PAT basic1026-1030（Python版）

最新推荐文章于 2024-07-03 17:31:11 发布

原创最新推荐文章于 2024-07-03 17:31:11 发布 · 886 阅读

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本文提供了四个PAT竞赛Python编程题目的解决方案，涉及时间计算、字符串处理、有效性验证及列表操作等，适合初学者学习和练习。

Python版的PAT，Python2环境。

1026：

from sys import exit
str = raw_input()

for i in range(len(str)):
	if str[i] == ' ':
		flag = i
		break

c1 = int(str[:flag])
c2 = int(str[flag+1:])

CTK = 100

sec_sum = (c2 - c1) / CTK

if (c2 - c1) % 100 >= 50:
	sec_sum += 1

hh = sec_sum / 3600
mm = (sec_sum % 3600) / 60
ss = sec_sum % 60

print "%02d:%02d:%02d" % (hh, mm, ss)
exit(0)

1027：

from sys import exit
str = raw_input()

for i in range(len(str)):
	if str[i] == ' ':
		flag = i
		break

c1 = int(str[:flag])
c2 = str[flag+1:]

c1 += 1
num = 0
i = 0
while 1:
	c1 -= 2 * ( 2 * i + 1 )
	i += 1
	num += 1

	if c1 < 2 *(2 * i + 1):
		break

for j in range(num):
	print "%s%s" % (j*' ', ( 2*(num-1-j) + 1 )*c2)

for j in range(1,num):
	print "%s%s" % ( (num-j-1)*' ', (2*j + 1)*c2)

print c1
exit(0)

1028：

from sys import exit

def isValid( birth ):
    if birth <= "2014/09/06" and birth >= "1814/09/06":
        return 0
    else:
         return 1

str = raw_input()
num = int(str)
max = []
min = []
validNum = 0

for i in range(num):
    tmp = raw_input().split()
    if isValid( tmp[1] ) == 0:
        validNum += 1                              
        if len(max) == 0 or max[1] > tmp[1]:
            max = tmp

        if len(min) == 0 or min[1] < tmp[1]:
            min = tmp
if len(max) != 0:
    print validNum, max[0], min[0]
else:
    print '0'

exit(0)

1029：

from sys import exit
str1 = raw_input()
str2 = raw_input()
str3 = ""

for ch in str1:
    if not ch in str2:
        if not (ch in str3 or ch.upper() in str3):
            str3 += ch.upper()

print str3

exit(0)

1030：

from sys import exit
str1 = raw_input()
str2 = raw_input()

Np = str1.split(' ')
N = int(Np[0])
p = int(Np[1])

strNum = str2.split(' ')
num = []

for i in strNum:
    num.append(int(i))

num.sort()       

start = 0
end = 0
maxLen = 0

while True:

    if start == end:
        end += 1
        continue

    if end >= N:
        break

    if num[end] <= p * num[start]:
        end += 1
    else:
        if maxLen < (end - start):
            maxLen = end - start
        start += 1
    
    if end == N:
        break

if maxLen < end - start:
    maxLen = end - start

print maxLen
exit(0)

刚学Python，代码写的可能不是很规范，请多多指教。