Author has gone out of the stories about Vasiliy, so here is just a formal task description.
You are given q queries and a multiset A, initially containing only integer 0. There are three types of queries:
- "+ x" — add integer x to multiset A.
- "- x" — erase one occurrence of integer x from multiset A. It's guaranteed that at least one x is present in the multiset A before this query.
- "? x" — you are given integer x and need to compute the value , i.e. the maximum value of bitwise exclusive OR (also know as XOR) of integer x and some integer y from the multiset A.
Multiset is a set, where equal elements are allowed.
The first line of the input contains a single integer q (1 ≤ q ≤ 200 000) — the number of queries Vasiliy has to perform.
Each of the following q lines of the input contains one of three characters '+', '-' or '?' and an integer xi (1 ≤ xi ≤ 109). It's guaranteed that there is at least one query of the third type.
Note, that the integer 0 will always be present in the set A.
For each query of the type '?' print one integer — the maximum value of bitwise exclusive OR (XOR) of integer xi and some integer from the multiset A.
10
+ 8
+ 9
+ 11
+ 6
+ 1
? 3
- 8
? 3
? 8
? 11
11
10
14
13
After first five operations multiset A contains integers 0, 8, 9, 11, 6 and 1.
The answer for the sixth query is integer — maximum among integers , , , and .
题目大意:给一定操作,(1)增加一个数x, (2)删除一个数x, (3)询问在该multiset的数中与x异或的最大值
个人感悟:看了多种做法, 有用字典树的, 还有直接用stl Multiset用位运算模拟, 相比之下我觉得后者过程更简洁,但更难理解些。
题目解析:下面将提供后者的做法及解释,如果有讲的错误和不懂的地方, 欢迎在下方评论
假设x的二进制形式为0000 1000(8),那么要想取得最大值, 则y应该与x的二进制形式相反才可以, 如0111 0111(119), x^y = 0111 1111(127)
所以我们尽可能使y的二进制形式 与x对应0位置 从左到右 存在1, (可能有点拗口)继续以x的二进制形式为0000 1000(8), 如果存在0100 0000(64), 0011 1111(63)这两个数,我们肯定会选64, 就是这个从左向右选择1的道理
那么我们怎么判断最高为是否为1呢, 我们可以用(1 << i)表示最高位为1, ( 注 i : 29 -> 0) 然后 ~x&(1 << i)就可以得到最高位了。最后再用个数ans记录当前能取到的最大值
ans = ans | (~x&(1 << i)), 判断这个ans是否在multimap中, 如果不存在就返回到之前的ans, ans = ans ^ (1 << i)
接下来就讲一下STLmultimap 的一些简单函数
首先要定义multimap : multimap<int>ms;
增加一个数: ms.insert(x);
删除所有键值为x的数:ms.earse(x)
删除一个数:ms.earse(ms.find(x))
二分查找:ms.lower_bound(x);
注:multimap中的数按照从小到大排序
然后可以贴代码了
#include <iostream> #include <set> #include <cstdio> using namespace std; multiset<int>s; int main() { ios::sync_with_stdio(false); int n, x; char op; cin >> n; s.insert(0); while(n--){ cin >> op >> x; if(op == '+'){ s.insert(x);//添加一个数 }else if(op == '-'){ s.erase(s.find(x));//删除一个数 }else if(op == '?'){ int ans = 0; for(int i = 29; i >= 0; i--){ ans |= (~x&(1 << i));//求出最高位的数,然后不断往下搜 auto it = s.lower_bound(ans);//二分搜索, 得到一个 if(it == s.end() || *it >= ans + (1 << i)){//没有搜到, 第一个是搜到末尾了,第二个是搜到了一个大于ans的数,具体可以查阅lower_bound函数 ans ^= 1<<i; } } x ^= ans; cout << x << endl; } } return 0; }