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要求:1s 262144k
公司用一个字符串来表示员工的出勤信息:
absent 缺勤
late 迟到
leaveearly 早退
present 正常上班
能获得全勤奖的条件:
缺勤次数不超过1次
不能连续迟到/早退
任意的连续7次考勤中,缺勤、迟到、早退的次数不超过3次
eg:
输入:
2
present
present present
输出:true true
输入:
2
present
leaveearly late present present
输出: true false
import sys def award(s): if s.count('absent') > 1: ##判断缺勤不超过1次 return 'false' if s.count('late late') > 0 or s.count('leaveearly leaveearly') > 0 or s.count('late leaveearly') or s.count('leaveearly late'): ##判断没有连续的迟到早退 return 'false' sl = s.split(" ") ##把每行字符串转换成列表 for i in range(len(sl)): ##判断任意连续7次考勤,迟到早退和缺勤不超过3次 if sl[i] == 'absent' or sl[i] == 'late' or sl[i] == 'leaveearly': num = 0 l = sl[i+1:i+7] num += l.count('absent') num += l.count('late') num += l.count('leaveearly') if num > 2: return 'false' return 'true' n = int(sys.stdin.readline()) ##标准输入 res = [] for i in range(n): line = sys.stdin.readline().strip() ##strip()以换行分割 res.append(award(line)) print(" ".join(res))
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