杭电oj~~1021

尝试了很多办法，用打表，数组遍历都不行，要不超时要不数据溢出，最后想，利用求余运算，终于AC了。

题目描述：

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
   
   
   
   

AC代码：

import java.util.Scanner;

public class Main {
public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		
		while(in.hasNext())
		{
			int a,b,c,t=0,i;
			a = 7;b = 11;
			c = in.nextInt();
			if(c==0||c==1)
			{
				System.out.println("no");
			}
			else
			{
			  for(i=2;i<=c;i++)
			 {
				t = (a+b)%3;
				a = b%3;
				b = t%3;
			 }
			  if(t%3==0)
			 {
				System.out.println("yes");
			 }
			 else
			 {
				System.out.println("no");
		 	 }
		 }
	   } 
	}
}