python编程题目
❶ python程序设计题
这是一个二分思想题。以1为例子。每次的二分点为50，25，13，7，8，4，2。现在就知道答案肯定是1了。
❷ 用python写编程题
第一个问题使用排序算法，有很多种，可以使用简单一点的冒泡排序。第二个问题为了确保是输入了5个整数，可以使用while循环+try。
❸ python编程题：
我用的是Python3.6.x ，刚才一个题和这个一样是你问的吗？
importos
importsqlite3
__ROOT__="C:/PythonFiels/"
ifnotos.path.exists(__ROOT__):
os.mkdir(__ROOT__)
__LINK__=sqlite3.connect(os.path.join(__ROOT__,'datas.db'))
__CURSOR__=__LINK__.cursor()
defstudents():
students=[]
foriinrange(1,31):
students.append(
{
'学号':i,
'姓名':'李荣浩',
'性别':'男',
'年龄':34,
}
)
returnstudents
definit():
try:
createTable()
except:
pass
forstudentinstudents():
studentPath=os.path.join(__ROOT__,str(student['学号']))
os.mkdir(studentPath)
studentAvatar=os.path.join(studentPath,'{}.jpg'.format(student['学号']))
withopen(studentAvatar,'wb')asf:
pass
studentExcel=os.path.join(studentPath,"{}.csv".format(student['姓名']))
withopen(studentExcel,'w')asf:
f.write("姓名,性别,年龄,照片 ")
f.write("{},{},{},{}".format(
student['姓名'],student['性别'],student['年龄'],'{}.jpg'.format(student['学号'])
))
sql="INSERTINTOstudents(id,name,gender,age,avatar)VALUES('{}','{}','{}','{}','{}')".format(
student['学号'],student['姓名'],student['性别'],student['年龄'],'{}.jpg'.format(student['学号'])
)
if__CURSOR__.execute(sql):
__LINK__.commit()
print('一条数据插入成功 ')
else:
print('一条数据插入失败 ')
returnmenu()
defcreateTable(tablename='students'):
sql="""CREATETABLE{}
(idINT(6)PRIMARYKEYNOTNULL,
nameVARCHAR(10)NOTNULL,
ageVARCAHR(10)NOTNULL,
genderVARCHAR(10)NOTNULL,
avatarVARCHAR(10)NOTNULL)""".format(tablename)
return__CURSOR__.execute(sql)
defread():
print('所有数据列表：')
forvalueinall():
print(value)
returnmenu()
defall():
sql="SELECT*FROMstudents"
r=__CURSOR__.execute(sql)
returnr
definfo():
sql="SELECTcount(*)FROMstudents"
r=__CURSOR__.execute(sql)
number=0
age=0
forvalueinr:
number=value[0]
sql="SELECTavg(age)FROMstudents"
r=__CURSOR__.execute(sql)
forvalueinr:
age=value[0]
print('学生共有{}人,年龄平均数为:{}'.format(number,age))
menu()
defupdate():
forvalueinall():
age=value[2]+1
id=value[0]
sql="UPDATEstudentsSETage='{}'WHEREid={}".format(age,id)
__CURSOR__.execute(sql)
__LINK__.commit()
print('更新完毕!')
returnmenu()defaction(id):
id=str(id)
ifid=='1':
returninit()
elifid=='2':
returnread()
elifid=='3':
returninfo()
elifid=='4':
returnupdate()
else:
print('请对照选项进行选择!')
returnmenu()
defmenu():
id=input("请选择:1.初始化数据2.读取所有数据(请先初始化数据)3.显示学生数量和平均年龄(请先初始化数据)4.每人年龄+1(请先初始化数据)");
action(id)
menu()
❹ 求助python编程的题目
def count(val, seq):
"""
>>> count(5, (1, 5, 3, 7, 5, 8, 5))
3
>>> count('s', 'Mississippi')
4
>>> count((1, 2), [1, 5, (1, 2), 7, (1, 2), 8, 5])
2
"""
return list(seq).count(val)
def reverse(seq):
"""
>>> reverse([1, 2, 3, 4, 5])
[5, 4, 3, 2, 1]
>>> reverse(('shoe', 'my', 'buckle', 2, 1))
(1, 2, 'buckle', 'my', 'shoe')
>>> reverse('Python')
'nohtyP'
"""
return seq[::-1]
def sort_sequence(seq):
"""
>>> sort_sequence([3, 4, 6, 7, 8, 2])
[2, 3, 4, 6, 7, 8]
>>> sort_sequence((3, 4, 6, 7, 8, 2))
(2, 3, 4, 6, 7, 8)
>>> sort_sequence("nothappy")
'ahnoppty'
"""
if type(seq) == str:
return ''.join(sorted(seq))
else:
return type(seq)(sorted(seq))
def recursive_min(nested_num_list):
"""
>>> recursive_min([2, 9, [1, 13], 8, 6])
1
>>> recursive_min([2, [[100, 1], 90], [10, 13], 8, 6])
1
>>> recursive_min([2, [[13, -7], 90], [1, 100], 8, 6])
-7
>>> recursive_min([[[-13, 7], 90], 2, [1, 100], 8, 6])
-13
"""
smallest = nested_num_list[0]
while type(smallest) == type([]):
smallest = smallest[0]
for element in nested_num_list:
if type(element) == type([]):
min_of_elem = recursive_min(element)
if smallest > min_of_elem:
smallest = min_of_elem
else: # element is not a list
if smallest > element:
smallest = element
return smallest
def recursive_count(target, nested_num_list):
"""
>>> recursive_count(2, [2, 9, [2, 1, 13, 2], 8, [2, 6]])
4
>>> recursive_count(7, [[9, [7, 1, 13, 2], 8], [7, 6]])
2
>>> recursive_count(15, [[9, [7, 1, 13, 2], 8], [2, 6]])
0
>>> recursive_count(5, [[5, [5, [1, 5], 5], 5], [5, 6]])
6
"""
count = 0
for element in nested_num_list:
if type(element) == type([]):
count += recursive_count(target, element)
else: # element is not a list
if element == target:
count += 1
return count
def flatten(nested_num_list):
"""
>>> flatten([2, 9, [2, 1, 13, 2], 8, [2, 6]])
[2, 9, 2, 1, 13, 2, 8, 2, 6]
>>> flatten([[9, [7, 1, 13, 2], 8], [7, 6]])
[9, 7, 1, 13, 2, 8, 7, 6]
>>> flatten([[9, [7, 1, 13, 2], 8], [2, 6]])
[9, 7, 1, 13, 2, 8, 2, 6]
>>> flatten([[5, [5, [1, 5], 5], 5], [5, 6]])
[5, 5, 1, 5, 5, 5, 5, 6]
"""
flat_list = []
for element in nested_num_list:
if type(element) == type([]):
flat_list += flatten(element)
else: # element is not a list
flat_list.append(element)
return flat_list
if __name__ == "__main__":
import doctest
doctest.testmod()
----------------------------------
6 items passed all tests:
3 tests in __main__.count
4 tests in __main__.flatten
4 tests in __main__.recursive_count
4 tests in __main__.recursive_min
3 tests in __main__.reverse
3 tests in __main__.sort_sequence
21 tests in 7 items.
21 passed and 0 failed.
Test passed.
-----------------------------------------
LZ在自学这个把
http://www.openbookproject.net/thinkCSpy/
确实是好书
❺ Python编程题
defformatit(x):
ifisinstance(x,int):
x=str(x)
return(''+x)[-2:]
defmain():
title=['S','M','T','W','Th','F','S']
days=eval(input('Enternumberofdays:'))
firstday=input('Thefirstdayoftheweek:')
fd=title.index(firstday)
wts=[]
wt=['','','','','','','']#一个星期的模板
foriinrange(days):
wt[(i+fd)%7]=i+1
if(i+fd)%7==6:#如果满一个星期，append当前模板，换新模板
wts.append(wt)
wt=['','','','','','','']
if(i+fd)%7!=6:#如果最后一天不是星期六，append当前板
wts.append(wt)
print(''.join(map(formatit,title)))#按预想的格式输出标题
forxinwts:
print(''.join(map(formatit,x)))#按预想的格式输出星期
main()
❻ python 编程几个小题目
#第一题
dicf={"chen":48,"li":72,"zhang":60,"jiang":92}
foritem,valueindicf.items():
ifvalue==max(dicf.values()):
printitem," ",value
#第二题
lista=[3,5,7,9]
print"initiallist:",lista#打印初始值
lista.insert(2,6)#插入操作
print"afterinsert:",lista
dellista[1]#删除操作
print"afterdelete:",lista
lista[0]=0#修改操作
print"afterchange:",lista
foriinlista:
printi#遍历打印
❼ python编程题
mons = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
def get_days(mon, day):
if mon == 1:
return mons[0], day
else:
count = sum(mons[:mon-1])
count = count + day
return mons[mon - 1], count
mon = int(input("请输入月份："))
day = int(input("请输入号数："))
result = get_days(mon, day)
print("{}月有{}天。".format(mon, result[0]))
print("{}月{}号是内该年的第容{}天".format(mon, day, result[1]))
❽ 一道简单的python编程题目
逆序切片-2，-4，-6...
❾ Python编程题
#缩进格式看图
t=eval(input())
for e in t:
if 1==t.count(e):
print(e)
❿ 一道python编程题
在第一个for的最后一行加个换行