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)==1)
{
m=(m-1)%8+1;
if(m==2||m==6)
cout<<"yes"<<endl;
else
cout<<"no\n";
}
return 0;
}

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