Thread: Factorial conversion

Hey guys I'm wondering how I could convert this to one that computes 3^n - 2^n....that is 3 to the power n minus 2 to the power n.

Code:

#include<iostream>
using namespace std;

template <int n> struct Fac // general case n! = n*(n-1)!
{
	enum { result = 3 ^ n - 2 ^ n }; // constant
};
//template <> struct Fac<0> // base case 0! = 1
//{
///enum { result = 1 }; // constant
//};
int main()
{
// Fac<c>::result is now a compile-time constant!
cout << Fac<5>::result << endl; 
cout << Fac<10>::result << endl;
cout << Fac<0>::result << endl; 
}

yeah i dont think C++ has the ^ (maybe with math.h, but idk)

it's not going to be as simple as you'd hope, but heres how you can do it:

Code:

int iIncrement = 0;   //increment used for the while loop
int iResult3 = 3;      // the number 3 to use in power
int iResult2 = 2;      //the number 2 to use in power
int iFinalResult;       //the result after we subtract 3 - 2
int iInput;   //this is the "n" that the user [i think thats what you want?] inputted

while (iIncrement <= iInput) {   //while the increment is still less than what the user inputted
     iIncrement++;      //add increment one
     iResult3 = iResult3 * iResult3;    //do 3 * 3
     iResult2 = iResult2 * iResult2;    //do 2 * 2
}          //it will do 3*3 and 2*2 n (or iInput) amount of times
//since, 3 ^ n is the same as 3 * 3 n ammount of times

iFinalResult = iResult3 - iResult2;  //and finally subtract the two powers

and, as a note, the increment way i used, its probably off a number or 2, but im hoping you can fix this

Thanks a lot bluehead greatly appreciated
but it has to be in a template class so you can compute at compile time, and you have to instantiate with Foo<n>::result

HOw does f(n+1) relate to f(n)?

Originally Posted by micha_mondeli

Thanks a lot bluehead greatly appreciated
but it has to be in a template class so you can compute at compile time, and you have to instantiate with Foo<n>::result

Template programming can be quite tricky, and judging from your rather naïve attempt

Code:

result = 3 ^ n - 2 ^ n ,

I don't think you're ready to do template programming. First, try to solve the problem using normal code. Then we will help you convert the code to template programming.