PAT 甲级 1062 Talent and Virtue

1062 Talent and Virtue

题目大意：给出一组人数，按照要求分类再排序输出。圣人是virtue和talent都超过h的人，君子是virtue超过h并且talent位于[l,h)区间内的人，笨蛋是两个都位于[l,h)区间内并且virtue>=talent的人，剩下的达到最低标准l的都是小人。PS：python日常超时，C++这里完美AC

class Person:def __init__(self,id,virtue,talent):self.id=idself.virtue=virtueself.talent=talentn,l,h=map(int,input().split())
sages=[]    #圣人
noblemen=[] #君子
foolmen=[]  #笨蛋
smallmen=[] #小人
cnt=0  #统计达标的总人数
for i in range(n):id,virtue,talent=map(int,input().split())if virtue<l or talent<l:continueelse:if virtue>=h and talent>=h: #圣人sages.append(Person(id,virtue,talent))elif virtue>=h and talent in range(l,h):    #君子noblemen.append(Person(id,virtue,talent))elif virtue in range(l,h) and talent in range(l,h) and virtue>=talent:  #笨蛋foolmen.append(Person(id,virtue,talent))else:smallmen.append(Person(id,virtue,talent))cnt+=1
sages.sort(key=lambda p:(-(p.virtue+p.talent),-p.virtue,p.id))
noblemen.sort(key=lambda p:(-(p.virtue+p.talent),-p.virtue,p.id))
foolmen.sort(key=lambda p:(-(p.virtue+p.talent),-p.virtue,p.id))
smallmen.sort(key=lambda p:(-(p.virtue+p.talent),-p.virtue,p.id))
print(cnt)
for i in sages:print(i.id,i.virtue,i.talent)
for i in noblemen:print(i.id,i.virtue,i.talent)
for i in foolmen:print(i.id,i.virtue,i.talent)
for i in smallmen:print(i.id,i.virtue,i.talent)

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PAT 甲级 1062 Talent and Virtue

1062 Talent and Virtue

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