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显示"石头"、"剪子"、"布",全靠运气，学习C++语言的小程序
显示"石头"、"剪子"、"布",全靠运气，学习C++语言的小程序-showed that the "stone", "scissors", "Bush," depends entirely on luck, the C language learning procedures for small
- 2022-03-12 21:27:46下载
- 积分：1
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tlv5636资料
tlv5636资料-tlv5636
- 2022-07-15 00:41:33下载
- 积分：1
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Beautiful Code is a useful book for programming.
代码之美，对于编程的学习非常有帮助。能够规范代码的编写，是一本值得一看的好书。-Beautiful Code is a useful book for programming.
- 2022-03-02 22:06:25下载
- 积分：1
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box frame _ _ combination of rolling bo x frame _ _ combination of rolling List...
列表框_组合框_滚动条
列表框_组合框_滚动条
-box frame _ _ combination of rolling bo x frame _ _ combination of rolling List Box _ _ rolling box combination of the list box frame _ _ combination of rolling box _ _ rolling portfolio of the box
- 2022-08-23 21:32:01下载
- 积分：1
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LEON3 GRlip SOC IP core. Memory controller.
LEON3 SOC GRlip IP core. Memory controller.-LEON3 GRlip SOC IP core. Memory controller.
- 2022-03-17 16:16:32下载
- 积分：1
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AutoCAD平台结构设计工具
按最新国家结构设计规范编写.
AutoCAD平台结构设计工具
按最新国家结构设计规范编写.-AutoCAD design tool based on the latest state standardized design prepared.
- 2023-04-26 14:05:04下载
- 积分：1
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介绍了NS2的详细安装过程，包括如何下载和安装的细节
介绍了NS2的详细安装过程，包括如何下载和安装的细节-Describes the details of NS2 installation process, including how to download and install the details of the
- 2022-01-26 02:28:43下载
- 积分：1
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收集来的，很不错的代码，一些小功能而已
收集来的，很不错的代码，一些小功能而已-collection of very good code, some small functions it
- 2022-07-02 16:07:59下载
- 积分：1
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商店存货系统，安装就能使用，功能很强大，很完整
商店存货系统，安装就能使用，功能很强大，很完整-OK
- 2022-06-28 19:10:05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
