在尝试比较两个代码的速度时，时间复杂度是唯一需要考虑的事情吗？

Question

Recently i wrote a code for finding primefactors of a number within the limits of "int" datatype in C. When i showed it to my friend he told that there are more optimised ways which have time complexities of O(sqrt(n)), O(log(n)). When I ran the code having time complexity O(sqrt(n)) and my code on Dev C++ i saw that the total time taken by both the codes was nearly the same. What is the time complexity of my code?

#include <stdio.h>
int main(void)
{
    int k = 2, n, m;
    printf("enter a +ve integer greater than 1:- ");
    scanf("%d", &n);
    m = n;
    do {
        if (n % k == 0) {
            n /= k;
            printf("%d,", k);
        }
        if (n == 1)
            break;
        while (n % k != 0) {
            if (k == 2)
                k = 3;
            else
                k += 2;
        }
    } while (n % k == 0);
    printf("prime factors of %d\n.", m);
    return 0;
}

Answer 1

The time-complexity analysis is made for the sake of large N . If you do not feel the effect yet, your N is not large enough. I changed the %d s to %lld and int s to long long int and factored the square of the Mersenne prime that Euler found, ie (2³¹-1)².

Results:

% echo 4611686014132420609 | time ./a.out
enter a +ve integer greater than 1:- 2147483647,2147483647,prime factors of 4611686014132420609
../a.out  12.15s user 0.00s system 99% cpu 12.154 total
% echo 4611686014132420609 | time factor
4611686014132420609: 2147483647 2147483647
factor  0.00s user 0.00s system 87% cpu 0.001 total

Ie even when compiled with GCC and -O3, your program required at least 12000 times as much CPU power to factor the number than the factor program.