ZOJ Problem Set - 2060 Fibonacci Again

Fibonacci Again

Time Limit: 2 Seconds Memory Limit: 65536 KB

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

Author: Leojay

Source: ZOJ Monthly, December 2003

分析：

题意：

有另一种斐波那契数列： F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2)

输入包括多组测试数据。每一组测试数据只有一个整数n（n<1000000）。如果F(n)能被3整除，输出“yes”；否则输出“no”。

首先要知道被3整除是一个周期函数，然后多写几个n，推一下周期就可以了。找到规律直接ac

ac代码：

#include<iostream>

#include<cstdio>

using namespace std;

int main()

{

int n;

while(scanf("%d",&n)!=EOF)

{

if((n-2)%4==0)

printf("yes\n");

else printf("no\n");

}

return 0;

}