【KTU Programming Camp (Day 3)】Queries

http://codeforces.com/gym/100739/problem/A

按位考虑，每一位建一个线段树。

求出前缀xor和，对前缀xor和建线段树。

线段树上维护区间内的0的个数和1的个数。

修改就修改p到最后的区间，进行区间取反。

回答询问时把总区间内0的个数和1的个数相乘即可。

时间复杂度\(O(n\log^2n)\)。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 100003;

int a[N], s[N], n, m;

struct node {
	node *l, *r;
	int sum0, sum1, mark;
	node() {
		sum0 = sum1 = mark = 0;
		l = r = NULL;
	}
	void pushdown() {
		if (mark) {
			mark = 0;
			if (l) {
				l->mark ^= 1;
				swap(l->sum0, l->sum1);
			}
			if (r) {
				r->mark ^= 1;
				swap(r->sum0, r->sum1);
			}
		}
	}
	void count_() {
		sum0 = sum1 = 0;
		if (l) {
			sum0 += l->sum0;
			sum1 += l->sum1;
		}
		if (r) {
			sum0 += r->sum0;
			sum1 += r->sum1;
		}
	}
} *rt[15];

node *build_tree(int l, int r, int x) {
	node *t = new node;
	if (l == r) {
		if ((s[l] >> x) & 1) ++t->sum1;
		else ++t->sum0;
		return t;
	}
	int mid = ((l + r) >> 1);
	t->l = build_tree(l, mid, x);
	t->r = build_tree(mid + 1, r, x);
	t->count_();
	return t;
}

void reserve(node *t, int l, int r, int L, int R) {
	if (L <= l && r <= R) {
		t->mark ^= 1;
		swap(t->sum0, t->sum1);
		return;
	}
	int mid = ((l + r) >> 1);
	t->pushdown();
	if (mid >= L) reserve(t->l, l, mid, L, R);
	if (mid < R) reserve(t->r, mid + 1, r, L, R);
	t->count_();
}

int S0, S1;

void count(node *t, int l, int r, int L, int R) {
	if (L <= l && r <= R) {
		S0 += t->sum0;
		S1 += t->sum1;
		return;
	}
	t->pushdown();
	int mid = ((l + r) >> 1);
	if (mid >= L) count(t->l, l, mid, L, R);
	if (mid < R) count(t->r, mid + 1, r, L, R);
}

int main() {
	//freopen("a.in", "r", stdin);
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
	for (int i = 1; i <= n; ++i) s[i] = (a[i] ^ s[i - 1]);
	for (int i = 0; i < 15; ++i)
		rt[i] = build_tree(0, n, i);

	int p, x, aa, bb, op;
	while (m--) {
		scanf("%d", &op);
		if (op == 1) {
			scanf("%d%d", &p, &x);
			for (int i = 0; i < 15; ++i)
				if (((a[p] >> i) & 1) != ((x >> i) & 1)) {
					reserve(rt[i], 0, n, p, n);
					a[p] ^= (1 << i);
				}
		} else {
			scanf("%d%d", &aa, &bb);
			int ans = 0;
			for (int i = 0; i < 15; ++i) {
				S0 = S1 = 0;
				count(rt[i], 0, n, aa - 1, bb);
				(ans += 1ll * S0 * S1 % 4001 * (1 << i) % 4001) %= 4001;
			}
			printf("%d\n", ans);
		}
	}

	return 0;
}