On the non recursive and recursive implementation of finding the nth Fibonacci number respectively

Fibonacci sequence ：1,1,2,3,5,8,13,21,34......

The first two numbers are 1, When n>=2 when , hinder Fibonacci The number is the sum of the first two Fibonacci numbers .

Non recursive method to achieve N Fibonacci Numbers

#include <stdio.h>
int main()
{
	int i = 0;
	int a = 1;
	int b = 1;
	int c = a + b;
	int n = 2;
	scanf("%d", &n);
	if (n > 2)//n<3 The output 1,n>3 Enter circulation 
	{
		for (i = 1; i <= n - 3; i++)// As the first 4 Fibonacci Numbers , Only bad circulation 1 Time .
		{
			a = b;		//a=1
			b = c;		//b=2
			c = a + b;	//c=3
		}
		printf("%d\n", c);// Output No 4 Fibonacci Numbers 3
	}
	else

		printf("%d\n", 1);//


	return 0;
}

The recursive method realizes the N Fibonacci Numbers

#include <stdio.h>
int Fibonacci_number(int n)
{
	if (n == 1)
	{
		return  1;
	}
	if (n == 2)
	{
		return 1;
	}
	else
		return Fibonacci_number(n - 1) + Fibonacci_number(n - 2);
	
}
int main()
{
	int n = 0;
	scanf("%d",&n);
	printf(" The first %d The Fibonacci number is %d\n", n, Fibonacci_number(n));
	return 0;
}

Recursive drawing demonstration

If scanf Input n yes 4, Get into Fibonacci_number(n) Function body

  

After calling the function ,return Return value , First return the last recursive function Fibonacci_number(n - 1) The value of is 1, Fibonacci_number(n - 2) yes 1,1+1=2; Return to the second recursive return 2、1, The final return value of the main function is 3.

Final output ： The first 4 The Fibonacci number is 3.