1021 Fibonacci Again

最新推荐文章于 2025-09-05 16:34:31 发布

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最新推荐文章于 2025-09-05 16:34:31 发布

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文章标签： java

原文链接：http://www.cnblogs.com/97-ly/p/6824291.html

本文介绍了一种特殊的斐波那契数列，其中F(0)=7,F(1)=11,之后每一项为前两项之和。通过观察规律发现，只有当输入值减2后为4的倍数时，该数列对应的项才能被3整除。文章提供了一个简单的C++程序来验证这一规律。

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60452 Accepted Submission(s): 28262

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

找规律的题

 1 #include<iostream>
 2 #include<stdio.h>
 3 using namespace std;
 4 int main()
 5 {
 6     int n;
 7     while(scanf("%d",&n)!=EOF)
 8     {
 9         if(n==0||n==1)
10         cout<<"no"<<endl;
11         else if((n-2)%4==0)
12         cout<<"yes"<<endl;
13         else 
14         cout<<"no"<<endl;
15     }
16     return 0;
17 }