1021 Fibonacci Again （hdoj）

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

 

1 #include<iostream>/*打表找规律*/
2 using namespace std;
3 int main(){
4    int n;
5    while(cin>>n){
6         if(n%4==2) cout<<"yes"<<endl;
7         else       cout<<"no"<<endl;
8    }return 0;
9 }