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遗传算法求函数最优解
遗传算法求解函数的最大值，且函数只有一个变量。求函数f(x)=x+10*sin(5x)+7*cos(4x)，x∈[0,9]。染色体长度为20，交叉概率为0.95，变异概率为0.08，迭代次数为25。 程序修改后可以用于多参数函数的求解。
- 2022-07-16 21:22:56下载
- 积分：1
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"工程师硬件使用手册"此书使你工作轻轻松松查询各种常用电子线路及电子元器件引脚符号!...
"工程师硬件使用手册"此书使你工作轻轻松松查询各种常用电子线路及电子元器件引脚符号!-"hardware engineers use manual," this book to make you work a light-hearted advice on various common electronic circuits and electronic devices yuan Symbol pieces pin!
- 2022-02-01 19:12:09下载
- 积分：1
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给人写的毕设，但没过
给人写的毕设，但没过-people wrote Albert located, but after
- 2022-05-29 09:21:08下载
- 积分：1
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用WindowsAPI访问INI文件
用WindowsAPI访问INI文件-with WindowsAPI visit INI file
- 2022-03-18 05:13:31下载
- 积分：1
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内存分配算法,用c实现内存的管理分配,属于操作系统范畴
内存分配算法,用c实现内存的管理分配,属于操作系统范畴-memory allocation algorithm, using c realization in the administration and distribution of memory, operating system areas
- 2022-05-31 21:56:07下载
- 积分：1
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POJ probably more difficult on 50 topics of the code, please share
大概POJ上50道比较难的题的代码，欢迎分享-POJ probably more difficult on 50 topics of the code, please share
- 2022-01-26 16:25:25下载
- 积分：1
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不错的particle filter的程序，c语言写的，适合对particle filer感兴趣的的初学者和编程人员。...
不错的particle filter的程序，c语言写的，适合对particle filer感兴趣的的初学者和编程人员。-Good particle filter procedure, c language, and suitable for the particle filer interested beginners and programmers.
- 2023-06-11 15:00:04下载
- 积分：1
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Win32 API 编写的
Win32 API 编写的-prepared by the Win32 API
- 2022-09-22 22:00:02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
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我自己做得小游戏
我自己做得小游戏-doing my own small game
- 2022-01-28 01:50:54下载
- 积分：1
