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

题目大意：

        数组 F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (2<=n<1,000,000).输入一个数n，求F(n)是否为3的倍数。

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int a,b,c,d;
    while(cin>>a)
    {
        if(a<=2)
        {
            if(a==2)
                cout<<"yes"<<endl;
            else
                cout<<"no"<<endl;
        }
        else
        {
            if((a-2)%4==0)
            {
                cout<<"yes"<<endl;
            }
            else
                cout<<"no"<<endl;
        }
    }
    return 0;
}