hoj1159-Fibonacci Again(斐波那契数列%3)

给出新的数列： another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2)

第n个斐波那契数列是否能被3整除...

找规律问题，找出规律后发现8个行程循环，第三个和第七个可以被三整除

#include <stdio.h>
 int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    if((n+1)%8==3||(n+1)%8==7)printf("yes\n");
    else printf("no\n");
    return 0;
}