hdu1021 Fibonacci Again

题目(http://acm.hdu.edu.cn/showproblem.php?pid=1021)

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

打表发现有规律：

#include <iostream>
#include <stdio.h>
using namespace std;

int f[1000005];
int main()
{
   f[0]=1; f[1]=2;
   int n;
   while(scanf("%d",&n)!=EOF)
   {
     for(int i=2;i<=n;i++)
     f[i]=(f[i-1]+f[i-2])%3;
     if(f[n]%3==0)
     printf("yes\n");
     else
     printf("no\n");
   }
  return 0;
}