ZOJ Problem Set - 2060 Fibonacci Again

Fibonacci Again

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

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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;
}