Fibonacci Again

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

Total Submission(s): 65456    Accepted Submission(s): 30363

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

 

= =。。。一开始用递归，内存超限。。。随后用这个递归得出规律，就是从第二个开始每隔4个就一次YES。。。AC

 

1 #include 
        
          2 using namespace std; 3 long FIB(long n); 4 int main() 5 { 6     long n; 7     while(cin>>n) 8     { 9         if(FIB(n)%3==0)10             cout<<"yes"<
        

找到规律后：

1 #include 
        
          2 using namespace std; 3 long FIB(long n); 4 int main() 5 { 6     long n; 7     while(cin>>n) 8     { 9         if(n%4==2)10             cout<<"yes\n";11         else12             cout<<"no\n";13     }14 15     return 0;16 }