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ADO address book management, is a VC+ access through the ADO connection, the dat...
ADO通讯录管理,是一个VC+access 通过ADO连接,操作数据库.-ADO address book management, is a VC+ access through the ADO connection, the database operation.
- 2022-07-10 15:01:38下载
- 积分:1
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本图为k160型万用表电路图,希望对大家有用
本图为k160型万用表电路图,希望对大家有用-the map k160- Multimeter circuit, and we hope to useful
- 2022-07-15 09:23:05下载
- 积分:1
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模仿港台《百万富翁》节目,具有很强的娱乐性和知识性,
模仿港台《百万富翁》节目,具有很强的娱乐性和知识性,-imitate Hong Kong and Taiwan "millionaire" program is highly entertaining and informative.
- 2022-11-21 23:30:03下载
- 积分:1
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Masters wrote classic second generation wavelet, we may as well take a look at,...
大师写的二代小波经典之作,大家不妨看看,也许有用-Masters wrote classic second generation wavelet, we may as well take a look at, perhaps useful
- 2022-05-21 04:06:15下载
- 积分:1
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完成双正交信号的数字通信系统的蒙特卡罗仿真。分析其在加性高斯白噪声下误码率与信噪比之间的关系,绘出误码率...
完成双正交信号的数字通信系统的蒙特卡罗仿真。分析其在加性高斯白噪声下误码率与信噪比之间的关系,绘出误码率-信噪比曲线图,并和理论曲线进行比较。-Complete pairs of orthogonal signals in Monte Carlo simulation of digital communication systems. Analysis of the additive white Gaussian noise bit error rate and signal to noise ratio under the relationship between the mapped BER- signal to noise ratio curve, and and the theoretical curves for comparison.
- 2022-08-22 03:43:30下载
- 积分:1
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A good game play is very appropriate because the 8 lines of code changed too lon...
一款不错的游戏 很有适玩8行 因为代码太长改了一部分 可以使用经过调试了-A good game play is very appropriate because the 8 lines of code changed too long after you can use the debugging
- 2022-11-20 02:25:03下载
- 积分:1
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coco2dx游戏引擎实现加减法练练看
项目使用coco2dx游戏引擎实现的加减法练练看,类似于水果练练看,通过完成加法和减法练习来达到消除的目的,游戏使用C++语言编写。
- 2023-07-18 14:15:07下载
- 积分:1
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基于ADS7846启动代码,用于医疗上面的一个设备.
基于ADS7846启动代码,用于医疗上面的一个设备.-ADS7846-based startup code for a medical equipment above.
- 2022-01-25 13:59:28下载
- 积分:1
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随机文件在视觉基础知识中的应用
Sample usage of Random Files in Visual Basics
- 2022-05-22 03:55:51下载
- 积分:1
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Find prime numbers prepared by screening the function: void sieve (bool isPrime...
编写筛选查找素数函数:
void sieve(bool isPrime[], int n)
其中isPrime[ ]为一个布尔型数组,n为数组大小。由于2是第一个素数,
所以设置isPrime[0]和isPrime[1]的值为false,并设置其余的元素初值为true。
然后对从4到n-1的每一个i,判断i是否能够被2整除,如果i能够被2整除,则设置isPrime[i]为false。
对从2到n/2的每一个可能的数值重复以上处理,当操作结束时,所有值为true的isPrime[i]所对应的i就是素数,它们将是从筛子中掉下来的。-Find prime numbers prepared by screening the function: void sieve (bool isPrime [], int n) which isPrime [] as a Boolean array, n is the size of the array. Because 2 is the first prime number, so set the isPrime [0], and isPrime [1] a value of false, and set the initial value of the remaining elements of true. And then from 4 to n-1 for each i, to determine whether i can be divisible by 2, if i can be divisible by 2, then set the isPrime [i] is false. From 2 to n/2 possible values for each one to repeat the above processing, when the operation ended, all the true value of the isPrime [i] corresponding to i is a prime number, they will be falling from the sieve in the past.
- 2023-03-02 04:45:03下载
- 积分:1