-
floating point multiplier
floating point multiplier
- 2022-08-25 05:32:46下载
- 积分：1
-
这是一个基本的调查系统，可用于任何网站。
这是一个基本的调查系统，可以在任何网站上使用。-This is a basic survey system which can be used in any website.
- 2022-02-12 12:20:11下载
- 积分：1
-
基于ODBC的数据库备份及恢复
基于ODBC的数据库备份及恢复 -based ODBC database backup and recovery
- 2022-04-21 10:07:20下载
- 积分：1
-
约瑟夫环，用顺序表实现的，也是可以用链表实现的
约瑟夫环，用顺序表实现的，也是可以用链表实现的-Joseph Wan, the implementation of the order form, but also can realize the list
- 2022-03-22 00:51:13下载
- 积分：1
-
1 summarizes the most current methods of ellipse fitting articles and in
一篇总结了当前大多数椭圆拟合方法的文章，并进行了深度分析，提出了改进的方法，适用于初学者进阶阅读。-1 summarizes the most current methods of ellipse fitting articles and in-depth analysis, put forward an improved method, applicable to advanced beginners to read.
- 2022-02-03 23:09:08下载
- 积分：1
-
一个简单的计算器
一个简单的计算器-a simple calculator
- 2022-06-27 12:06:32下载
- 积分：1
-
MATLAB实现的伽罗华域乘法，在AES的安全算法中应用到
MATLAB实现的伽罗华域乘法，在AES的安全算法中应用到-MATLAB implementation Galois field multiplication, in the AES security algorithm applied to the
- 2022-03-11 18:10:44下载
- 积分：1
-
对seg2格式数据处理的一个包装类，标准c++编写，测试过没问题，希望和同仁共享...
对seg2格式数据处理的一个包装类，标准c++编写，测试过没问题，希望和同仁共享-Seg2 format data to deal with a wrapper class, standards c++ Preparation, testing had no problem, hope and sharing Tongren
- 2022-06-29 08:07:36下载
- 积分：1
-
* 程序名称: NET
* 程序名称: NET-OA网络办公系统 版本：3.0 2003.08.29
*
* 版权所有: Net-OA.com
*
* 程序设计: 江南行客-procedures* Name : NET- OA Network Office System Version : 3.0 2003.08.29** Copyright : Net-OA.com** Design : Jiangnan Customers
- 2022-08-22 17:20:09下载
- 积分：1
-
解决以下问题：
For solving the following problem:
"There is No Free Lunch"
Time Limit: 1 Second Memory Limit: 32768 KB
One day, CYJJ found an interesting piece of commercial from newspaper: the Cyber-restaurant was offering a kind of "Lunch Special" which was said that one could "buy one get two for free". That is, if you buy one of the dishes on their menu, denoted by di with price pi , you may get the two neighboring dishes di-1 and di+1 for free! If you pick up d1, then you may get d2 and the last one dn for free, and if you choose the last one dn, you may get dn-1 and d1 for free.
However, after investigation CYJJ realized that there was no free lunch at all. The price pi of the i-th dish was actually calculated by adding up twice the cost ci of the dish and half of the costs of the two "free" dishes. Now given all the prices on the menu, you are asked to help CYJJ find the cost of each of the dishes.
- 2022-12-06 21:00:03下载
- 积分：1
