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1求区间第k大

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 2e5 + 5;
const ll MOD = 100000007;
ll a[MAXN], b[MAXN], n, m, q, sz;
int lc[MAXN << 5], rc[MAXN << 5], rt[MAXN << 5], sum[MAXN << 5];
void init() {
    sz = 0;
    rt[0] = 0;
}
void build(int &rt, int l, int r) {
    rt = ++sz;
    sum[rt] = 0;
    if (l == r) return ;
    int mid = l + r >> 1;
    build(lc[rt], l, mid);
    build(rc[rt], mid + 1, r);
}
int update(int o, int l, int r, int p) {
    int oo = ++sz;
    lc[oo] = lc[o], rc[oo] = rc[o], sum[oo] = sum[o] + 1;
    if (l == r) return oo;
    int mid = l + r >> 1;
    if (mid >= p) {
        lc[oo] = update(lc[oo], l, mid, p);
    } else {
        rc[oo] = update(rc[oo], mid + 1, r, p);
    }
    return oo;
}

int query(int u, int v, int l, int r, int k) {
    int mid = l + r >> 1;
    if (l == r) return l;
    // 改变左右即可求区间第k大
    // 当前为区间第k小
    // int x = sum[lc[v]] - sum[lc[u]];
    // if(x >= k) return query(lc[u], lc[v], l, mid, k);
    // else return query(rc[u], rc[v], mid + 1, r, k - x);
    // 当前为区间第k大
    int x = sum[rc[v]] - sum[rc[u]];
    if(x >= k) query(rc[u], rc[v], mid + 1, r, k);
    else return query(lc[u], lc[v], l, mid, k - x);
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    cin >> n >> m;
    init();
    for(int i = 1; i <= n; i++) {
        cin >> a[i];
        b[i] = a[i];
    }
    // 主席树有序节点
    sort(b + 1, b + 1 + n);
    // 不同元素个数
    q = unique(b + 1, b + 1 + n) - b - 1;
    // 构建0为根的一颗线段树
    build(rt[0], 1, q);
    for(int i = 1; i <= n; i++) {
        // 将元素一个一个插入进去
        int tmp = lower_bound(b + 1, b + 1 + q, a[i]) - b;
        rt[i] = update(rt[i - 1], 1, q, tmp);
    }
    while(m--) {
        int l, r, k;
        cin >> l >> r >> k;
        // 查询区间第k小
        cout << b[query(rt[l - 1], rt[r], 1, q, k)] << endl;
    }
    return 0;
}

2.求区间前k大的和

void build(int &rt, int l, int r) {
    rt = ++sz;
    sum[rt] = 0;
    ans[rt] = 0;
    if (l == r) return ;
    int mid = l + r >> 1;
    build(lc[rt], l, mid);
    build(rc[rt], mid + 1, r);
}

int update(int o, int l, int r, int p) {
    int oo = ++sz;
    lc[oo] = lc[o], rc[oo] = rc[o], sum[oo] = sum[o] + 1, ans[oo] = ans[o] + b[p];
    if (l == r) return oo;
    int mid = l + r >> 1;
    if (mid >= p) {
        lc[oo] = update(lc[oo], l, mid, p);
    } else {
        rc[oo] = update(rc[oo], mid + 1, r, p);
    }
    return oo;
}
ll querySum(int u, int v, int l, int r, int k) {
    int mid = l + r >> 1;
    if (l == r) return b[l] * k;
    // 当前为区间第k大
    int x = sum[rc[v]] - sum[rc[u]];
    if(x >= k) return querySum(rc[u], rc[v], mid + 1, r, k);
    else return querySum(lc[u], lc[v], l, mid, k - x) + ans[rc[v]] - ans[rc[u]];
}