斐波那契数列

解题思路:用一个变量存储上一个数

注意事项:无

参考代码:

 #includde<stdio.h>

  int main(){

         int n;

         int t;

         int q;

         int p=0;

         scanf_s("%d", &n);

         int a[40] = {1};

         if (n>0&&n <= 40) {

         for (int i = 0; i < n-1; i++) {

         q = a[i];

         a[i + 1] = a[i] + p;

         p = q;

         }

         for (int i = 0; i < n; i++) {

         printf("%d ", a[i]);

         }

         }

         else {

         printf("error");

          }

      return 0;

     }