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杭电1021--Fibonacci Again

时间:2015-06-15 00:01:33      阅读:132      评论:0      收藏:0      [点我收藏+]

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Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43441    Accepted Submission(s): 20755


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
// 7 11 18 29 47 76 123..........模3取余后为1 2 0 2 2 1 0..... , 多写几组会发现 只要满足n%4==2; F[n]就会被3整除;
 1 #include<stdio.h>
 2 int main()
 3 {
 4     int n;
 5     while(~scanf("%d",&n))
 6     {
 7         if(n%4==2)
 8         printf("yes\n");
 9         else
10         printf("no\n");
11     }
12     return 0;
13 }

 

 

杭电1021--Fibonacci Again

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原文地址:http://www.cnblogs.com/fengshun/p/4575924.html

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