python练习题怎么做？ 这个python 编程练习题，怎么做？头疼
Python\u4e2d\u57fa\u7840\u7ec3\u4e60\u9898\uff1f\u6cd5\u4e00:\u5229\u7528set()\u51fd\u6570\u7684\u53bb\u91cd\u529f\u80fd\uff0c\u53bb\u91cd\u540e\u518d\u4f7f\u7528list()\u51fd\u6570\u5c06\u96c6\u5408\u8f6c\u6362\u4e3a\u6211\u4eec\u60f3\u8981\u7684\u5217\u8868
list1 = [11,22,33]
list2 = [22,33,44]
list3 = list(set(list1 + list2))
list3.sort()
print(list3)
-------------
\u6cd5\u4e8c\uff1a\u5229\u7528if\u548cfor,\u5148\u904d\u5386list1\u6240\u6709\u5143\u7d20\u8ffd\u52a0\u5230list3\u4e2d\uff0c\u7136\u540e\u904d\u5386list2,\u6761\u4ef6\u5224\u65adlist2\u4e2d\u5f53\u524d\u5143\u7d20\u662f\u5426\u5728list3\u4e2d\uff0c\u5982\u679c\u4e0d\u5728\u5219\u8ffd\u52a0\u5230list3\u4e2d
list1 = [11,22,33]
list2 = [22,33,44]
list3 = []
for ele1 in list1:
list3.append(ele1)
for ele2 in list2:
if ele2 not in list3:
list3.append(ele2)
print(list3)
\u793e\u4ea4\u7f51\u7edc\u4e2d\u6211\u4eec\u7ed9\u6bcf\u4e2a\u4eba\u5b9a\u4e49\u4e86\u4e00\u4e2a\u201c\u6d3b\u8dc3\u5ea6\u201d\uff0c\u73b0\u5e0c\u671b\u6839\u636e\u8fd9\u4e2a\u6307\u6807\u628a\u4eba\u7fa4\u5206\u4e3a\u4e24\u5927\u7c7b\uff0c\u5373\u5916\u5411\u578b\uff08outgoing\uff0c\u5373\u6d3b\u8dc3\u5ea6\u9ad8\u7684\uff09\u548c\u5185\u5411\u578b\uff08introverted\uff0c\u5373\u6d3b\u8dc3\u5ea6\u4f4e\u7684\uff09\u3002\u8981\u6c42\u4e24\u7c7b\u4eba\u7fa4\u7684\u89c4\u6a21\u5c3d\u53ef\u80fd\u63a5\u8fd1\uff0c\u800c\u4ed6\u4eec\u7684\u603b\u6d3b\u8dc3\u5ea6\u5dee\u8ddd\u5c3d\u53ef\u80fd\u62c9\u5f00\u3002
\u8f93\u5165\u683c\u5f0f\uff1a
\u8f93\u5165\u7b2c\u4e00\u884c\u7ed9\u51fa\u4e00\u4e2a\u6b63\u6574\u6570N\uff082 <= N <= 105\uff09\u3002\u968f\u540e\u4e00\u884c\u7ed9\u51faN\u4e2a\u6b63\u6574\u6570\uff0c\u5206\u522b\u662f\u6bcf\u4e2a\u4eba\u7684\u6d3b\u8dc3\u5ea6\uff0c\u5176\u95f4\u4ee5\u7a7a\u683c\u5206\u9694\u3002\u9898\u76ee\u4fdd\u8bc1\u8fd9\u4e9b\u6570\u5b57\u4ee5\u53ca\u5b83\u4eec\u7684\u548c\u90fd\u4e0d\u4f1a\u8d85\u8fc7231\u3002
\u8f93\u51fa\u683c\u5f0f\uff1a
\u6309\u4e0b\u5217\u683c\u5f0f\u8f93\u51fa\uff1a
Outgoing #: N1
Introverted #: N2
Diff = N3
\u5176\u4e2d N1 \u662f\u5916\u5411\u578b\u4eba\u7684\u4e2a\u6570\uff1bN2 \u662f\u5185\u5411\u578b\u4eba\u7684\u4e2a\u6570\uff1bN3 \u662f\u4e24\u7fa4\u4eba\u603b\u6d3b\u8dc3\u5ea6\u4e4b\u5dee\u7684\u7edd\u5bf9\u503c\u3002
\u8f93\u5165\u6837\u4f8b1\uff1a
10
23 8 10 99 46 2333 46 1 666 555
\u8f93\u51fa\u6837\u4f8b1\uff1a
Outgoing #: 5
Introverted #: 5
Diff = 3611
\u8f93\u5165\u6837\u4f8b2\uff1a
13
110 79 218 69 3721 100 29 135 2 6 13 5188 85
\u8f93\u51fa\u6837\u4f8b2\uff1a
Outgoing #: 7
Introverted #: 6
Diff = 9359```
\u6211\u7684\u4ee3\u7801\uff1a
n = int(input('\u6570\u636e\u4e2a\u6570'))
a = []
for i in range(n):
a.append(int(input('\u8bf7\u8f93\u5165\u7b2c %d \u4e2a\u6570\u636e'% (i+1))))
\u6ce8\u610fa/n\u7684\u6570\u636e\u7c7b\u578b\uff0c\u4e00\u5b9a\u8981\u8f6c\u6362\uff0c\u5426\u5219\u51fa\u9519
a\u4e0d\u53ef\u4ee5\u76f4\u63a5list(input(''))\uff0c\u5426\u5219\u51fa\u6765\u7684\u662f\u4e00\u4e2a\u5e26\u5f15\u53f7\u548c\u9017\u53f7\u7684list
\u6211\u89c9\u5f97\u53ef\u4ee5\u7528\u5207\u7247\uff0c\u4f46\u662f\u4e5f\u633a\u9ebb\u70e6\u7684\uff0c\u60f3\u77e5\u9053\u6709\u6ca1\u6709\u7b80\u5355\u4e00\u70b9\u7684\u65b9\u6cd5
a.sort()
sum1 = 0
sum2 = 0
sum3 = 0
\u4e0a\u9762\u7684sum1,sum2,sum3\u90fd\u8981\u5b9a\u4e49\uff0c\u56e0\u4e3a\u7b2c\u4e00\u6b21\u51fa\u73b0\u7684\u5f0f\u5b50\u53f3\u8fb9\u5b58\u5728
if n % 2 == 0: #\u53d6\u4f59\u6570\u4e24\u79cd\u65b9\u6cd51. x%y 2. divmod(x,y)
x = n/2
for i in range(int(x)): #\u6ce8\u610fx\u8981int, range\u7684\u53c2\u6570\u53ea\u80fd\u662fint\u7c7b\u578b\uff0c\u5426\u5219\u51fa\u9519
sum1 = sum1 + a[i]
for i in range(
stds_list= [
{"id": 1, "name": "小明", "c_s": 85, "python_s": 78},
{"id": 2, "name": "小花", "c_s": 69, "python_s": 88},
{"id": 3, "name": "小东", "c_s": 79, "python_s": 83},
]
# 1) 显示学生信息：“学生id：学生姓名：小明，C语言成绩：85, Python成绩：78”。
for ind in range(len(stds_list)):
if stds_list[ind]['name'] == '小明':
print('学生id：{id}，学生姓名：{name}，C语言成绩：{c_s}, Python成绩：{python_s}'.format(**stds_list[ind]))
# 2) 修改“小明”的Python成绩为90
for ind in range(len(stds_list)):
if stds_list[ind]['name'] == '小明':
stds_list[ind]['python_s'] = 90
break
# 3) 删除“小东”的信息
for ind in range(len(stds_list)):
if stds_list[ind]['name'] == '小东':
del stds_list[ind]
break
# 2. 定义一个空列表，用于保存5个学生信息，一个学生信息包括三个属性：id、姓名、年龄
# 提示：列表元素是字典、向列表中添加数据用append()
stds_list2 = []
for i in range(5):
print('第{}个学生信息：')
stds_list2.append({})
for j in ['id','姓名','年龄']:
stds_list2[-1][j] = input('{}:'.format(j))
print(stds_list2)
代码缩进
第一题：
1
>>> stds_list[0]
{'id': 1, 'name': '小明', 'c_s': 85, 'python_s': 78}
2
>>> stds_list[0]["python_s"]=90
3
>>> stds_list.pop()
第二题
student_info=[]
student_info.append({"id":1,"name":"A","age":10})
student_info.append({"id":2,"name":"B","age":10})
student_info.append({"id":3,"name":"C","age":10})
student_info.append({"id":4,"name":"D","age":10})
student_info.append({"id":5,"name":"E","age":10})
print(student_info)
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