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52 lines
1.3 KiB
52 lines
1.3 KiB
#include <bits/stdc++.h>
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using namespace std;
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//为什么这个平衡树也可以用树状数组来做呢？
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const int N = 1e5 + 10;
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int n, tot, sz;
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int op[N], a[N], t[N], c[N];
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int pos(int x) {
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return lower_bound(t + 1, t + 1 + sz, x) - t;
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}
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void add(int x, int k) {
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for (x; x <= sz; x += x & -x) c[x] += k;
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}
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int query(int x) {
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int tmp = 0;
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for (x; x; x -= x & -x) tmp += c[x];
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return tmp;
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}
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int kth(int x) {
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int rs = 0;
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for (int i = 17; i >= 0; --i) {
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rs += (1 << i);
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if (rs > sz || c[rs] >= x)
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rs -= (1 << i);
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else
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x -= c[rs];
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}
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return t[rs + 1];
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}
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int main() {
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scanf("%d", &n);
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//树状数组为啥需要离线处理呢
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for (int i = 1; i <= n; i++) {
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scanf("%d%d", &op[i], &a[i]);
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if (op[i] != 4) t[++tot] = a[i];
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}
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//离散化
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sort(t + 1, t + 1 + tot);
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sz = unique(t + 1, t + 1 + tot) - t - 1;
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for (int i = 1; i <= n; i++) {
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int x = a[i];
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if (op[i] == 1) add(pos(x), 1);
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if (op[i] == 2) add(pos(x), -1);
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if (op[i] == 3) printf("%d\n", query(pos(x) - 1) + 1);
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if (op[i] == 4) printf("%d\n", kth(x));
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if (op[i] == 5) printf("%d\n", kth(query(pos(x) - 1)));
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if (op[i] == 6) printf("%d\n", kth(query(pos(x)) + 1));
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}
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return 0;
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} |