【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
#include
#include
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;
}

转载于:https://www.cnblogs.com/abclzr/p/9191561.html