C language jump step problem solution

There are n steps in total. If you can jump 1 step at a time, you can also jump 2 steps. Find the total number of jumping methods, and analyze the time complexity of the algorithm.
Answer: a function f(n) is used to represent the total jump of n steps.
1. If there is only one step, f(1) = 1;
2. If there are 2 steps, f(2) = 2;
3, when there are n steps, if the first jump 1 level, there are f(n-1) jumping method, if the first jump 2 levels, there are f(n-2) jumping method, that is, f(n) = f(n-1) + f(n-2).
Is the Fibonacci sequence.

#include "stdafx.h"
#include <iostream>
using namespace std;
//cycle
int TotalStep(int n)
{
    if (n <= 0)
    {
        return 0;
    }
    else if (1 == n || 2 == n)
    {
        return n;
    }
    int first = 1;
    int second = 2;
    int total = 0;
    for (int i = 3; i <= n; i++)
    {
        total = first + second;
        first = second;
        second = total;
    }
    return total;
}
//recursive
int RecurTotalStep(int n)
{
    if (n <= 0)
    {
        return 0;
    }
    else if (n == 1 || n == 2)
    {
        return n;
    }
    else
    {
        return RecurTotalStep(n - 1) + RecurTotalStep(n - 2);
    }
}
int _tmain(int argc, _TCHAR* argv[])
{
    cout<<TotalStep(20)<<endl;
    cout<<RecurTotalStep(20)<<endl;
    return 0;
}

The operation interface is as follows:

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