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散斑资料
说明： 有关散斑资料及其matlab编程(Speckle data and MATLAB programming)
- 2020-05-01 17:02:29下载
- 积分：1
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数字信号处理算法。用于多采样率系统
数字信号处理算法。用于多采样率系统-digital signal processing algorithms. For more sampling rate system
- 2022-04-12 01:49:18下载
- 积分：1
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Fert2000
说明： 这是一个变电站的监控程序，数据库采用的是sql server7.0(This is a substation of the monitoring program, the database is used in sql server7.0)
- 2006-03-24 11:54:56下载
- 积分：1
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ELM
说明： 极限学习机的基础分类，从基础了解极限学习机的用途(The basic classification of extreme learning machine, understand the use of extreme learning machine from the basis)
- 2020-11-27 11:03:14下载
- 积分：1
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距离最短路由
用于多跳路由，寻找最小距离路径的一种算法。(An algorithm for multi-hop routing looking for the path with the smallest distance.)
- 2020-07-08 17:48:55下载
- 积分：1
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容积卡尔曼ckf
说明： 超宽带室内定位采用容积卡尔曼滤波算法，对初步了解超宽带定位的学生有很大帮助(UWB indoor positioning, using volume Kalman filter algorithm, is very helpful for students who have a preliminary understanding of UWB positioning)
- 2021-04-27 17:58:44下载
- 积分：1
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《电动机的DSP控制——TI公司DSP应用》——王晓明——配套光盘程序
《电动机的DSP控制——TI公司DSP应用》——王晓明——配套光盘程序，北航出版社出版
- 2019-04-19下载
- 积分：1
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基地的转换过程可以2
进制转换的程序，可由二至十的互换-base for the conversion process can be 2-10 fungible
- 2022-02-25 10:10:14下载
- 积分：1
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ac_dc_ac
说明： 将单相交流输入整流直流电，再经过三相逆变电路转变为三相交流电。(The single-phase AC input rectifier DC, and then through the three-phase inverter circuit into three-phase AC.)
- 2019-12-02 10:00:29下载
- 积分：1
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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
