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：

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cmath>
 5 #include <cstring>
 6 #include <queue>
 7 #include <stack>
 8 #include <map>
 9 #include <vector>
10
11 #define PI acos((double)-1)
12 #define E exp(double(1))
13 using namespace std;
14
15 int main (void)
16 {
17     int m;
18     while(scanf("%d",&m)==1)
19     {
20         m=(m-1)%8+1;
21         if(m==2||m==6)
22             cout<<"yes"<<endl;
23         else
24             cout<<"no\n";
25     }
26     return 0;
27 }