题解 | #密码验证合格程序#
密码验证合格程序
https://www.nowcoder.com/practice/184edec193864f0985ad2684fbc86841
import sys Charmi=['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'] charmi=list(map(lambda x:x.lower(),Charmi)) Num=['0','1','2','3','4','5','6','7','8','9'] # print(charmi) for line in sys.stdin: a = line.strip() LEN=len(a)>8 if not LEN: #使用not 和布尔判断,注意continue和break,continue说明换了一行 print('NG') continue daxie=not set(a).isdisjoint(set(Charmi)) #isdisjoint函数返回没有交集的判例 xiaoxie=not set(a).isdisjoint(set(charmi)) shuzi=not set(a).isdisjoint(set(Num)) for x in a: if x in Charmi or x in charmi or x in Num: qita=0 else: qita=1 break if int(xiaoxie)+int(daxie)+int(shuzi)+qita<3: print('NG') continue # da=bool(set(a)&set(Charmi))#测试是否交集 # if not Len: # print('NG') # if not (daxie and xiaoxie and shuzi and LEN) i=0 DD={} while(i+3<len(a)): if a[i:i+3] not in DD: DD[a[i:i+3]]=0 else: DD[a[i:i+3]]+=1 i+=1 if sum(DD.values())>=1: print('NG') else: print('OK') # print(chong)