Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44035    Accepted Submission(s): 21020

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 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
  

 

http://acm.hdu.edu.cn/showproblem.php?pid=1021

 先打出前50个的数的答案发现他有一种规律，要推得话也是可以的，7%3=1,11%3=2，那么（7+11）%3一定是0，后面余数分别是2，2，1，0,1,1,2,0.。。。。。。

so代码如下：

#include <cstdio>
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        if(n==0||n==1)printf("no\n");
        else if(n%4==2)printf("yes\n");
        else printf("no\n");
    }
    return 0;
}