Fibonacci Again

最新推荐文章于 2024-07-25 21:08:56 发布

原创最新推荐文章于 2024-07-25 21:08:56 发布 · 189 阅读

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本文介绍了一种特殊的斐波那契数列，其初始值为7和11，而非传统的0和1。通过C语言实现了一个函数，用于计算该数列中任意项是否为3的倍数。输入为一系列整数，输出则是对应项是否能被3整除的判断。

Problem
DescriptionThere 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

#include<stdio.h>
long long  Fibonacci( long long  n )
{
 	long long a=7,b=11;
 	if(n==0)
  		return a;
 	else if(n==1)
  		return b;
 	int i=2,x,y;
 	while(i<=n)
 	{
  		x=a+b;
  		a=b%3;
  		b=x%3;
  		i++;
  		x=x%3;
 	}
	 return x;
}
int main()
{
 	long long  n,f;
 	while(scanf("%lld",&n)!=EOF)
 	{
  		f=Fibonacci(n);
  		if(f%3==0)
   			printf("yes\n");
 		 else 
   			printf("no\n");
 	}
 	return 0;
 }