HDU 1021 Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 56132 Accepted Submission(s): 26291

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
写过类似的题目，找规律，列出前十几个数，发现出现循环规律，第二项，第六项，第十项均是3的倍数，即4个一组循环，下面就很简单了
AC代码：

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