hdu 1021 Fibonacci Again

斐波那契的递推；

有规律，但是我打表过了；

f[0]=7;

f[1]=11;   

实际对他们%3 就是 f[0]=1. f[1]=2;正规的斐波那契了；

 则有n>2  f[n]=(f[n-1]+f[n-2])%3;

递推出斐波那契数列；

则问输入n，f[n]%3==0 yes . 否则 no；

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 
     
     
     
     

#include <bits/stdc++.h>
using namespace std ;
long long dp[10000000];
int main()
{
	dp[0]=7%3;
	dp[1]=11%3;
	for(int i = 2;i<=1000000;i++)
	{
		dp[i]=(dp[i-1]+dp[i-2])%3;
	}
	int n ;
	while(cin>>n)
	{
		if(dp[n]%3==0) printf("yes\n");
		else printf("no\n");
		}	
		return 0 ;
}