HDU1021

最新推荐文章于 2025-09-10 22:21:01 发布

weixin_30564901

最新推荐文章于 2025-09-10 22:21:01 发布

阅读量61

点赞数

CC 4.0 BY-SA版权

文章标签： java

原文链接：http://www.cnblogs.com/BOW1203/p/8000641.html

本文介绍了一种特殊的Fibonacci序列，该序列以7和11为初始值，并探讨了如何判断序列中某项是否能被3整除的方法。通过观察和分析，找到了一个简单规律，即从第二项开始，每间隔4项就会出现一个可以被3整除的数。

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 65456 Accepted Submission(s): 30363

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

Sample Output

yes

= =。。。一开始用递归，内存超限。。。随后用这个递归得出规律，就是从第二个开始每隔4个就一次YES。。。AC

 1 #include <iostream>
 2 using namespace std;
 3 long FIB(long n);
 4 int main()
 5 {
 6     long n;
 7     while(cin>>n)
 8     {
 9         if(FIB(n)%3==0)
10             cout<<"yes"<<endl;
11         else
12             cout<<"no"<<endl;
13     }
14 
15     return 0;
16 }
17 
18 long FIB(long n)
19 {
20     if(n==0)  return 7;
21     if(n==1)  return 11;
22     return FIB(n-1)+FIB(n-2);
23 }

找到规律后：

 1 #include <iostream>
 2 using namespace std;
 3 long FIB(long n);
 4 int main()
 5 {
 6     long n;
 7     while(cin>>n)
 8     {
 9         if(n%4==2)
10             cout<<"yes\n";
11         else
12             cout<<"no\n";
13     }
14 
15     return 0;
16 }