def orderBy(): number1 = input("请输入第1个数字:") number2 = input("请输入第2个数字:") number3 = input("请输入第3个数字:") number = [int(number1),int(number2),int(number3)] i = number.sort() numer = [j for j in number] print(numer) orderBy()
时间: 2024-02-23 19:00:06 浏览: 136
这段代码存在一个问题:`number.sort()` 会改变原列表 `number` 的顺序,但是它不会返回任何值。因此,`i` 的值将为 `None`,而 `numer` 中的数字顺序仍然是原始顺序。可以使用以下代码来修复这个问题:
```
def orderBy():
number1 = input("请输入第1个数字:")
number2 = input("请输入第2个数字:")
number3 = input("请输入第3个数字:")
number = [int(number1), int(number2), int(number3)]
number.sort()
numer = [j for j in number]
print(numer)
orderBy()
```
或者,可以使用 Python 内置的 `sorted()` 函数来创建一个新的已排序列表,而不改变原始列表的顺序:
```
def orderBy():
number1 = input("请输入第1个数字:")
number2 = input("请输入第2个数字:")
number3 = input("请输入第3个数字:")
number = [int(number1), int(number2), int(number3)]
numer = sorted(number)
print(numer)
orderBy()
```
相关问题
实现功能: 用户在第一个页面输入数字一,在第二个页面输入数字二,第三个页面显示:“数字一 + 数字二 = 和”。 要求: 求和计算功能必须使用SQL技术实现
好的,根据你的需求,我可以为你提供一个简单的实现方案:
1. 创建一个名为 "sum_app" 的 Django 应用程序。
2. 在 "sum_app" 应用程序下的 "models.py" 文件中定义一个名为 "Number" 的模型,代码如下:
```
from django.db import models
class Number(models.Model):
num1 = models.IntegerField()
num2 = models.IntegerField()
```
3. 在 "sum_app" 应用程序下的 "views.py" 文件中定义一个名为 "calculate_sum" 的视图函数,代码如下:
```
from django.shortcuts import render
from django.db import connection
from .models import Number
def calculate_sum(request):
if request.method == 'POST':
num1 = request.POST.get('num1')
num2 = request.POST.get('num2')
# 将 num1 和 num2 存入数据库中
Number.objects.create(num1=num1, num2=num2)
# 使用 SQL 查询计算和
with connection.cursor() as cursor:
cursor.execute("SELECT num1 + num2 FROM sum_app_number ORDER BY id DESC LIMIT 1")
result = cursor.fetchone()[0]
context = {'result': result}
return render(request, 'sum_app/result.html', context)
return render(request, 'sum_app/index.html')
```
4. 在 "sum_app" 应用程序下的 "urls.py" 文件中定义一个名为 "calculate_sum" 的 URL,代码如下:
```
from django.urls import path
from . import views
urlpatterns = [
path('', views.calculate_sum, name='calculate_sum'),
]
```
5. 在 "sum_app" 应用程序下的 "templates" 文件夹中创建两个 HTML 文件,分别为 "index.html" 和 "result.html",代码如下:
index.html:
```
<!DOCTYPE html>
<html>
<head>
<title>Calculate Sum</title>
</head>
<body>
<h1>Enter two numbers to calculate their sum:</h1>
<form method="post">
{% csrf_token %}
<label for="num1">Number 1:</label>
<input type="number" name="num1" id="num1">
<br>
<label for="num2">Number 2:</label>
<input type="number" name="num2" id="num2">
<br>
<input type="submit" value="Calculate Sum">
</form>
</body>
</html>
```
result.html:
```
<!DOCTYPE html>
<html>
<head>
<title>Sum Result</title>
</head>
<body>
<h1>The sum of the two numbers is:</h1>
<p>{{ result }}</p>
</body>
</html>
```
6. 最后,在项目的 "urls.py" 文件中包含 "sum_app" 应用程序的 URL 配置,代码如下:
```
from django.contrib import admin
from django.urls import path, include
urlpatterns = [
path('admin/', admin.site.urls),
path('', include('sum_app.urls')),
]
```
这样,当用户访问应用程序的首页时,将会看到一个表单,要求输入两个数字。当用户提交表单后,将会计算这两个数字的和,并显示在另一个页面上。计算和的过程中使用了 SQL 查询,从数据库中获取最新的两个数字并计算它们的和。
You are given a positive integer nn. In this problem, the MEXMEX of a collection of integers c1,c2,…,ckc1,c2,…,ck is defined as the smallest positive integer xx which does not occur in the collection cc. The primality of an array a1,…,ana1,…,an is defined as the number of pairs (l,r)(l,r) such that 1≤l≤r≤n1≤l≤r≤n and MEX(al,…,ar)MEX(al,…,ar) is a prime number. Find any permutation of 1,2,…,n1,2,…,n with the maximum possible primality among all permutations of 1,2,…,n1,2,…,n. Note: A prime number is a number greater than or equal to 22 that is not divisible by any positive integer except 11 and itself. For example, 2,5,132,5,13 are prime numbers, but 11 and 66 are not prime numbers. A permutation of 1,2,…,n1,2,…,n is an array consisting of nn distinct integers from 11 to nn in arbitrary order. For example, [2,3,1,5,4][2,3,1,5,4] is a permutation, but [1,2,2][1,2,2] is not a permutation (22 appears twice in the array), and [1,3,4][1,3,4] is also not a permutation (n=3n=3 but there is 44 in the array).
To find a permutation of 1, 2, ..., n with the maximum possible primality, we can use the following approach:
1. If n is 1, return [1] as the permutation.
2. If n is 2, return [2, 1] as the permutation. This guarantees that the MEX is 1, which is a prime number.
3. If n is 3, return [3, 1, 2] as the permutation. Again, this guarantees that the MEX is 1, which is a prime number.
4. If n is greater than 3, we can construct the permutation as follows:
- Start with [2, 3, 1], which guarantees that the MEX of the first three elements is 4.
- Append the remaining numbers from 4 to n in any order.
Here's an example implementation in Python:
```python
def max_prime_permutation(n):
if n == 1:
return [1]
elif n == 2:
return [2, 1]
elif n == 3:
return [3, 1, 2]
else:
permutation = [2, 3, 1] + list(range(4, n + 1))
return permutation
n = int(input("Enter a positive integer: "))
permutation = max_prime_permutation(n)
print("Permutation:", permutation)
```
This approach ensures that the MEX of any consecutive subarray of the permutation is always a prime number.
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