This task is to find the number of 'For' loops present in the given cpp program.

Example Input: int varieties=0,p;
int dfs(int dep,int last,int k)    //???????????k????? ?? 
{
    if (k == 1)              //??k=1 
    {
        varieties++;            //???+1 
        return 0;
    }
    for (int i = last; i <= k; i++)      
    if (k % i == 0)                    //??k??i????????????k=1 
    {
        dfs(dep+1,i,k/i);
    }
}
int main()
{
    int i,n,t;
    cin>>n;            //?????? 
    int m[n];          //????n???????? 
    for(i=0;i<n;i++)
    {
        cin>>t;            
        dfs(0,2,t);
        m[i]=varieties;
        varieties=0;
    }
    for(i=0;i<n;i++)
    cout<<m[i]<<endl;      //??????????? 
    return 0;
}
Example Output: 3

Example Input: int shuliang(int m,int n);
main()
{
    int n;
    scanf("%d",&n);
    for (int i = 1;i <= n;i ++)
    {
        int x,q,answer=1;
        scanf("%d",&x);
        q=sqrt(x)+1;
        for (int j = 2; j < q; j ++)
        {
            if (x%j==0)
                answer+=shuliang(j,x/j);
                
        }
        printf("%d\n\n",answer);
    }
}

int shuliang(int m,int n)
{
    int q,answer=0,y=0;
    q=sqrt(n)+1;
    for (int j = m;j < q;j ++)
    {
        if (n%j==0)
            answer+=shuliang(j,n/j),y=1;
    }
    if(y==1)
        return answer+1;
    else
        return 1;
}
Example Output: 3

Example Input: int factors(int n,int min)
{
     int varieties=0,i;
     if(n==1)  return 1;
     else
     {
            for(i=min;i<=n;i++)
            {
                if(n%i==0)
                {
                    varieties+=factors(n/i,i);
                }
            }
        }
        return varieties;
}
int main()
{
    int n;
    cin>>n;
    int a[n],b[n];
    int i;
    for(i=0;i<n;i++)
    {
        cin>>a[i];
        b[i]=factors(a[i],2);
    }
    for(i=0;i<n;i++)
    cout<<b[i]<<endl;
    return 0;
}

Example Output:
3