hdoj 1021 Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40734    Accepted Submission(s): 19513

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
   
   
   
   

 

以前剩的题，今天轻松干掉。

附上两种代码：

 

常规思路：每次都对3求余。。。下面不用多说了吧

 

#include<stdio.h>
#define max 1000000+10
int f[max];
int main()
{
    int i,n;
    f[0]=7;f[1]=11;
    for(i=2;i<max;i++)
    {
        f[i]=(f[i-1]%3+f[i-2]%3)%3;
    }
    while(scanf("%d",&n)!=EOF)
    {
        if(f[n]%3==0)
        printf("yes\n");
        else
        printf("no\n");
    }
    return 0;
}

规律：该数列的数对3取余后 8个一循环，循环体 为｛1,2,0,2,2,1,0,1｝

#include<stdio.h>
int main()
{
    int i,n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n%8==2||n%8==6)
        printf("yes\n");
        else
        printf("no\n");
    }
    return 0;
}