华为od题 56题 考勤问题--python

'''
 公司用一个字符串来标识员工的出勤信息

  absent:    缺勤
  late:      迟到
  leaveearly:早退
  present:   正常上班

  现需根据员工出勤信息,判断本次是否能获得出勤奖,
  能获得出勤奖的条件如下：
      1.缺勤不超过1次
      2.没有连续的迟到/早退
      3.任意连续7次考勤 缺勤/迟到/早退 不超过3次
    如：
     2
     present
     present absent present present leaveearly present absent

    输出描述：
    根据考勤数据字符串
    如果能得到考勤奖输出true否则输出false
    对于输出示例的结果应为
     true false

    示例二
     输入：
      2
      present
      present absent present present leaveearly present absent
     输出：
      true false
'''
'''
  absent:    缺勤
  late:      迟到
  leaveearly:早退
  present:   正常上班
'''
n = int(input())
kq_list = []
for i in range(n):
    kq_list.append(input().split())

# kq_list = [['present'],['present','present','leaveearly','absent','leaveearly','present','present']]

l = []
for kq in kq_list:
    # 1.缺勤不能超过1次
    if kq.count('absent') >1:
        l.append('false')
        continue

    #2.没有连续的迟到/早退
    for i in range(len(kq)-1) :
        a= kq[i]
        b =kq[i+1]
        if a in ['late','leaveearly'] and b in ['late','leaveearly']:
            l.append('false')
            break
    else:
        #3.任意连续7次考勤 缺勤/迟到/早退 不超过3次
        for i in range(len(kq)-6):
            temp_list = kq[i:i+7]
            d,e,f =0,0,0
            for j in temp_list:
                if j == 'absent':
                    d +=1
                elif j=='late':
                    e +=1
                elif j == 'leaveearly':
                    f +=1
            if d+e+f > 3:
                l.append('false')
                break
        else:
            l.append('true')
print(l)