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VHDL语言设计的出租车计费系统,他们的指导下完成…
VHDL语言设计的出租车计费系统,在老师的指导下完成的,并在实验箱上测试通过。-VHDL language design taxi billing system, completed under the guidance of their teachers, and the experiment on the test box.
- 2023-07-08 20:35:02下载
- 积分:1
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检测 10 种 C 语言编译环境的程序
检测 10 种 C 语言编译环境的程序-10 C language compiler environment procedures
- 2022-08-13 06:05:03下载
- 积分:1
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移动 t 检验技术
时间序列变异诊断,趋势分析
- 2022-03-18 12:42:58下载
- 积分:1
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数字图像矩阵数据的显示及其傅立叶变换等一些代码及例子
数字图像矩阵数据的显示及其傅立叶变换等一些代码及例子
-Digital image data display matrix and its Fourier transform and some code and examples
- 2022-03-05 10:33:51下载
- 积分:1
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equally wonderful, not to miss
同样很精彩,不要错过-equally wonderful, not to miss
- 2022-02-04 04:36:09下载
- 积分:1
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汇编下的环境建立和的运行,调用汇编,和汇编调用程序
汇编下的环境建立和的运行,调用汇编,和汇编调用程序-Compilation of environmental establishment and operation, call the compilation, and compile the procedure call
- 2022-03-24 19:52:51下载
- 积分:1
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extaspnet extaspnet extaspnet
ExtAspNet ExtAspNet ExtAspNet- ExtAspNetExtAspNetExtAspNet
- 2022-03-11 05:21:51下载
- 积分:1
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图像编码 如H263 264 AVS等编码都要用DCT变换,DCT变换IP核很有用...
图像编码 如H263 264 AVS等编码都要用DCT变换,DCT变换IP核很有用-Image coding, such as H263 264 AVS must be used, such as DCT transform coding, DCT transform IP core very useful
- 2022-09-20 09:55:03下载
- 积分:1
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Matrices, including matrix addition, subtraction, multiplication, there is inver...
矩阵计算,包括矩阵相加、相减,相乘,还有求逆、转置,可直接调用-Matrices, including matrix addition, subtraction, multiplication, there is inverse, transpose, can be directly invoked
- 2022-06-03 08:39:10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