HDU

题目链接

题目大意

  询问区间内满足(c xor a < b)的不同的数的个数。

解题思路

  求区间不同的数可以用树状数组离线来做，求满足(c xor a < b)的数可以用字典树来做，树状数组的每一个节点开一个字典树即可。算是两种思路的结合吧。

const int maxn = 1e5+10;
const int maxm = maxn*400;
int n, m, a[maxn];
int tr[maxm][2], cnt[maxm], idx;
int c[maxn], rt[maxn], lst[maxn], ans[maxn];
struct INFO {
    int l, a, b, i;
};
vector<INFO> q[maxn];
void insert(int &p, int v, int x) {
    if (!p) p = ++idx;
    int now = p; //不能用p往下走，因为上面是引用
    for (int i = 20; i>=0; --i) {
        int t = x>>i&1;
        if (!tr[now][t]) tr[now][t] = ++idx;
        now = tr[now][t];
        cnt[now] += v;
    }
}
void add(int x, int v, int y) {
    while(x<=n) {
        insert(rt[x], v, y);
        x += x&-x;
    }
}
int query(int p, int a, int b) {
    int sum = 0;
    for (int i = 20; i>=0; --i) {
        int t1 = a>>i&1;
        int t2 = b>>i&1;
        /*
            a   b
            1   1   1++ 0
            0   1   0++ 1
            1   0   1
            0   0   0
        */
        if (t2) {
            if (t1) sum += cnt[tr[p][1]], p = tr[p][0];
            else sum += cnt[tr[p][0]], p = tr[p][1];
        }
        else {
            if (t1) p = tr[p][1];
            else p = tr[p][0];
        }
        if (!p) break;
    }
    return sum+cnt[p];
}
int ask(int x, int a, int b) {
    int sum = 0;
    while(x) {
        sum += query(rt[x], a, b);
        x -= x&-x;
    }
    return sum;
}
int main() {
    IOS;
    cin >> n;
    for (int i = 1; i<=n; ++i) cin >> a[i];
    cin >> m;
    for (int i = 1; i<=m; ++i) {
        int l, r, c, d; cin >> l >> r >> c >> d;
        q[r].push_back({l, c, d, i});
    }
    for (int i = 1; i<=n; ++i) {
        if (lst[a[i]]) add(lst[a[i]], -1, a[i]);
        add(i, 1, a[i]);
        //for (int j = 1; j<=n; ++j) cout << rt[j] << endl;
        lst[a[i]] = i;
        for (auto v : q[i]) ans[v.i] = ask(i, v.a, v.b)-ask(v.l-1, v.a, v.b);
    }
    for (int i = 1; i<=m; ++i) cout << ans[i] << endl;
    return 0;
}