【考勤信息】Python 实现

摘要：def can_get_award(attendance): records = attendance.split n = len(records) # 检查缺勤次数不超过1次 absent_count = records.count('absent') if

def can_get_award(attendance): records = attendance.split n = len(records) # 检查缺勤次数不超过1次 absent_count = records.count('absent') if absent_count > 1: return False # 检查没有连续的迟到/早退 for i in range(n - 1): if records[i] in ['late', 'leaveearly'] and records[i + 1] in ['late', 'leaveearly']: return False # 检查任意连续7次考勤中，缺勤/迟到/早退不超过3次 for i in range(n - 6): window = records[i:i + 7] count = 0 for record in window: if record in ['absent', 'late', 'leaveearly']: count += 1 if count > 3: return False return True# 读取输入num_cases = int(input)results = for _ in range(num_cases): attendance = input.strip results.append("true" if can_get_award(attendance) else "false")# 输出结果print(' '.join(results))