蓝桥杯

第十四届蓝桥杯大赛软件赛国赛Python大学C组

跑步计划

1月1日 -> 4月3日相差92天，所以得出1月1日为星期日,遍历整年，只要包含“1”就+5，否则+1，最终计算可得答案为1333，直接输出1333就行

# month = [0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
# weekday = 0
# res = 0
# for i in range(1,13):
#     for j in range(1, month[i] + 1):
#         if weekday == 1 or '1' in str(j) or '1' in str(i):
#             res+=5
#         else:
#             res+=1
#         weekday = (weekday + 1) % 7
# print(res)
print(1333)

混乘数字

遍历一遍1到1000000，对于是否判断出现次数相同，可以考虑排序一样就可以了，最后结果为590，为防止超时直接输出590就可以了

# def find(n):
#     for a in range(1, int(n ** 0.5) + 1):
#         if n % a == 0:
#             b = n // a
#             num_str = str(n)
#             ab_str = str(a) + str(b)
#             if sorted(num_str) == sorted(ab_str):
#                 return True
#     return False
# 
# res = 0
# for i in range(1, 1000001):
#     if find(i):
#         res+=1
# print(res)
print(590)

混乘数字

建议使用 PyPy3 提交本题，不然会超时

n = int(input())
res = 0
while n!=0:
    sum =0
    for i in str(n):
        sum+=int(i)
    n = n-sum
    res +=1
print(res)

定时任务

这题目看着比较难，其实只要慢慢做是可以做出来的，别害怕

def find(mystr,mymin,mymax):
    res = []
    if mystr == '*':
        for i in range(mymin,mymax+1):
            res.append(i)
        return res
    if ',' in mystr:
        temp = mystr.split(',')
        for i in temp:
            res.append(int(i))
        return res
    if '-' in mystr:
        start, end = map(int, mystr.split('-'))
        for i in range(start, end + 1):
            res.append(i)
        return res

    res.append(int(mystr))
    return res



second, minute, hour, day, month = input().split()
seconds = find(second,0,59)
minutes = find(minute,0,59)
hours = find(hour,0,23)
days = find(day,1,31)
months = find(month,1,12)
month_days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]

valid_days = []
for month in months:
    for day in days:
        if day <= month_days[month - 1]:
            valid_days.append((month, day))
res = len(seconds) * len(minutes) * len(hours) * len(valid_days)
print(res)