Question

用Python解决下列问题 谢了

7-18 统计字符（15 分）

本题要求编写程序，输入10个字符，统计其中英文字母、空格或回车、数字字符和其他字符的个数。
输入格式:

输入为10个字符。最后一个回车表示输入结束，不算在内。
输出格式:

在一行内按照

letter = 英文字母个数, blank = 空格或回车个数, digit = 数字字符个数, other = 其他字符个数

的格式输出。
输入样例:

aZ &
09 Az

输出样例:

letter = 4, blank = 3, digit = 2, other = 1

Answer 1

废话不说，Python2、Python3上代码（Python3代码经过测试，Python2的环境损坏了，正准备重装）
Python3代码：
# Python 3 Code
letter = 0
blank = 0
digit = 0
other = 0

while True:
s = input()
if s != '':
blank = blank + 1
for i in s:
if '0' <= i and i <= '9':
digit = digit + 1
elif 'a' <= i and i <= 'z':
letter = letter + 1
elif 'A' <= i and i <= 'Z':
letter = letter + 1
elif i == ' ':
blank = blank + 1
else:
other = other + 1
else:
blank = blank - 1
break

print("letter = %s" % letter, end = ',')
print("blank = %s" % blank, end = ',')
print("digit = %s" % digit, end = ',')
print("other = %s" % other, end = ',')
# Finish Code

Python2代码：
# Python 2 Code
letter = 0
blank = 0
digit = 0
other = 0

while True:
s = raw_input()
if s <> '':
blank = blank + 1
for i in s:
if '0' <= i and i <= '9':
digit = digit + 1
elif 'a' <= i and i <= 'z':
letter = letter + 1
elif 'A' <= i and i <= 'Z':
letter = letter + 1
elif i == ' ':
blank = blank + 1
else:
other = other + 1
else:
blank = blank - 1
break

print "letter = %d,blank = %d,digit = %d,other = %d" % (letter, blank, digit, other)
# Finish Code

Answer 2