ACM HDU 1021 Fibonacci Again

#include<iostream>
using namespace std;
int main()
{
    int n;
    while(cin>>n)
    {
        if((n+1)%4==3)
            cout<<"yes"<<endl;
        else
            cout<<"no"<<endl;
    }
    return 0;
}

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

 

 

看到 Fibonacci Again 的第一眼就像用递归，感觉挺水的，敲代码，妥妥的超时了。  没办法，看看大神的，一致觉得水，水在那呢，看了之后发现，这题是找规律的，4为一个循环，1,2,3,4,分别是no，yes，no，no，所以就用（n+1）%4==3  >> no