Python实现 Pat1019 数字黑洞

def number(N):    if int(N)>0 and int(N)<10000:        if 0<int(N)<10:            N='000'+N        elif 10<=int(N)<100:            N='00'+N        elif 100<=int(N)<1000:            N='0'+N        else:            N=N    return Ndef operation(N):    N=number(N)    while True:        L = []        for i in range(len(N)):            L.append(N[i])            L.sort(reverse=True)            n1 = ''.join(L)            L.sort()            n2 = ''.join(L)        if int(n1)==int(n2):            print('%s - %s = 0000' % (n1,n2))            break        result=int(n1)-int(n2)        if result<1000:            result='0'+str(result)        print('%s - %s = %s'%(n1,n2,result))        if result!=6174:            N=str(result)            continue        else:            breakif __name__=='__main__':    N = input()    operation(N)