int getNthFib(int n) { // assume n is positive if (n <= 2) return n; int i = 1; int j = 2; int k; count = 2; while (count < n) { k = i + j: count ++; i = j; j = k; } return k;

I've been asked this question several times at interviews, and it's usually associated with recursion, so the answer most would like to see is like: int getFib(int n) { return n <= 1 ? n : getFib(n - 1) + getFib(n - 2); } The interviewer will also want to know you understand that recursion would be slow for high values of n.

#include using namespace std; int getFib(int n) { if(n==1||n==2) return n; return getFib(n-1)+getFib(n-2); } int main(){ int n=6; cout<