-
包括VC++时尚小游戏20个,包括象棋等小游戏的源代码
包括VC++时尚小游戏20个,包括象棋等小游戏的源代码
- 2022-10-10 14:00:03下载
- 积分:1
-
MALLOC调试工具(附源码
MALLOC调试工具(附源码-Malloc debugging tools (with source code
- 2023-01-31 19:40:03下载
- 积分:1
-
ocr
ocr-optical character recognition-it is a very useful souece code for optical chARACTER recognition
- 2022-01-26 07:53:56下载
- 积分:1
-
Prim算法完整实现代码
用Prim算法求无向图的最小生成树
Prim算法完整实现代码
用Prim算法求无向图的最小生成树
-Prim algorithm to achieve a complete code with Prim algorithm try to map the minimum spanning tree
- 2023-03-18 20:15:04下载
- 积分:1
-
极好的屏幕保护程序 有需要的可以下载 打雷下雨
极好的屏幕保护程序 有需要的可以下载 打雷下雨-Excellent screen saver needy can download thunder rain
- 2022-08-06 18:46:48下载
- 积分:1
-
他是一个非常好的娱乐小游戏,可以很好的打发你的寂寞时光,通过磊满一行消去后得分,直到磊满整个空间游戏结束...
他是一个非常好的娱乐小游戏,可以很好的打发你的寂寞时光,通过磊满一行消去后得分,直到磊满整个空间游戏结束-He is a very good fun little game, you can kill a good time for your loneliness, through the Lei Man and his entourage, after elimination score, until the lei over the entire space the game is over
- 2022-03-19 14:46:05下载
- 积分:1
-
鳄鱼论坛,自己在网上找的,测试了,数据库是mysql的,方便用户使用,有需要的下载吧...
鳄鱼论坛,自己在网上找的,测试了,数据库是mysql的,方便用户使用,有需要的下载吧-Crocodile Forum, the Internet looking for their own testing, and database is mysql, user-friendly to use, there is a need to download bar. .
- 2022-03-20 13:52:21下载
- 积分:1
-
打印以下图案: *
***
*****
*******
好郁闷人的作业,原来这么简单...
打印以下图案: *
***
*****
*******
好郁闷人的作业,原来这么简单-Print the following pattern:
*
***
*****
*******
- 2022-03-09 18:32:28下载
- 积分:1
-
内容
Readme.txt
BcbChess 中国象棋程序实例的源代码所在的文件夹
BcbExe 中国象棋程序实例的可执行程序所在的文件夹...
内容
Readme.txt
BcbChess 中国象棋程序实例的源代码所在的文件夹
BcbExe 中国象棋程序实例的可执行程序所在的文件夹
BcbTest 关于时间测试的项目所在的文件夹
-content Readme.txt BcbChess Chinese chess program is the source code where the text Annex folder BcbExe Chinese chess procedures examples of where the executable folder BcbTest test of time Examination of the project folder
- 2023-08-23 02:20:04下载
- 积分:1
-
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