天天看點

HDOJ1021題 Fibonacci Again 應用求模公式

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

1

2

3

4

5

Sample Output

no

yes

應用求模公式

(1) (a + b) % p = (a % p + b % p) % p

(2) (a - b) % p = (a % p - b % p) % p

(3) (a * b) % p = (a % p * b % p) % p

(4) a ^ b % p = ((a % p)^b) % p

如果不用的話會溢出。

代碼:

繼續閱讀