[C language] Fibonacci sequence (Fibonacci sequence) recursive implementation and non-recursive implementation

Fibonacci sequence leads

The Fibonacci sequence is widely used and will not be listed here.

C language code implementation - recursion

At the same time, there are many kinds of code implementation methods, here are only the implementation methods in C language:

The core idea is to use "recursion".

//C语言：递归实现 Fibonacci数列
//输入n,求 Fibonacci[n]   （在数学中，常常写作Fibonacci(n) ） 
#include <stdio.h>
#define N 100
long Fibonacci(int n);
int main(){
    
    
	int n;
	long k;
	printf("请输入n:");
	scanf("%d",&n);
	k=Fibonacci(n);
	putchar('\n');
	printf("Fibonacci(%d)=%ld\n",n,k);
	
	
	return 0;
	
}
long Fibonacci(int n){
    
    
	if(n==1||n==2)
	{
    
    
		return 1;
	}	
	else{
    
    
		return Fibonacci(n-1)+Fibonacci(n-2);
	}		
}

A few tests:

In the following example, when the data is too large, an error should occur: due to precision issues, representation range issues.
Positive solution: Fibonacci(50) = 12 586 269 025
and pay attention to the time-consuming below, which is also very long.

Looking at the textbook, it can be seen that long intthe range of representation is:
-2 147 483 648 ~ 2 147 483 647(10 digits). And the result of Fibonacci(50) above is 11 digits, of course it will go wrong!

Also, the following result is incorrect:
also because of representation range issues.
Correct solution: Fibonacci(49) = 7 778 742 049

Let's test 正确the results below: ↓ ↓ ↓
In the local test, the 46th item, Fibonacci(46), should be the maximum value that long int can test.
It took 8.822 seconds.

C language code implementation - non-recursive

//非递归实现Fibonacci数列
#include <stdio.h>
long Fibonacci(int n);
int main(){
    
    
	int n;
	long k;
	printf("请输入n:");
	
	scanf("%d",&n);
	k=Fibonacci(n);
	
	putchar('\n');
	printf("Fibonacci(%d)=%ld\n",n,k);
	
	return 0;
}
//非递归实现Fibonacci数列 
long Fibonacci(int n)
{
    
    
	int n1=1,n2=1;
	int t=0,i=0;
	if (n<3)
	{
    
    	//即如果n为1或者2的话，那么Fibonacci(1) = Fibonacci(2)=1 
		return 1;
	}
	else  //下面举个例子，便于理解： 
	{
    
    //例如n为6，我们知道Fibonacci数列为：1、1、2、3、5、8...
	//此时对于下面for循环,则i=0,i<4。开始时n1和n2都是1,即第一和第二项 
		for(i=0; i<n-2; i++)
		{
    
    	
			t = n1 + n2; //t就是数列第三项，为 1+1=2 
			n1 = n2; //注意:令n2赋值给n1,即n1和n2分别 后移一位,这时新的n1即第二项F(2)=1 
			n2 = t;	//这里同理,t作为新的n2，即F(3)=2 
			//依次类推，即可实现求每一项，然后得到 Fibonacci(6),即 Fibonacci(n) 
		}//注意：如果还不好理解，那么动笔写一下就会理解啦！ 
		return t;
	}
}

I think I have written the comments clearly. If you don't understand, take a look at the following picture, it should be well understood:

Then there are a few tests:



it turns out that the above results are correct.

(In addition, for the first 100 items of the Fibonacci sequence, I will upload it to " 资源" in a txt format, which can be downloaded directly.)

As for which one to choose, it is recommended for beginners to use the recursive method, which is easier to understand.