// Chap 2, p 71, p 72

// Demonstrates Pow1, Pow2, and Pow3.
#include <iostream.h>

int Pow1(int X, int N)
// ---------------------------------------------------
// Exponentiation function -- ITERATIVE SOLUTION
// Precondition: X is an integer; N is a nonnegative
// integer.
// Postcondition: Returns X raised to the Nth power.
// ---------------------------------------------------
{
   int Temp = 1;
   for (int Exponent = 1; Exponent <= N; ++Exponent)
      Temp *= X;
   return Temp;
}  // end Pow1

int Pow2(int X, int N)
// Exponentiation function -- RECURSIVE SOLUTION
{
   if (N == 0)
      return 1;
   else
      return X * Pow2(X, N-1);
}  // end Pow2

int Pow3(int X, int N)
// Exponentiation function -- MORE EFFICIENT
// RECURSIVE SOLUTION
{
   if (N == 0)
      return 1;

   else
   {  int HalfPower = Pow3(X, N/2);
      if (N % 2 == 0)
	 return HalfPower * HalfPower;      // even N

      else
	 return X * HalfPower * HalfPower;  // odd N
   }  // end else
}  // end Pow3

// ******SAMPLE MAIN PROGRAM******
main()
{  int  Number, Result1, Result2, Result3, Power;
   char Response;

   do
   {  cout << "\n\nEnter an integer, a space, and an integer exponent: ";
      cin >> Number >> Power;

      Result1 = Pow1(Number, Power);
      Result2 = Pow2(Number, Power);
      Result3 = Pow3(Number, Power);

      cout << "\n" << Number << " to the " << Power << " is\n";
      cout << "  " << Result1 << " using Pow1\n";
      cout << "  " << Result2 << " using Pow2\n";
      cout << "  " << Result3 << " using Pow3\n\n";

      cout << "Continue? ";
      cin >> Response;
   }  while ((Response != 'n') && (Response != 'N'));
   return(0);
} // end program