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1021 Fibonacci Again (hdoj)

时间:2015-05-12 01:36:11      阅读:93      评论:0      收藏:0      [点我收藏+]
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
 
1 #include<iostream>/*打表找规律*/
2 using namespace std;
3 int main(){
4    int n;
5    while(cin>>n){
6         if(n%4==2) cout<<"yes"<<endl;
7         else       cout<<"no"<<endl;
8    }return 0;
9 }

 

1021 Fibonacci Again (hdoj)

原文:http://www.cnblogs.com/a1225234/p/4496184.html

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