// version B
int powerB(int x, int n)
{
if (n == 0)
return 1;
if (n == 1)
return x;
if (n%2 == 0)
return powerB(x*x, n/2);
else
return powerB(x*x, n/2)*x;
}
// Pre-condition: base raised to the power exp is not greater than 2^31-1, and
// exp is non-negative.
// Post-condition: base raised to the power exp is returned.
int powerB(int base, int exp) {
// Everything raised to the 0 power is 1.
if (exp == 0)
return 1;
// Anything raised to the 1st power is itself.
else if (exp == 1)
return base;
// Utlize the even powered exponent and the rules of exponentiation.
else if (exp%2 == 0)
return powerB(base*base, exp/2);
// If we can't, then just use our regular solution.
else
return base*powerB(base, exp-1);
}
int main() {
// Very basic, NOT comprensive, tests of the recursive methods.
int vals[4];
printf("7! = %d\n", fact(7));
printf("31^2 + 32^2 + ... + 200^2 = %d\n", sumsq(31, 200));
vals[0] = 37; vals[1] = 48; vals[2] = 56; vals[3] = 63;
printf("vals has %d Odd values.\n", ArrayOdd(vals, 4));
print_reverse("writethisbackwards", 18);
printf("\n");
printf("3^10 = %d\n", powerA(3,10));
printf("3^11 = %d\n", powerB(3,11));
if (Rbinary(33, vals, 0, 3) == -1)
printf("33 was not found in vals.\n");
dectobin(179);
printf("\n");
if (check("madamimadam", 11))
printf("madamimadam is a palindrome.\n");
printf("The 27th Fibonacci number is %d\n", fibonacci(27));
// Test of fast exponentiation vs. regular version
int start = time(0);
int ans1 = slowModPow(6874, 1000000000, 13713);
int end1 = time(0);
int ans2 = modPow(6874, 1000000000, 13713);
int end2 = time(0);
printf("ans1 = %d, ans2 = %d.\n", ans1, ans2);
printf("slowModExp took %d sec.\n", end1-start);
printf("modPow took %d sec.\n", end2-end1);
system("PAUSE");
return 0;
}
