;
			a[root].ans.zl = a[root].ans.zr = !x;
			a[root].ans.znl = a[root].ans.znr = x;
			return;
		}
		int mid = (l + r) / 2;
		build(a[root].lc, l, mid);
		build(a[root].rc, mid + 1, r);
		a[root].ans = a[a[root].lc].ans + a[a[root].rc].ans;
	}
	void init(int x) {
		n = x;
		root = size;
		build(root, 1, n);
	}
	void modify(int root, int l, int r, int pos) {
		if (l == r) {
			int x = !a[root].ans.sum;
			a[root].ans.sum = x;
			a[root].ans.len = 1;
			a[root].ans.l = a[root].ans.r = x;
			a[root].ans.nl = a[root].ans.nr = !x;
			a[root].ans.zl = a[root].ans.zr = !x;
			a[root].ans.znl = a[root].ans.znr = x;
			return;
		}
		int mid = (l + r) / 2;
		if (mid >= pos) modify(a[root].lc, l, mid, pos);
		else modify(a[root].rc, mid + 1, r, pos);
		a[root].ans = a[a[root].lc].ans + a[a[root].rc].ans;
	}
	void modify(int pos) {
		modify(root, 1, n, pos);
	}
	info query(int root, int l, int r, int ql, int qr) {
		if (l == ql && r == qr) return a[root].ans;
		int mid = (l + r) / 2;
		if (mid >= qr) return query(a[root].lc, l, mid, ql, qr);
		else if (mid + 1 <= ql) return query(a[root].rc, mid + 1, r, ql, qr);
		else return query(a[root].lc, l, mid, ql, mid) + query(a[root].rc, mid + 1, r, mid + 1, qr);
	}
	long long query(int l, int r) {
		info ans = query(root, 1, n, l, r);
		return ans.sum;
	}
} ST;
int n, m;
int main() {
	read(n);
	ST.init(n);
	read(m);
	for (int i = 1; i <= m; i++) {
		int opt; read(opt);
		if (opt == 1) {
			int x; read(x);
			ST.modify(x);
		} else {
			int l, r; read(l), read(r);
			int len = r - l + 1;
			long long ans = (len + 1ll) * len / 2;
			ans -= ST.query(l, r);
			writeln(ans);
		}
	}
	return 0;
}