http://acm.hdu.edu.cn/showproblem.php?pid=1021

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 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 找出循环节（8） #include #include #include using namespace std; int a[50]; int main() { int n; a[0]=7; a[1]=11; for(int i=2;i<=8;i++) a[i]=a[i-1]+a[i-2]; while(~scanf("%d",&n)) { n=n%8; n=a[n]%3; if(n==0) printf("yes "); else printf("no "); } return 0; }

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 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