1021 Fibonacci Again （hdoj）
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.
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 }
