*汉诺塔问题是一个著名的问题，初始模型如图所示。其来源据说是在约19世纪末欧洲的商店中出售一种智力玩具，在一块铜板上有三根杆，最左边的杆自上而下、由小到大顺序串着64个圆盘构成的塔，游戏的目的是将最左边A杆上的圆盘，借助最右边的C杆，全部移动到中间的B杆上，条件是一次仅能移动一个盘，且不允许大盘放在小盘的上面。 **输入格式要求："%d" 提示信息："Please enter the number of discs:" **输出格式要求："\tTotal:%d\n" "%2d-(%2d):%c==>%c\n" 程序运行示例如下： Please enter the number of discs: Please enter the number of discs: 5 1-( 1):a==>b 2-( 2):a==>c 3-( 1):b==>c 4-( 3):a==>b 5-( 1):c==>a 6-( 2):c==>b 7-( 1):a==>b 8-( 4):a==>c 9-( 1):b==>c 10-( 2):b==>a 11-( 1):c==>a 12-( 3):b==>c 13-( 1):a==>b 14-( 2):a==>c 15-( 1):b==>c 16-( 5):a==>b 17-( 1):c==>a 18-( 2):c==>b 19-( 1):a==>b 20-( 3):c==>a 21-( 1):b==>c 22-( 2):b==>a 23-( 1):c==>a 24-( 4):c==>b 25-( 1):a==>b 26-( 2):a==>c 27-( 1):b==>c 28-( 3):a==>b 29-( 1):c==>a 30-( 2):c==>b 31-( 1):a==>b Total:31

请看下面的代码实现：

def hanoi(n, a, b, c):
    if n == 1:
        print("%2d-(%2d):%c==>%c" % (1, 1, a, b))
        return 1
    else:
        count = hanoi(n - 1, a, c, b)
        print("%2d-(%2d):%c==>%c" % (n, n - count, a, b))
        count += 1
        count += hanoi(n - 1, c, b, a)
        return count


print("Please enter the number of discs:")
n = int(input().strip())
print()
count = hanoi(n, 'a', 'b', 'c')
print("\tTotal:%d" % count)

运行结果如下：

Please enter the number of discs:
5

 1-( 1):a==>b
 2-( 2):a==>c
 3-( 1):b==>c
 4-( 3):a==>b
 5-( 1):c==>a
 6-( 2):c==>b
 7-( 1):a==>b
 8-( 4):a==>c
 9-( 1):b==>c
10-( 2):b==>a
11-( 1):c==>a
12-( 3):b==>c
13-( 1):a==>b
14-( 2):a==>c
15-( 1):b==>c
16-( 5):a==>b
17-( 1):c==>a
18-( 2):c==>b
19-( 1):a==>b
20-( 3):c==>a
21-( 1):b==>c
22-( 2):b==>a
23-( 1):c==>a
24-( 4):c==>b
25-( 1):a==>b
26-( 2):a==>c
27-( 1):b==>c
28-( 3):a==>b
29-( 1):c==>a
30-( 2):c==>b
31-( 1):a==>b
	Total:31