R

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

012345

Sample Output

nonoyesnonono

 刚开始觉的数据挺大，要用大数打表，先把1到N的斐波契那全搞出来，然后再求余；但想到之前做过的一道题，发现像这种大数和求余结合在一起的题，其实可以利用求余将

大数给化解，用普通的数组就能解决，也就是一边斐波契那，一边求余；

#include <iostream>#include <cstdio>#include <cstring>#include<algorithm>#include<string>#include<map>using namespace std;int a[1000001];int main(){    int T;    //freopen("1.txt","r",stdin);    memset(a,0,sizeof(a));a[0]=1,a[1]=2;    for(int i=2;i<=1000000;i++){        a[i]=a[i-1]+a[i-2];a[i]%=3;    }    while(scanf("%d",&T)!=EOF)    {        if(a[T]==0)cout<<"yes\n";        else cout<<"no\n";    }    return 0;}