There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).
Input
Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).
Output
Print the word "yes" if 3 divide evenly into F(n).
Print the word "no" if not.
Sample Input
0 1 2 3 4 5
Sample Output
no no yes no no no
题解：
刚开始做的时候是利用打表法求出1,000,000以内的所有f(n)的值，然后判断是否是3的倍数，但是提交上去是错误的。
后来有往后推了几个数，找的了规律。余数是循环的，8个一循环【1,2,0,2,2,1,0,1】，然后在判断。
代码：
#include<stdio.h>
int main()
{
int a[8]={1,2,0,2,2,1,0,1};//存取余于3的余数
int n;
while(scanf("%d",&n)!=EOF)
{
if(a[n%8]==0)
printf("yes\n");
else
printf("no\n");
}
return 0;
}
