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很好的c语言习题解答
很好的c语言习题解答-good language to answer the questions
- 2022-07-14 14:42:49下载
- 积分:1
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VB编写的计算器
VB编写的计算器-VB prepared by the Calculator
- 2022-02-07 08:12:19下载
- 积分:1
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djvu格式的电子书阅读器,解压缩后可以直接运行使用。
djvu格式的电子书阅读器,解压缩后可以直接运行使用。-DjVu format e-book reader, unzip it using direct running.
- 2022-03-02 19:17:19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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这是I2C总线的一个经典,能够与大家一起分享,希望大家的也多上传些...
这是I2C总线的一个经典,能够与大家一起分享,希望大家的也多上传些-This is the I2C bus is a classic, and everyone can share the hope that the U.S. also upload more
- 2023-09-04 16:40:03下载
- 积分:1
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VB编的一个计算器程序,考虑到绝对的计算器功能。
VB编的一个计算器程序,绝对考虑到了计算器的每个细微功能,由于是初学者,所以语法可能有点不成熟,但对于初学者的学习很有帮助!-VB series a calculator program that takes into account the absolute calculator function of every nuance, as is the starter So syntax may be a bit premature, but for beginners learning helpful!
- 2022-06-17 10:02:09下载
- 积分:1
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The shortest path solution Freudian [] Dev
最短路径求解Fuudie[] DEV-C++弗洛依德最小路径数组的方式为最短路径计算数组解决方案
- 2022-09-30 06:25:03下载
- 积分:1
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Achieved a convolution coding and Viterbi decoding, but also a simple BPSK modul...
实现了卷积编码和维特比译码,也有简单的BPSK调制解调过程。-Achieved a convolution coding and Viterbi decoding, but also a simple BPSK modulation and demodulation process.
- 2022-01-31 01:09:18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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一个自己作的FALSH导航式的文件,如果做的技术上有问题请多多包含...
一个自己作的FALSH导航式的文件,如果做的技术上有问题请多多包含-a FALSH themselves for the type of navigation, so if it is technically you will contain
- 2023-02-24 09:25:04下载
- 积分:1