一晚上做了好几家的，头疼，这些捞b公司贼烦
我自己没有参加笔试，就是纯粹A着玩，纯属娱乐
"""百词斩第一题"""
"""自我实现版本"""
def get_time_diff(time1, time2):
res = 0
res += (int(time2[:2]) - int(time1[:2])) * 3600
res += (int(time2[3:5]) - int(time1[3:5])) * 60
res += int(time2[-2:]) - int(time1[-2:])
return res
time1 = input()
time2 = input()
sec = get_time_diff(time1, time2)
print(int(30*sec//3600))
print(int(6*sec//60))
print(int(sec*6))
"""工业版本
可惜笔试不支持
"""
from dateutil.parser import parse
def get_time_diff(time1, time2):
time1 = parse(time1)
time2 = parse(time2)
return (time2-time1).total_seconds()
time1 = '2018-09-20/' + input()
time2 = '2018-09-20/' + input()
sec = get_time_diff(time1, time2)
print(int(30*sec//3600))
print(int(6*sec//60))
print(int(sec*6))
"""
14:52:11
21:41:14
00:00:00
18:00:00
"""
"""百词斩第二题"""
"""百词斩第二题"""
n = int(input())
nums = [int(i) for i in input().split()]
res = ''
# tmp = nums[0]
i = 0
start = 0
while i <= n-1:
i += 1
tmp = 0
while i <= n-1 and nums[i] == nums[i-1]+1:
i += 1
tmp += 1
# print(tmp, i)
if tmp >= 2:
# print(111)
res += str(nums[start]) + '-' + str(nums[i-1]) + ' '
# start = i
elif tmp == 0:
# print(222)
res += str(nums[start]) + ' '
else:
# print(333)
# print(nums[start:i+1])
res += ' '.join([str(i) for i in nums[start:i]]) + ' '
start = i
# tmp = 0
print(res[:-1])
"""
6
1 2 3 4 6 7
6
1 2 4 5 6 7
"""
"""不知道谁的题目"""
lookup = {}
inputs = []
flag = True
while True:
user_input = input()
if user_input == 'END':
break
inputs.append(user_input)
user_input = user_input.split('#')
n, m = user_input[0], user_input[1]
num = int(m, int(n))
lookup[num] = lookup.get(num, 0) + 1
res = []
for i in range(len(inputs)):
tmp = inputs[i].split('#')
n, m = tmp[0], tmp[1]
num = int(m, int(n))
if lookup[num] == 1:
res.append(inputs[i])
if not res:
print('None')
else:
for i in res:
print(i)
print(int('33', 4))
"""
10#15
4#32
4#33
8#17
END
"""
“”“不知道谁的题目
三种颜色的球
”“”
class Solution(object):
res = 0
def cal_next(self, p, q, r, prev):
tmp = [p, q, r]
cur_max = max(tmp)
# 此时前面排好了且都满足，想要不相邻，数量最多的球中间的所有空格
# 必须小于或等于另外两个球的数量，否则说明此时排列不合法直接return
if cur_max - 1 > sum(tmp) - cur_max:
return
if p == q == r == 0:
self.res += 1
return
if p > 0 and prev != 'p':
self.cal_next(p-1, q, r, 'p')
if q > 0 and prev != 'q':
self.cal_next(p, q-1, r, 'q')
if r > 0 and prev != 'r':
self.cal_next(p, q, r-1, 'r')
s = Solution()
p, q, r = [int(i) for i in input().split()]
s.cal_next(p, q, r, '')
print(s.res)