'commit'

TangDou/AcWing/BalanceTree/TODO/253_Szsz.cpp

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