Fibonacci Again

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 39   Accepted Submission(s) : 31

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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

Author

Leojay

找规律的题目。

把f（n）分裂开就能够得到规律。

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
    long n;
    while(scanf("%ld",&n) != EOF)
       if (n%8==2 || n%8==6)
           printf("yes\n");
       else
           printf("no\n");
    return 0;
}

运行结果：