1-S-Fibonacci Again

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

 

 

 

/*
#include<stdio.h>
long fibonacci(long n);
int main()
{
    long n;
    while(scanf("%ld",&n)!=EOF&&n<1000000&&n>=0)
    {
        //printf("%ld\n",fibonacci(n));
            if(fibonacci(n)%3==0)printf("yes\n");
        else printf("no\n");
    }
    return 0;
}

long fibonacci(long n)
{
    if(n==0) return 7;
    else if(n==1) return 11;
    else return fibonacci(n-1)+fibonacci(n-2);
}




//以上是超时代码，以下是百度来的正确答案
#include<stdio.h>
int main()
{
    int n;
    while (scanf("%d",&n)==1)
    if(n%8==2||n%8==6)
    printf("yes\n");
    else printf("no\n");
    return 0;
}
*/

//以下是我明白了用递归会超时的真理后按照百度自己又写了一遍

#include<stdio.h>
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n%8==2||n%8==6)//但我还是不懂为什么要这样取余！！！！！！！！！
        //终于懂了，为什么这样取余
        /*
        首先一个个找规律，发现
        f(0)=F(0)%3=7%3=1
        f(1)=F(1)%3=11%3=2
        f(2)=F(2)%3=18%3=0
        f(3)=F(3)%3=29%3=2
        f(4)=F(4)%3=47%3=1
        f(5)=F(5)%3=76%3=1
        f(6)=F(6)%3=123%3=0
        f(7)=F(7)%3=199%3=1
        f(8)=F(8)%3=322%3=1
        f(9)=F(9)%3=521%3=2
        f(10)=F(10)%3=843%3=0
        f(11)=F(11)%3=1364%3=2
        f(12)=F(12)%3=2207%3=2//算错了不想算了

        总之这个Fibonacci数列对3取余后余数8个一组循环
        然后如果输入一个数n它对8取余余数为2和6的都是所求答案！
                */
        printf("yes\n");//presentation error！=v= 又忘记回车转义字符了
        else printf("no\n");
    }
    return 0;
}