HDU 1021 Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22217    Accepted Submission(s): 10727

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

 

Author

Leojay

 

Recommend

JGShining

 

分析：输入N，找下F(n)是不是能被3整除，找规律f(0)/3余1,f(1)余2,f(2)余0,f(3)余2,f(4)余2,f(5)余1,f(6)余0,f(7)余1,f(8)余1,f(9)余2,这里就看出循环了每8个数一个循环，在每个循环里只有2个数可以被3整除。

代码：

#include<stdio.h>

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
       if(n%8==2||n%8==6)
           printf("yes\n");
       else
           printf("no\n");
       }
    return 0;
}