区间不同元素个数(D-query) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

struct Query {
    int l, r, id;
    bool operator<(const Query& o) const {
        return r < o.r;
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    int q;
    cin >> q;
    vector<Query> qs(q);
    for (int i = 0; i < q; ++i) {
        cin >> qs[i].l >> qs[i].r;
        qs[i].id = i;
    }
    sort(qs.begin(), qs.end());
    vector<int> bit(n + 1, 0), lastPos(1000005, 0), ans(q);
    auto add = [&](int idx, int v) {
        for (int i = idx; i <= n; i += i & -i) bit[i] += v;
    };
    auto query = [&](int idx) {
        int s = 0;
        for (int i = idx; i > 0; i -= i & -i) s += bit[i];
        return s;
    };
    int curR = 0;
    for (const auto& qu : qs) {
        while (curR < qu.r) {
            ++curR;
            int v = a[curR];
            if (lastPos[v]) add(lastPos[v], -1);
            lastPos[v] = curR;
            add(curR, 1);
        }
        ans[qu.id] = query(qu.r) - query(qu.l - 1);
    }
    for (int i = 0; i < q; ++i) cout << ans[i] << "\n";
    return 0;
}