用python编程，给出‘a’-‘e’这五个英文字母，基于这5个字符输出所有长度为4的单词及这些单词的总个数 python中有一个字符串为，line=”name=%E6%...
Python\u6c42\u5927\u4f6c\u6025\u6551\u3002\u9996\u5148\uff0c\u8fd9\u662f\u4e2a\u6570\u5b66\u9898\uff01\uff01\uff01
\u5176\u6b21\u4e2a\u7528\u6392\u5217\u7ec4\u5408\u5c31\u80fd\u89e3\u51b3
import urllib.parse
urllib.parse.unquote(\u201dname=%E6%9D%8E%E9%9B%B7")
unquote\u4e2d\u4e5f\u6709\u5176\u4ed6\u53c2\u6570\uff0c\u5177\u4f53\u53c2\u89c1
http://docs.python.org/py3k/library/urllib.parse.html
import requests,re
def download():
str2="abcde"
l_word=[]
for i in str2:
for j in str2:
for k in str2:
for l in str2:
if "a"in (i,j,k,l):
if "d" in (i,j,k,l):
if i!=j and i!=k and i!=l and j!=k and j!=l and k!=l:
word=i+j+k+l
l_word.append(word)
print(l_word)
print("单词的总个数为%s" % len(l_word))
if __name__ == "__main__":
while True:
download()
break
绛旓細鏄竴涓瓧绗︿覆鏄'寮犻,鍏崇窘'锛岃杈撳嚭'寮犵窘,鍏抽'鏄悧锛熷彧瑕佺敤绱㈠紩灏辫浜嗭紙鏈涢噰绾硚锛a = '寮犻,鍏崇窘'print(a[0] + a[4] + a[2] + a[3] + a[1])
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绛旓細鎴戝垰鍋氬畬杩欓銆傛垜鐨勭瓟妗堬細import matha = float(raw_input())b = float(raw_input())c = float(raw_input())degree = math.degrees(math.acos((a**2 + b**2 - c**2) / (2 * a * b)))print '{0:.{1}f}'.format(degree, 1)OJ娴嬭瘯閫氳繃銆
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绛旓細a = []for i in ['w','c','li']:for j in ['z','y','liu']:if i == 'w' and j == 'z':print('not match',i,j)elif i == 'li' and j == 'z' or i == 'li' and j == 'liu':print('not match',i,j)elif i == 'li' and j == 'y':print('...
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