Using C + + recursion to solve jump-step problem _c language

Topic:

A step has a total of n-level, if you can jump 1, you can jump 2 level. How many jumps are there in total?

Analysis:

It is also a relatively basic topic, which can be easily solved by recursion.
The FIB (n) is used to denote the number of jumps of N-step steps, the frog jumps up to the N-Step step number 1 (n-step Jump) and sets the FIB (0) = 1;
When n = 1 o'clock, there is only one method of jumping, that is, 1-step Jump: Fib (1) = 1;
When n = 2 o'clock, there are two ways of jumping, first and second order jumps: Fib (2) = Fib (1) + Fib (0) = 2;
When n = 3 o'clock, there are three kinds of ways to jump, the first step out of one, followed by FIB (3-1) in the Jump method; The first time out of second order, followed by FIB (3-2) in the Jump method; The first time out of the third Order, followed by the FIB (3-3) in the Jump method
Fib (3) = Fib (2) + FIB (1) +fib (0) = 4;
When n = n, there are n ways to jump, the first step out of a there is also fib (n-1) in the back of the Jump method, the first time out of the second order, there are fib (n-2) in the Jump method ... ..... The first time to jump out of the N-order, there are Fib (n-n) in the back of the jump method.
Fib (n) = Fib (n-1) +fib (n-2) +fib (n-3) +..........+fib (n-n) =fib (0) +fib (1) +fib (2) +.......+fib (n-1)
Also because of Fib (n-1) =fib (0) +fib (1) +fib (2) +.......+fib (n-2)
Two-type subtraction: Fib (n)-fib (n-1) =fib (n-1) ===== "Fib (n) = 2*fib (n-1) n >= 2
The recursive equation is as follows:

The code is implemented as follows (GCC compiles through):

#include "stdio.h"
#include "stdlib.h"
 
int function (int n);
 
int main (void)
{
  int tmp;
   
  TMP = function (5);
  printf ("%3d\n", TMP);
 
  return 0;
}
 
int function (int n)
{
  if (n = 1) return
    1;
  else if (n = = 2) return
    2;
  else return  
    function (n-1) + function (n-2);
}