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87 lines
2.3 KiB
87 lines
2.3 KiB
2 years ago
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##[$HDU$ $2852$ $KiKi's$ $K-Number$](http://acm.hdu.edu.cn/showproblem.php?pid=2852)
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### 一、题目大意
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给定三种操作:
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`0 x` 表示把`x`插入容器 ;
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`1 x` 表示删除一个`x`如果没有`x`则输出 `No Elment!` ;
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`2 a k` 表示比`a`大的数中的第`k`大的数 如果没有输出`No Find!`
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### 二、解题思路
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* 树状数组维护元素出现次数(权值)前缀和即可
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* 操作$0$即修改
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* 操作$1$先查询$x$是否存在，如果存在删除一个即可
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* 操作$2$可以用二分逐渐逼近答案
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### 三、二分+树状数组+维护个数
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**复杂度：$O(logN * logN)$**
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```cpp {.line-numbers}
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#include <iostream>
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#include <algorithm>
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#include <cstdio>
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#include <cstring>
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typedef long long LL;
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using namespace std;
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const int N = 100010;
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// 树状数组模板
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#define lowbit(x) (x & -x)
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typedef long long LL;
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int c[N];
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void add(int x, int v) {
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while (x < N) c[x] += v, x += lowbit(x);
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}
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LL sum(int x) {
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LL res = 0;
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while (x) res += c[x], x -= lowbit(x);
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return res;
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("HDU2852.in", "r", stdin);
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#endif
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// 加快读入
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ios::sync_with_stdio(false), cin.tie(0);
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int m;
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while (cin >> m) {
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memset(c, 0, sizeof c);
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while (m--) {
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int op;
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cin >> op;
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if (op == 0) {
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int x;
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cin >> x;
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add(x, 1);
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} else if (op == 1) {
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int x;
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cin >> x;
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if (sum(x) - sum(x - 1) == 0)
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puts("No Elment!");
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else
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add(x, -1);
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} else {
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int a, k;
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cin >> a >> k;
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int x = sum(a);
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if (sum(N - 1) - x < k) // 剩余数量不足K个
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puts("Not Find!");
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else {
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int l = 1, r = N - 1;
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while (l < r) {
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int mid = (l + r) >> 1;
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if (sum(mid) - x >= k)
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r = mid;
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else
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l = mid + 1;
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}
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printf("%d\n", l);
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}
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}
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}
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}
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return 0;
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}
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```
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