water- Fibonacci Again-优快云博客

本文探讨了一种特殊的斐波那契数列F(n)，其中F(0)=7, F(1)=11，并研究了当n小于1,000,000时，F(n)是否能被3整除。通过查找规律，提供了一个简单的算法来判断F(n)的性质。

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25156 Accepted Submission(s): 12182

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

***********************************************************************************************************

View Code

 1 #include <stdio.h>
 2 #define A 1000005
 3 long ch[A];
 4 void tool()
 5 {
 6     long i,cnt=0;
 7     ch[0]=7;
 8     ch[1]=11;
 9     for(i=2;cnt<1000000;i++)
10     {
11         ch[i]=((ch[i-1]%3)+ch[i-2]%3)%3;
12         cnt++;
13     }
14 }
15 int main()
16 {
17     tool();
18     long int n;
19     while(scanf("%ld",&n)!=EOF)
20     {
21         if(ch[n]==0)
22             printf("yes\n");
23         else
24             printf("no\n");
25     }
26     return 0;
27 }

找规律求解：

View Code

 1 #include<stdio.h>
 2 int main()
 3 {
 4     int n;
 5     while(scanf("%d",&n)!=EOF)
 6     {
 7         if(n%4==2)
 8             printf("yes\n");
 9         else
10             printf("no\n");
11     }
12     return 0;
13 }