C语言训练-斐波纳契数列-题解（C语言代码）

解题思路:

注意事项:

参考代码:

#include<stdio.h>

#include<math.h>

int main()

{

     int n,t = 0, t0 = 1, t1 = 1;

     scanf("%d",&n);

     if(n == 1)

          printf("%d\n",t0);

     else if(n == 2)

          printf("%d %d\n",t0,t1);

     else

     {

         printf("%d %d ",t0,t1);

         for(int i = 3; i <= n; i++)

         {

             t = t0 + t1;

             t0 = t1;

             t1 = t;

             if(i != n)

                  printf("%d ",t);

             else

                  printf("%d\n",t);

         }

     }

     return 0;

}