hdu1021Fibonacci Again找规律

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.

私以为找规律比构造矩阵好写==

#include <iostream>
#include <stdio.h>
#include <math.h>
#define Mod 3
using namespace std;

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n%8==2||n%8==6) puts("yes");
        else puts("no");
    }
    return 0;
}