fib = {} f = [1, 1] fib[1] = True while f[-1] < 1e10: # 不断的计算,然后加在尾部,最后比对“in” f.append(f[-1]+f[-2]) fib[f[-1]] = True T = input() for i in xrange(T): N = input() if N in fib: print 'IsFibo' else: print 'IsNotFibo' Link: https://www.hackerrank.com/cha…
Fibonacci again and again Problem Description 任何一个大学生对菲波那契数列(Fibonacci numbers)应该都不会陌生,它是这样定义的:F(1)=1;F(2)=2;F(n)=F(n-1)+F(n-2)(n>=3);所以,1,2,3,5,8,13……就是菲波那契数列.在HDOJ上有不少相关的题目,比如1005 Fibonacci again就是曾经的浙江省赛题.今天,又一个关于Fibonacci的题目出现了,它是一个小游戏,定义如下:1. …
Fibonacci again and again Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8074 Accepted Submission(s): 3357 Problem Description 任何一个大学生对菲波那契数列(Fibonacci numbers)应该都不会陌生,它是这样定义的:F(1)=1;F(2)=2;…
1.Thread.join([timeout]) Wait until the thread terminates. This blocks the calling thread until the thread whose join() method is called terminates – either normally or through an unhandled exception – or until the optional timeout occurs. 等待进程结束.也就是…