I solved this problem using a method similar to ‘search pruning’. I found that the expected complexity of this method is guaranteed by running the code.
How can I prove it mathematically rigorous?
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+10;
struct{int s[2];}t[maxn * 64];
int tot;
void ins(unsigned long long x){for(int i=64,u=0;~--i;u=t[u].s[x>>i&1]?:t[u].s[x>>i&1]=++tot);}
void dfs(int u,int v,int _d){
if(!_d)cout<<"YES",exit(0);
for(int c:{0,1})for(int d:{0,1})
if(!(c&d))if(t[u].s[c]&&t[v].s[d])
dfs(t[u].s[c],t[v].s[d],_d-1);
}
int main() {
mt19937_64 rng(random_device{}());
int n;cin>>n;
for(int i=1;i<=n;++i)ins(rng());
dfs(0,0,64);cout<<"NO";
}
Can you explain the problem you are solving and the code itself?
The problem is determine if the bitwise OR of two numbers is all 1 in binary.
He uses a Trie to keep track of all the numbers, and then traverses the Trie almost brute force and checks.
It's something $$$O(\min((\frac{4}{3})^w,n^2))$$$ under random data, nothing new.
您能解释一下 4/3 的来历吗？如果 $$$n=O(2^w)$$$，那么该做法复杂度显然不是 $$$O((\frac43)^w)$$$ 的啊？
不知道这个回答为啥被踩这么多，codeforces 人都他妈瞎的？一群屁都不懂废物在这瞎骂，换个红名号来评论就老实了。
随机数据下随便两个零一矢量在某一维度不正交概率是 (1/2)^2=1/4，正交概率为 3/4，那么总共 w 位每一位正交概率就是 (3/4)^w，于此期望在找 (4/3)^w 个随机矢量对之后找到一个正交的矢量对。
然后暴力查找所有的是 n^2 的，然后因为找到第一个矢量对之后就直接弹出了，所以是和 (4/3)^w 取 min（不过这里是期望）。
然后你这个 n=O(2^w) 算是我不严谨？不过计算机也不太会允许你放 2^64 个数字进去吧，稍微改一下的话就是 O(max(n log n,min(n^2,(4/3)^w))?
和 trie 没啥关系，直接全存到 vector 里然后跑 n^2 暴力和你这个差不多快。