3 solutions

  • 12
    @ 2024-5-15 13:58:30
    #include<bits/stdc++.h>
    using namespace std;
    int main()
    {
    	int a;
    	cin>>a;
    	int s=0; //约数的个数
    	for(int i=1;i<=a;i++)
    	{
    		if(a%i==0) //小的约数存在
    		{
    			s++;
    		}
    	}
    	cout<<s;
    	return 0;
    }
    
    
    • 0
      @ 2026-8-24 20:18:18
      #include<bits/stdc++.h>
      using namespace std;
      #define int long long
      int n,ans;
      signed main()
      {
      	cin>>n;
      	for(int i=1;i<=sqrt(n);i++)
      	{
      		if(n%i==0)
      		{
      			if(i*i==n)
      			{
      				ans++;
      			}
      			else
      			{
      				ans+=2;
      			}
      		}
      	}
      	cout<<ans;
      	return 0;
      }
      
      
      #include<bits/stdc++.h>
      using namespace std;
      #define int long long
      int n,ans;
      signed main()
      {
      	cin>>n;
      	for(int i=1;i<=sqrt(n);i++)
      	{
      		if(n%i==0)
      		{
      			if(i*i==n)
      			{
      				ans++;
      			}
      			else
      			{
      				ans+=2;
      			}
      		}
      	}
      	cout<<ans;
      	return 0;
      }
      
      
      • 0
        @ 2026-8-10 14:20:25
        #include<bits/stdc++.h>
        using namespace std;
        int main()
        {
            int a;
            cin>>a;
        	int cnt=0;
        	for(int b=1;b<=a;b++)
        	{
        		if(a%b==0)
        		{
        			cnt++;
        		}
        	}
        	cout<<cnt;
        	return 0; 
        }
        
        • 1

        Information

        ID
        33
        Time
        1000ms
        Memory
        256MiB
        Difficulty
        1
        Tags
        (None)
        # Submissions
        328
        Accepted
        89
        Uploaded By