Fibonacci sequence C language three implementation methods

Fibonacci series of mathematical formulas

Include: 1, 1, 2, 3, 5, 8, 13 ,,,

The third term is equal to the first term and the second term

The use of an array of solving

#include <stdio.h>

int main (void)
{
    int i, i_num, F_num[10000];
    F_num[1] = F_num[2] = 1;//下标为1时，时斐波那契而数列数列的第1项 
    scanf("%d", &i_num);//n表示要求的第n项
    if(i_num == 1 || i_num == 2)
         printf("%d\n", F_num[i_num]);
    else if(i_num >= 3)
    {
         for(i = 3; i <= i_num; i ++)
         F_num[i] = F_num[i-1] + F_num[i-2];
         printf("%d\n", F_num[i_num]);
    }
    return 0;
}

The use of iterative solution

#include <stdio.h>

int main (void)
{
    int i, index, i_num;
    int F_num1 = 1, F_num2 = 1; 
    scanf("%d", &i_num);//n表示要求的第n项 
    if(i_num == 1 || i_num == 2)
        printf("%d\n", F_num1);
    else if(i_num >= 3)
    {
          for(i = 3; i <= i_num; i ++)
          {
                 index = F_num2;
                 F_num2 = F_num1 + F_num2;
                 F_num1 = index;
          }
   
          printf("%d\n", F_num2);
    }
   
   return  0;
}

Recursive method to solve

#include <stdio.h>
int  fun(int n)
{
	if( n == 1 || n == 2)
  		return 1;	 
        else if(n >= 3)
                return fun(n-1) + fun(n-2);
}
int main(void)
{
	 int i_num;
 	scanf("%d", &i_num);//n表示要求的第n项 
 	printf("%d\n", fun(i_num));
	return 0;
}

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