n2

integrate CMS with an existing ASP.NET web site

由 孤街醉人 提交于 2019-12-05 02:55:52
I am trying to integrate a CMS into an existing ASP.NET Web site built with .aspx files and .aspx.cs code behind files (not compiled). My research indicated that my best candidate is N2 CMS. My ideal approach is to configure the editor with various editable content types and have my existing pages and user controls use the API to retrieve the content and place it pro grammatically into containers. I couldn't find how this can be done, anyone have any idea how and if can this be done ? Is there any other CMS that is better suited for the task? We have had this exact setup existing web

小学四则运算编程实践

由 孤街醉人 提交于 2019-12-05 00:16:34
代码 import random from fractions import Fraction ##两个整数的四则运算 def c1(q, ans): symbol = random.choice(['+', '-', '*', '/']) # 生成随机符号 if symbol == '+': n1 = random.randint(0, 20) n2 = random.randint(0, 20) q.append(str(n1) + '+' + str(n2) + '=') ans.append(n1 + n2) elif symbol == '-': n1 = random.randint(0, 20) n2 = random.randint(0, 20) n1,n2 = max(n1,n1),min(n1,n2)#防止出现负数 q.append(str(n1) + '-' + str(n2) + '=') ans.append(n1 - n2) elif symbol == '*': n1 = random.randint(0, 20) n2 = random.randint(0, 20) q.append(str(n1) + '×' + str(n2) + '=') ans.append(n1 * n2) else: n1 = random.randint(0, 20)

共轭对称序列的快速傅里叶反变换

由 偶尔善良 提交于 2019-12-04 16:34:12
一、功能 计算共轭对称复序列的快速傅里叶反变换，其变换结果是实数。 二、方法简介 序列 \(x(n)\) 的离散傅里叶变换为 \[ X(k)=\sum_{n=0}^{N-1}x(n)W_{N}^{nk}, \ k=0,1,...,N-1 \] 序列 \(X(k)\) 的离散傅里叶反变换为 \[ x(n)=\frac{1}{N}\sum_{n=0}^{N-1}X(k)W_{N}^{-nk}, \ n=0,1,...,N-1 \] 共轭对称复序列的傅里叶反变换，可用复序列快速傅里叶反变换算法进行计算。但考虑到 \(X(k)\) 是共轭对称的，其傅里叶反变换 \(x(n)\) 是实数，因此，为进一步提高计算效率，需要对一般的复序列IFFT算法进行一定的修改。 共轭对称序列 \(X(k)\) 具有如下性质： \(X(0)\) 和 \(X(\frac{N}{2})\) 都是实数，且有 \[ X(k)=X^{*}(N-k), \ 1\leqslant k\leqslant \frac{N}{2}-1 \] 即 \(X(k)\) 的实部是偶对称，虚部是奇对称。在计算傅里叶反变换时，利用这种共轭对称性，我们就可以不必计算和存储 \(X(k)(\frac{N}{2}+1\leqslant k\leqslant N-1)\) 以及 \(X(0)\) 和 \(X(\frac{N}{2})\) 的虚部

count

由 落爺英雄遲暮 提交于 2019-12-04 15:46:59
str.count(x, _start, _end) x:需要查找的字符串（包含单个字符）。 _start:查找开始的位置，默认为0，也就是str开始的位置。 _end:查找结束的位置，默认为-1，也就是str结束的位置。 sentence = 'I can because I think I can' n1 = sentence.count('i') n2 = sentence.count('can') print('n1', n1) print('n2', n2) 来源： https://www.cnblogs.com/qianslup/p/11828947.html

Web site projects vs Web Application - which is better?

由 ￣綄美尐妖づ 提交于 2019-12-04 10:44:26
问题 I've just started using a cms (N2) which has a template basic implementation using a web site project template in vs2008. I've noticed that when compiling it it takes a lot longer than using a web application project which I am more use to. My questions are: Why does it seem to take a lot longer to compile? Which is better to use? Should I convert it into being a web application? I apologise if this is a duplicate but I couldn't find a similar question. Cheers 回答1: The Major differences are:

Coq equality implementation

由 匿名 (未验证) 提交于 2019-12-03 08:46:08
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I'm writing a toy language where nodes in the AST can have any number of children ( Num has 0, Arrow has 2, etc). You might call these operators. Additionally, exactly one node in the AST might be "focused". We index the data type with Z if it has a focus, or H if it doesn't. I need advice on a few parts of the code. Hopefully it's alright to ask all of these at once, since they're related. How would you define the type of internal nodes with one focus, InternalZ ? Right now I say "we have S n children -- n of them are unfocused and one (at

How to generate two different random numbers?

由 匿名 (未验证) 提交于 2019-12-03 08:41:19
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I need to generate two different random numbers, they can't be equal to each other or to a third number. I tried to use a lot of if's to cover every possibility but, it seems my algorithm skills are not that good. Can anyone help me on this? var numberOne = Math.floor(Math.random() * 4); var numberTwo = Math.floor(Math.random() * 4); var numberThree = 3; // This number will not always be 3 if((numberOne == numberThree) && (numberOne + 1 < 3)) { numberOne++; } else if ((numberOne == numberThree) && (numberOne + 1 == 3)) { numberOne = 0; } if

Copy contents of one textbox to another

由 匿名 (未验证) 提交于 2019-12-03 02:48:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Suppose an entry is made in a textbox. Is it possible to retain the same entered text in a second text box? If so, how is this done? <html> <label>First</label><input type="text" name="n1" id="n1"> <label>Second</label><input type="text" name="n1" id="n1"/> </html> Thanks. 回答1: <script> function sync() { var n1 = document.getElementById('n1'); var n2 = document.getElementById('n2'); n2.value = n1.value; } </script> <input type="text" name="n1" id="n1" onkeyup="sync()"> <input type="text" name="n2" id="n2"/> 回答2: More efficiently it can be

TypeScript sorting an array

由 匿名 (未验证) 提交于 2019-12-03 02:45:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I've been trying to figure out a very strange issue I ran into with typescript. It was treating an inline Boolean expression as whatever the first value's type was instead of the complete expression. So if you try something simple like the following: var numericArray:Array<number> = [2,3,4,1,5,8,11]; var sorrtedArray:Array<number> = numericArray.sort((n1,n2)=> n1 > n2); TryIt You will get an error on your sort method saying the parameters do not match any signature of the call target, because your result is numeric and not Boolean. I guess I

1009 Product of Polynomials (25)（25 point(s)）

由 匿名 (未验证) 提交于 2019-12-03 00:39:02
problem This time, you are supposed to find A*B where A and B are two polynomials. Input Specification: Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 a~N1~ N2 a~N2~ ... NK a~NK~, where K is the number of nonzero terms in the polynomial, Ni and a~Ni~ (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000. Output Specification: For each test case you should output the product of A and B in one line, with the same format as the input. Notice
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