Fibonacci Again

Fibonacci Again
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

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源代码：

#include<iostream>using namespace std;int main(){    int n,arr[50],i;//找出规律//  arr[0]=7;arr[1]=11;//  for(i=2;i<50;i++){//      arr[i]=arr[i-1]+arr[i-2];//      arr[i]=arr[i]%3;//  }//     //  for(i=0;i<50;i++){ //  if(arr[i]%3==0)cout<<"yes"<<"   ";//  else cout<<"no"<<"  ";//  cout<<i<<"  "<<arr[i]<<endl;  //  }         while(cin>>n ){              if(n%4==2)cout<<"yes"<<endl;        else cout<<"no"<<endl;    }    return 0;}