HDU 1021 Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 27899    Accepted Submission(s): 13476

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

 

直接找规律即可，和NEFU那道斐波那契一样，代码就是这么短~~

 

#include<iostream>

using namespace std;

int main()
{
    long n;
    while(cin>>n)
        cout<<(n%4==2?"yes":"no")<<endl;
    return 0;
}