"Python" Hanoi tower problem

Count = 0
def Hannuota (N,src,dst,mid): #n是圆盘数, SRC is the start, DST is the target, mid is over
    Global Count
    if n = = 1:
        Print (' {}:{}->{} '. Format (1,SRC,DST)) #当圆盘是1时, move from the starting column to the target column
        Count + = 1
    Else
        Hannuota (N-1,SRC,MID,DST) #剩余的圆盘从A移到B柱子
        Print (' {}:{}->{} '. Format (N,SRC,DST)) #最大的圆盘从A移到C柱子
        Count + = 1
        Hannuota (N-1,MID,DST,SRC) #剩余的圆盘从B柱子到C柱子
Hannuota (3, ' A ', ' C ', ' B ')
Print (count)

#我现在理解的是, imagine a 1,2,3,4,5 disc, first move the above four discs from a to B pillars, then move 5 from A to C pillars, then move the 1234 from B to the C pillar, and then break these actions into n=1.
#我也搞不懂, it's so hard!

"Python" Hanoi tower problem