Orsay information through a 1211: Analyzing element exists

1211: Analyzing element exists
[Title] Description
has generated a set of M is: (1) k is a known set of element M; (2) if y is an element of M, then, 2y + 1 and 3y + 1 It is the element M; (3) in addition to the above two cases, nothing else can become a number of element M.

Question: any given k and x, please determines whether x is an element of M. where k is an unsigned integer, x is not greater than 100,000, and if so, output YES, otherwise, output NO.

[INPUT]
input integer k and x, separated by commas.

[Output]
If so, then the output YES, otherwise output NO.

[Sample input]
0,22
[sample] output
YES
obviously recursive

#include<iostream>
using namespace std;
bool judge(int x,int y)
{
    //cout<<x<<" "<<y<<endl;
    if(x==y)
        return true;
    if(x>y)
        return false;
    if(x<y)
    {
        int c=x;
        (x*=3)++;
        (c*=2)++;
        return (judge(x,y)||judge(c,y));//两个条件满足其一
    }
}
int main()
{
    char d;
    int k,n;
    cin>>k>>d>>n;
    if(judge(k,n))
    {
        cout<<"YES";
    }
    else cout<<"NO";
}