C. She would have to resort to trial and errors to find a matching exponent
Answer:
#include <stdio.h>
int fib(int n) {
if (n <= 0) {
return 0;
}
if (n <= 2) {
return 1;
}
return fib(n-1) + fib(n-2);
}
int main(void) {
for(int nr=0; nr<=20; nr++)
printf("Fibonacci %d is %d\n", nr, fib(nr) );
return 0;
}
Explanation:
The code is a literal translation of the definition using a recursive function.
The recursive function is not per se a very efficient one.
I believe it's A. online meeting
Answer:
Answer explained
Explanation:
From the previous question we know that while searching for n^(1/r) we don't have to look for guesses less than 0 and greater than n. Because for less than 0 it will be an imaginary number and for rth root of a non negative number can never be greater than itself. Hence lowEnough = 0 and tooHigh = n.
we need to find 5th root of 47226. The computation of root is costlier than computing power of a number. Therefore, we will look for a number whose 5th power is 47226. lowEnough = 0 and tooHigh = 47226 + 1. Question that should be asked on each step would be "Is 5th power of number < 47227?" we will stop when we find a number whose 5th power is 47226.
If it's a holiday, you should be getting a year-end bonus. They hope for you to have a new year when you aren't getting a bonus!
