1021: 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

//  i    i=0 i=1 i=2 i=3   i=4 i=5 i=6 i=7   i=8 i=9 i=10 1=11//f(i)%3  1   2   0   2     2   1   0   1     1   2   0    2   (按规律循环出现：2101 1202/ 2101 1202/...... )//        no  no yes  no    no  no yes  no    no  no  yes//由上规律可知yes出现位置为2 6 10 14 ......即i%4==2处！！！#include <iostream>using namespace std;int main(){    int i;    while(cin>>i)    {        if(i%4==2)            cout<<"yes"<<endl;        else            cout<<"no"<<endl;    }    return 0;}