6 solutions

  • 1
    @ 2022-12-27 19:51:28

    相当炸裂啊

    #include<bits/stdc++.h>
    using namespace std;
    const int N=1e5+6;
    int prime[N];
    bool vis[N];
    int k=0;
    int seive()
    {
    	memset(vis,true,sizeof(vis));
    	vis[0]=vis[1]=false;
    	for(int i=2;i<N;i++)
    	{
    		if(vis[i])
    			prime[k++]=i;
    		for(int j=0;i<k;j++)
    		{
    			if(i*prime[j]>N) break;
    			vis[i*prime[j]]=false;
    			if(!i%prime[j])break;
    		}
    	}
    	return k;
    }
    int main()
    {
    	seive();
    	int n;
    	cin>>n;
    	for(int i=0;i<k;i++)
    	{
    		if(n%prime[i]==0)
    		{
    			printf("%d",n/prime[i]);
    			break;
    		}	
    			
    	}
    	return 0;
    }
    

    Information

    ID
    44
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    7
    Tags
    # Submissions
    772
    Accepted
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