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this is a game, has completed the hope that the exhibitions
这个是斗地主的游戏,还没有全部完成,希望多多指教-this is a game, has completed the hope that the exhibitions
- 2022-02-05 11:54:58下载
- 积分:1
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automatically remove documents automatically remove documents
自动清除文件 -automatically remove documents automatically remove documents
- 2022-07-21 10:34:21下载
- 积分:1
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can bus control usin pic assembly code
can bus control usin pic assembly code
- 2022-04-08 04:45:51下载
- 积分:1
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俄罗斯方块对战版VC源代码
俄罗斯方块对战版VC源代码-screen version of Tetris VC source code
- 2023-04-18 08:15:04下载
- 积分:1
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李刚 请赶快放开我的权限 我上载的源码都是很精湛的,请查阅,qing请加我 li_xue_ming@msn.com必要的话可以和我在线沟通...
李刚 请赶快放开我的权限 我上载的源码都是很精湛的,请查阅,qing请加我 li_xue_ming@msn.com必要的话可以和我在线沟通-Gang quickly open my competence, I uploaded the source are very exciting, accessible, qing requested increase, I li_xue_ming@msn.com necessary so I can communicate online
- 2022-03-25 10:19:49下载
- 积分:1
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Several good js calendar, personal recommendation from the Internet to find U.S.
几款不错js日历, 个人从网上找到的推荐给大家 -Several good js calendar, personal recommendation from the Internet to find U.S.
- 2022-03-18 15:58:32下载
- 积分:1
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用简单的语言来增强权威的行政长官与一个模块的源代码…
用易语言做的CE源码带提升权限模块用易语言做的CE源码带提升权限模块-With easy language to enhance the authority of the CE source code with a module for easy language to enhance the authority of the CE source code with a module
- 2023-06-16 09:05:03下载
- 积分:1
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一个物流公司的物流配送系统,比较简单,数据库为sqlserver,压缩包里的databak就是。...
一个物流公司的物流配送系统,比较简单,数据库为sqlserver,压缩包里的databak就是。-a logistics company"s logistics and distribution system, a relatively simple database for SQLServer, compressed bag databak is.
- 2022-03-19 03:25:52下载
- 积分:1
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英培特嵌入式开发系统IDE软件包,包括软件仿真功能,是ARM开发的好帮手...
英培特嵌入式开发系统IDE软件包,包括软件仿真功能,是ARM开发的好帮手-British tissue culture embedded system development IDE package, including software simulation function, the development of the ARM is a good helper
- 2022-10-02 18:50:03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