Fibonacci AgainTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 44439    Accepted Submission(s): 21214Problem DescriptionThere are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).InputInput consists of a sequence of lines, each containing an integer n. (n < 1,000,000).OutputPrint the word "yes" if 3 divide evenly into F(n).Print the word "no" if not.Sample Input012345Sample Outputnonoyesnonono

解题思路：实在没啥说的，就是找寻环节，直接上代码吧，其实还可以用矩阵乘法做，感觉有点麻烦，但是为了学会矩阵乘法，我觉定要用矩阵做一下，当然这是待会的事情了，嘿嘿：

/*2015 - 8 - 13Author: ITAK今日的我要超越昨日的我，明日的我要胜过今日的我，以创作出更好的代码为目标，不断地超越自己。*/#include 
     
      #include 
      
       using namespace std;int data[30];int main(){    /*    打表看一下，找寻环节    data[0] = 7;    data[1] = 11;    for(int i=2; i<30; i++)        data[i] = data[i-1]+data[i-2];    for(int i=0; i<30; i++)        if(data[i]%3 == 0)            cout<
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