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 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
5
Sample Output
no
Sample Input
2
Sample Output
yes



我的代码：
#include<stdio.h>
int main()
{
    int a[100]={7,11},n,i;
    scanf("%d",&n);
    for(i=2;i<100;i++)
        a[i]=a[i-1]+a[i-2];
    if(a[n]%3==0)
        printf("yes");
    else
        printf("no");
   

   

return 0;
}

你的数组大小只有100你怎么找到a[100000]

醉了，我真是笨 啊，谢

找规律吧

#include <stdio.h>
int main()
{
    int n;

    scanf("%d", &n);
    puts((n - 2) % 4 ? "no" : "yes");

    return 0;
}