hdu1021 Fibonacci Again

Problem Description

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 integern. (n < 1,000,000).

 

Output

Print the word "yes" if 3 divide evenly into F(n).

Print the word "no" if not.

 

Sample Input

012345

 

Sample Output

nonoyesnonono

 

题目大意：
斐波那契数

F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).

F（n）能被3整除输出“yes”，不能输出“no”。

可能想到中国剩余定理，但是如果多写几组数据就会发现有规律

源代码：

根据规律
#include<stdio.h>int main(){ int n; while(scanf("%d",&n)!=EOF) {  if(n%4==2) printf("yes\n");  else printf("no\n"); } return 0;}