Converting A Basic Program to A C++ Program

[bsmath]

I wrote the following program in Basic. It works in Basic, but I cannot get it
converted to C++. Any help in that direction will be appreciated. The values for
a, S, and R can be changed.

a=3
for S=1 to 5 step 2
R=2
N=((2^R)*S)+1
if int(((a^S)-1)/((2^R)*S+1))=((a^S)-1)/N then print "yes"; "S="; S; "R="; R; "a="; a else print "no"
next S

[newb16]

If you know the algirithm, reimplement it in C. If A^B is raisong to power, use pow(a,b) function for it.

[randysimes]

int a(3), S, R;
for (S=1; S <=5; S+=2)
{
R=2
N=((2^R)*S)+1
if (int(((a^S)-1)/((2^R)*S+1))==((a^S)-1)/N)
 std::cout<< "yes\n"
              << "S= " << S
              << "\nR= " << R
               << "\na= " << a;
 else 
   std::cout << "no";
}

[whodgson]

You could write something like the following:

#include<iostream>
#include<cmath>
using namespace std;
 
int main()
{
    double a=3.0,R=2;
    for(int S=1;S<11;S+=2)
    {
        double N=pow(2,R)+1;
        double X=pow(a,S)-1;
        double Y=pow(2,R)*S+1;
        if(X/Y==X/N)cout<<"yes";
        else cout<<"no";
      cout<<"\tS: "<<S<<"\tR: "<<R<<"\ta: "<<a<<endl;      
      }
char q;
cout<<"press any key to continue...";
cin.get(q);
return 0;
} 

//output
/*yes S: 1 R: 2 a: 3
no S: 3 R: 2 a: 3
no S: 5 R: 2 a: 3
no S: 7 R: 2 a: 3
no S: 9 R: 2 a: 3
press any key to continue...*/

[whodgson]

Note: The above code emulates the BASIC sphagetti style which would not attract the lowest level of approbation from a C++ professional. In this case two functions should have been used; one a bool type to test the expression and one called to print the yes/no and R,S and a values

[bsmath]

Thank you for your answer.