107 lines
3.1 KiB
C++
107 lines
3.1 KiB
C++
#include <algorithm>
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#include <cstdio>
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#include <cmath>
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#include <cstring>
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using namespace std;
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//快读
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int read() {
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int x = 0, f = 1;
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char ch = getchar();
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while (ch < '0' || ch > '9') {
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if (ch == '-') f = -1;
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ch = getchar();
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}
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while (ch >= '0' && ch <= '9') {
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x = (x << 3) + (x << 1) + (ch ^ 48);
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ch = getchar();
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}
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return x * f;
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}
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const int N = 400010;
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int n, m;
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int a[N]; //维护的序列
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int ans[N]; // 结果
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struct Node {
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// op=1 插入, op=-1 删除, op=2 查询
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int op;
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// 插入：l:插入的值,用于二分与mid PK,r:位置
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// 删除：l:删除的值,用于二分与mid PK,r:位置
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// 查询 [l,r]：查询范围
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// k:第k小
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int l, r, k;
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int id;
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} q[N], lq[N], rq[N];
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//树状数组
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int tr[N];
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void add(int x, int c) {
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for (; x < N; x += x & -x) tr[x] += c;
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}
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//查询x的个数和
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int getSum(int x) {
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int sum = 0;
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for (; x; x -= x & -x) sum += tr[x];
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return sum;
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}
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void solve(int l, int r, int ql, int qr) {
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if (ql > qr) return; //防止RE，递归边界
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if (l == r) { //递归到叶子结点，找到答案
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for (int i = ql; i <= qr; i++)
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if (q[i].op == 2) ans[q[i].id] = l; //所有查询请求，标识答案
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return;
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}
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int mid = (l + r) >> 1;
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int nl = 0, nr = 0;
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for (int i = ql; i <= qr; i++) {
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if (q[i].op < 2) {
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//插入、删除操作二分
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if (q[i].l <= mid)
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add(q[i].r, q[i].op), lq[++nl] = q[i]; //根据q[i].op的不同，通过一样的add函数，在位置q[i].r处，完成插入+1，删除-1的操作
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else
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rq[++nr] = q[i];
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} else {
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//查询数组二分
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int t = getSum(q[i].r) - getSum(q[i].l - 1);
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if (q[i].k <= t)
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lq[++nl] = q[i];
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else {
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q[i].k -= t;
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rq[++nr] = q[i];
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}
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}
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}
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// 有借有还，再借不难
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for (int i = ql; i <= qr; i++)
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if (q[i].op < 2)
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if (q[i].l <= mid) add(q[i].r, -q[i].op);
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for (int i = 1; i <= nl; i++) q[ql - 1 + i] = lq[i]; //将左儿子操作数组，拷贝到q中，原来的下标是[0~ql-1],现在无缝接上[ql-1+1~ql-1+nl]
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for (int i = 1; i <= nr; i++) q[ql - 1 + nl + i] = rq[i]; //将右儿子操作数组，拷贝到q中，原来的下标是[ql-1+nl+i]
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//继续二分左右儿子
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solve(l, mid, ql, ql + nl - 1), solve(mid + 1, r, ql + nl, qr);
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}
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int main() {
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int l, r, k, c = 0, qc = 0;
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n = read(), qc = read();
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for (int i = 1; i <= n; i++) a[i] = read(), q[++c] = {1, a[i], i}; // op=1:插入,a[i]:值，i:位置
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for (int i = 1; i <= qc; i++) {
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l = read(), r = read(), k = read();
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q[++c] = {2, l, r, k, i}; // op=2：查询,[l,r]:查询范围,k:第k小,i:查询序号
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}
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//整体二分
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solve(-1e9, 1e9, 1, c); //值域[-1e9,1e9],二分查询区间[1,n+m]
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//结果
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for (int i = 1; i <= qc; i++) printf("%d\n", ans[i]);
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return 0;
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}
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