杭电1021Fibonacci Again

地址：http://acm.hdu.edu.cn/showproblem.php?pid=1021

题目：

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

 

Sample Output

no

no

yes

no

no

no

思路：（a+b)%c=a%c+b%c，所以可以把F(0)看做1，F(1)看做2，F（3）看做0，可以看出该数列一定会循环，（且最大循环节为9，因为3*3=9)。。。。。。以此类推.

　　得到数列 （从0开始)1 2 0 2 2 1 0 1....(后面重复）；

　　所以。。。用n对8取模就好了。

　　取模公式n=(n-1)%8+1;

ac代码：

 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <queue>
 #include <stack>
 #include <map>
 #include <vector>

 #define PI acos((double)-1)
 #define E exp(double(1))
 using namespace std;

 int main (void)
 {
     int m;
     while(scanf("%d",&m)==)
     {
         m=(m-)%+;
         if(m==||m==)
             cout<<"yes"<<endl;
         else
             cout<<"no\n";
     }
     return ;
 }