CF1325F Ehab's REAL Number Theory Problem

Ehab's REAL Number Theory Problem

You are given an array \(a\) of length \(n\) that has a special condition: every element in this array has at most \(7\) divisors. Find the length of the shortest non-empty subsequence of this array product of whose elements is a perfect square.

\(1 \le n \le 10^5,1 \le a_i \le 10^6\)

题解

https://codeforces.ml/blog/entry/74235

Notice that for each element in the array, if some perfect square divides it, you can divide it by that perfect square, and the problem won't change. Let's define normalizing a number as dividing it by perfect squares until it doesn't contain any. Notice than any number that has \(3\) different prime divisors has at least \(8\) divisors, so after normalizing any element in the array, it will be \(1\), \(p\), or \(p*q\) for some primes \(p\) and \(q\). Let's create a graph where the vertices are the prime numbers (and \(1\),) and the edges are the elements of the array. For each element, we'll connect \(p\) and \(q\) or \(p\) and \(1\) if it's a prime after normalizing (只有 \(1\) 说明答案就是它一个数). What's the significance of this graph? Well, if you take any walk from node \(p\) to node \(q\), multiply the elements on the edges you took, and normalize, the product you get will be \(p*q\)! That's because every node in the path will be visited an even number of times, except p and q. So the shortest subsequence whose product is a perfect square is just the shortest cycle in this graph!

这题的主要难点在于如何找环。

The shortest cycle in an arbitrary graph takes \(O(n^2)\) to compute: you take every node as a source and calculate the bfs tree, then you look at the edges the go back to the root to close the cycle. That only finds the shortest cycle if the bfs source is contained in one. The graph in this problem has a special condition: you can't connect 2 nodes with indices greater than \(\sqrt{\max A_i}\). That's because their product would be greater than \(\max A_i\). So that means ANY walk in this graph has a node with index \(\le\sqrt{\max A_i}\). You can only try these nodes as sources for your bfs.

时间复杂度 \(O(A\ln A+\sqrt{A}n)\)

CO int N=1e6;
vector<int> pr,d[N+10];
int lp[N+10];
array<int,2> e[N];
vector<int> to[N+10];
int dis[N+10];

int main(){
	pr.push_back(1);
	for(int i=2;i<=N;++i){ // prime factors
		if(!lp[i]){
			pr.push_back(i);
			for(int j=i;j<=N;j+=i) lp[j]=i; // last prime
		}
		d[i]=d[i/lp[i]];
		if(d[i].size() and d[i].back()==lp[i])
			d[i].pop_back(); // p^2 => 1
		else d[i].push_back(lp[i]);
	}
	int n=read<int>();
	for(int i=1;i<=n;++i){
		int a=read<int>();
		if(d[a].empty()){
			puts("1");
			return 0;
		}
		if(d[a].size()==1) d[a].push_back(1);
		e[i]={d[a][0],d[a][1]};
		to[d[a][0]].push_back(i),to[d[a][1]].push_back(i);
	}
	int ans=1e9;
	for(int i:pr){
		if(i*i>N) break;
		for(int j:pr) dis[j]=0;
		deque<pair<int,int> > Q;
		for(int j:to[i]){
			Q.push_back({j,e[j][0]==i});
			dis[e[j][0]^e[j][1]^i]=1;
		}
		while(Q.size()){
			pair<int,int> p=Q.front();Q.pop_front();
			int x=e[p.first][p.second];
			for(int u:to[x])if(u!=p.first){
				pair<int,int> q={u,e[u][0]==x};
				int y=e[q.first][q.second];
				if(!dis[y] and y!=i){
					dis[y]=dis[x]+1;
					Q.push_back(q);
				}
				else ans=min(ans,dis[x]+dis[y]+1);
			}
		}
	}
	printf("%d\n",ans==1e9?-1:ans);
	return 0;
}