<I believe> song _verilog code for any device.
标签： believe verilog device code
上传时间： 2013-12-22
上传用户：wys0120
酒吧老板价格表vbkkljl个ipfgik皮带 东风康帕斯飞 帕斯克哦佛i 日减肥假话骗人 欧冠欧赔健康的飞机
上传时间： 2014-01-09
上传用户：WMC_geophy
Java is the first language to provide a cross-platform I/O library that is powerful enough to handle all these diverse tasks. Java is the first programming language with a modern, object-oriented approach to input and output. Java s I/O model is more powerful and more suited to real-world tasks than any other major language used today. Java I/O is the first and still the only book to fully expose the power and sophistication of this library.
标签： cross-platform language powerful provide
上传时间： 2014-01-07
上传用户：pompey
89C51 单片机 I/O 口模拟串行通信的实现方法
上传时间： 2016-12-28
上传用户：youke111
详细介绍了LPC2132的I/O实验中蜂鸣器实验的原程序以及引脚的设置等
上传时间： 2013-12-22
上传用户：z754970244
包含三个CC2430的I/O测试程序。程序基于IAR软件编写，用来控制片上LED的开关及闪烁。通过适当修改同样可以用于CC1110的开发，估计对开发无线传感器网络的相关人士比较重要。
上传时间： 2014-08-30
上传用户：hasan2015
The AVRcam source files were built using the WinAVR distribution (version 3.3.1 of GCC). I haven t tested other versions of GCC, but they should compile without too much difficulty. * The source files for the AVRcam had the author name and copyright information added back into them after the judging of the project, since it states in the competition rules that the author s name can not be present during their inspection. * The included source files are the ones that were submitted for the entry into the Circuit Cellar contest. I have continued to develop the AVRcam, and have added several new features (such as ignoring objects that aren t larger than a minimum size, removing tracked objects that overlap with each, and some general optimizations). If you are interested in the latest source, email me at john@jrobot.net * For more info about the AVRcam, check out http://www.jrobot.net John Orlando August 20, 2004
标签： distribution version AVRcam source
上传时间： 2016-12-30
上传用户：GavinNeko
Euler函数: m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n Euler函数： 定义：phi(m) 表示小于等于m并且与m互质的正整数的个数。 phi(m) = p1^(r1-1)*(p1-1) * p2^(r2-1)*(p2-1) * …… * pn^(rn-1)*(pn-1) = m*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pn) = p1^(r1-1)*p2^(r2-1)* …… * pn^(rn-1)*phi(p1*p2*……*pn) 定理：若(a , m) = 1 则有 a^phi(m) = 1 (mod m) 即a^phi(m) - 1 整出m 在实际代码中可以用类似素数筛法求出 for (i = 1 i < MAXN i++) phi[i] = i for (i = 2 i < MAXN i++) if (phi[i] == i) { for (j = i j < MAXN j += i) { phi[j] /= i phi[j] *= i - 1 } } 容斥原理：定义phi(p) 为比p小的与p互素的数的个数 设n的素因子有p1, p2, p3, … pk 包含p1, p2…的个数为n/p1, n/p2… 包含p1*p2, p2*p3…的个数为n/(p1*p2)… phi(n) = n - sigm_[i = 1](n/pi) + sigm_[i!=j](n/(pi*pj)) - …… +- n/(p1*p2……pk) = n*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pk)
上传时间： 2014-01-10
上传用户：wkchong
//Euler 函数前n项和 /* phi(n) 为n的Euler原函数 if( (n/p) ％ i == 0 ) phi(n)=phi(n/p)*i else phi(n)=phi(n/p)*(i-1) 对于约数：divnum 如果i|pr[j] 那么 divnum[i*pr[j]]=divsum[i]/(e[i]+1)*(e[i]+2) //最小素因子次数加1 否则 divnum[i*pr[j]]=divnum[i]*divnum[pr[j]] //满足积性函数条件 对于素因子的幂次 e[i] 如果i|pr[j] e[i*pr[j]]=e[i]+1 //最小素因子次数加1 否则 e[i*pr[j]]=1 //pr[j]为1次 对于本题： 1. 筛素数的时候首先会判断i是否是素数。 根据定义，当 x 是素数时 phi[x] = x-1 因此这里我们可以直接写上 phi[i] = i-1 2. 接着我们会看prime[j]是否是i的约数 如果是，那么根据上述推导，我们有：phi[ i * prime[j] ] = phi[i] * prime[j] 否则 phi[ i * prime[j] ] = phi[i] * (prime[j]-1) (其实这里prime[j]-1就是phi[prime[j]]，利用了欧拉函数的积性) 经过以上改良，在筛完素数后，我们就计算出了phi[]的所有值。 我们求出phi[]的前缀和 */
上传时间： 2016-12-31
上传用户：gyq
（一） 求a~b 之间各个数的约数个数之和。(其中包括a和b在内) ans = sigma(f(i)) , (a <= i <= b) , 其中f(i)表示i的约数的个数
上传时间： 2016-12-31
上传用户：daoxiang126
