1,使用Python解决数学问题。
ABCD乘9 = DBCA 那么 A=?,B=?,C? D=?
for A in range(1,10): for B in range(0,10): for C in range(0,10): for D in range(1,10): if (A*1000 + B * 100 + C * 10 +D)* 9 == (D*1000 + C*100 +B*10 + A): print('{0}{1}{2}{3}*9 == {3}{2}{1}{0}'.format(A,B,C,D))
2,使用python求阶乘的和。
0!+1!+2!+3!+4!+....+n!
def fact(n): if n == 0: return 1 return n * fact(n-1) while True: result = 0 n = input('Please input number:') if not n.isdigit(): print('你输入的不是纯数字,请重新输入!') continue for i in range(0,int(n)+1): result+=fact(i) print(result)
3,计算输入字符串有多少个空格、字符串、数字、特殊符号的个数。
while True: n = input('pelase input somthing:') num = strings = space = other = 0 for i in n: if str(i).isalpha(): strings+=1 elif str(i).isdigit(): num +=1 elif str(i).isspace(): space +=1 else: other+=1 print('数字有{0}个,字母有{1}个,空格有{2}个,其他字符{3}个'.format(num,strings,space,other))
4,乘法口诀
for h in range(1,10): for l in range(1,h+1): print('{0}*{1} = {2} '.format(l,h,l*h),end = '') if h == l: print('')
5,/etc/passwd排序,按照uid排序(思路:创建2个列表,一个列表存放所有的uid并排序,一个存放所有的行,然后存放uid的列表排序,最后将文件中uid与uid比较,追加到新的文件即可。)
import codecs file = 'passwd' sortfile = 'sortpasswd' filecontext = [] sortuid = [] with codecs.open('sortfile','wb') as fd: with codecs.open(file,encoding='utf-8') as f: filecontext += f.readlines() for line in filecontext: sortuid.append(int(line.split(':')[2])) sortuid.sort() for uid in sortuid: for line in filecontext: if str(uid) == line.split(':')[2]: fd.write(line.encode('utf-8'))