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HDU Fibonacci Again

时间:2015-05-27 10:14:07      阅读:146      评论:0      收藏:0      [点我收藏+]

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Fibonacci Again

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 39   Accepted Submission(s) : 31

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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
找规律的题目。
把f(n)分裂开就能够得到规律。
AC代码:
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
    long n;
    while(scanf("%ld",&n) != EOF)
       if (n%8==2 || n%8==6)
           printf("yes\n");
       else
           printf("no\n");
    return 0;
}

运行结果:
技术分享

HDU Fibonacci Again

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原文地址:http://blog.csdn.net/zp___waj/article/details/46041703

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