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do a simple, but it can easily expand. Time is the RAS API, I used the platform...
做的比较简单,但可以方便扩充。计时用的是RAS API,我用的是最新的platform sdk,如果没有可能编译不过,可以上MS网站去更新一下-do a simple, but it can easily expand. Time is the RAS API, I used the platform is the latest sdk, if it is not possible compiler can, however, on the MS site to update the
- 2023-02-23 20:25:04下载
- 积分:1
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CUDA FFT,实现OcenFFT参考代码
代码用于实现OcenFFT,GPU实现FFT的过程,相比CPU的C语言速度至少提升一个数量级,可实现二位的FFT,得到实时动态的二维海面
- 2022-02-25 10:06:42下载
- 积分:1
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this Program can get memory information,is good
this Program can get memory information,is good
- 2022-03-24 15:21:22下载
- 积分:1
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include Polynomialh
用链表实现多项式的加减乘法,c源代码。
用类封装。-include Polynomialh
- 2023-04-22 20:30:03下载
- 积分:1
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Problem A:
Problem A:编辑距离问题
Time Limit:1000MS Memory Limit:65536K
Total Submit:157 Accepted:79
Description
设A 和B 是2 个字符串。要用最少的字符操作将字符串A 转换为字符串B。这里所说的字符操作包括
(1)删除一个字符;
(2)插入一个字符;
(3)将一个字符改为另一个字符。
将字符串A变换为字符串B 所用的最少字符操作数称为字符串A到B 的编辑距离,记为d(A,B)。试设计一个有效算法,对任给的2 个字符串A和B,计算出它们的编辑距离d(A,B)。
编程任务:
对于给定的字符串A和字符串B,编程计算其编辑距离d(A,B)。
Input
输入由多组测试数据组成。
每组测试数据输入的第1 行是字符串A,第2行是字符串B。
Output
对应每组输入,输出的每行中的数是编辑距离d(A,B)。
Sample Input
fxpimu
xwrs
Sample Output
5-Problem A:
- 2022-03-16 09:31:03下载
- 积分:1
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C# 1 on the mapping process. . Very simple. . Help you learn C# tools for the us...
一个关于C#的作图程序。。很简单的。。帮助你学习C#工具栏的使用-C# 1 on the mapping process. . Very simple. . Help you learn C# tools for the use of column.
- 2022-03-26 11:59:52下载
- 积分:1
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Week Ligong Singlechip developed manual, the specific data manual see Zhou Ligon...
周立功公司开发的单片机手册,具体的数据手册见周立功网站下载。-Week Ligong Singlechip developed manual, the specific data manual see Zhou Ligong website.
- 2022-06-14 12:38:57下载
- 积分:1
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Case Study Reversi, Reversi based on artificial intelligence programming.
黑白棋实例研究,基于人工智能的黑白棋程序设计。-Case Study Reversi, Reversi based on artificial intelligence programming.
- 2023-05-24 11:25:03下载
- 积分:1
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Not is to provide C/C++ programmer job interview guide, which aims to examinatio...
并不在于提供C/C++程序员求职面试指导,而旨在从技术上分析面试题的内涵。文中的大多数面试题来自各大论坛,部分试题解答也参考了网友的意见。
许多面试题看似简单,却需要深厚的基本功才能给出完美的解答。企业要求面试者写一个最简单的strcpy函数都可看出面试者在技术上究竟达到了怎样的程度,我们能真正写好一个strcpy函数吗?我们都觉得自己能,可是我们写出的strcpy很可能只能拿到10分中的2分。读者可从本文看到strcpy函数从2分到10分解答的例子,看看自己属于什么样的层次。此外,还有一些面试题考查面试者敏捷的思维能力。
分析这些面试题,本身包含很强的趣味性;而作为一名研发人员,通过对这些面试题的深入剖析则可进一步增强自身的内功。
-Not is to provide C/C++ programmer job interview guide, which aims to examination questions from a technical analysis of face connotations. Most of the text surface examination questions from the major forums, part of the answer in reference to the Questions User s views.
Many face examination questions seem simple, yet profound basic skills needed in order to give the perfect answer. Company asked interviewers to write a simple strcpy function of the interview can be seen in the technology of how to achieve the extent of what we can really written a strcpy function do? We all feel that you can, but we write the strcpy is likely to receive only 10 points in 2
- 2022-03-01 08:24:29下载
- 积分:1
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数据结构中关于四则运算的实验,主要涉及到栈的操作,在运行窗口中选择1,进入运算操作.输入你所想要的式子,注意负数不要参与运算,但式子能产生负数.式子例如:...
数据结构中关于四则运算的实验,主要涉及到栈的操作,在运行窗口中选择1,进入运算操作.输入你所想要的式子,注意负数不要参与运算,但式子能产生负数.式子例如:
输入9*(8-6/2)+9/2然后按Enter键,可以得到式子的值是49.5.
或者输入9-10*2然后按Enter键,可以得到式子的值是-11.
-Data structure on the four computing experiments, mainly related to the operation of the stack, in the run window, select one, enter the computing operation. Enter the formula you want to pay attention to the negative not to take part in computing, but the formula can produce negative. Style son such as: Enter the 9* (8-6/2)+ 9/2 and then press Enter key, you can get the formula value is 49.5. or enter a 9-10* 2 and then press Enter key, you can get the value of formula is-11.
- 2022-03-17 18:39:04下载
- 积分:1