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1021 Fibonacci Again

时间:2017-05-08 13:02:34      阅读:229      评论:0      收藏:0      [点我收藏+]

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 }

 

1021 Fibonacci Again

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

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