Fibonacci Again之解题思路

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
解题思路：
  如果采用递归，对于小的数据n可以实现，而当n<1000000时，我们发现找不到一个数据类型来储存f（n）。因此当我们尝试列出n从0-10时便能发现规律；对n%==2时便能被三整除。于是程序也不难写出。
代码如下：
#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;

}