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酒吧老板價格表vbkkljl個ipfgik皮帶 東風康帕斯飛 帕斯克哦佛i 日減肥假話騙人 歐冠歐賠健康的飛機
上傳時間： 2014-01-09
上傳用戶：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.
標簽： cross-platform language powerful provide
上傳時間： 2014-01-07
上傳用戶：pompey
IMAGE and VIDEO COMPRESSION for MULTIMEDIA ENGINEERING Fundamentals, Algorithms, and Standards,by Yun Q. Shi
標簽： Fundamentals COMPRESSION ENGINEERING Algorithms
上傳時間： 2016-12-28
上傳用戶：myworkpost
89C51 單片機 I/O 口模擬串行通信的實現方法
上傳時間： 2016-12-28
上傳用戶：youke111
詳細介紹了LPC2132的I/O實驗中蜂鳴器實驗的原程序以及引腳的設置等
上傳時間： 2013-12-22
上傳用戶：z754970244
包含三個CC2430的I/O測試程序。程序基于IAR軟件編寫，用來控制片上LED的開關及閃爍。通過適當修改同樣可以用于CC1110的開發，估計對開發無線傳感器網絡的相關人士比較重要。
上傳時間： 2014-08-30
上傳用戶：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
標簽： distribution version AVRcam source
上傳時間： 2016-12-30
上傳用戶：GavinNeko
Euler函數: m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n Euler函數： 定義：phi(m) 表示小于等于m并且與m互質的正整數的個數。 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 則有 a^phi(m) = 1 (mod m) 即a^phi(m) - 1 整出m 在實際代碼中可以用類似素數篩法求出 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：定義phi(p) 為比p小的與p互素的數的個數 設n的素因子有p1, p2, p3, … pk 包含p1, p2…的個數為n/p1, n/p2… 包含p1*p2, p2*p3…的個數為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)
上傳時間： 2014-01-10
上傳用戶：wkchong
//Euler 函數前n項和 /* phi(n) 為n的Euler原函數 if( (n/p) ％ i == 0 ) phi(n)=phi(n/p)*i else phi(n)=phi(n/p)*(i-1) 對于約數：divnum 如果i|pr[j] 那么 divnum[i*pr[j]]=divsum[i]/(e[i]+1)*(e[i]+2) //最小素因子次數加1 否則 divnum[i*pr[j]]=divnum[i]*divnum[pr[j]] //滿足積性函數條件 對于素因子的冪次 e[i] 如果i|pr[j] e[i*pr[j]]=e[i]+1 //最小素因子次數加1 否則 e[i*pr[j]]=1 //pr[j]為1次 對于本題： 1. 篩素數的時候首先會判斷i是否是素數。 根據定義，當 x 是素數時 phi[x] = x-1 因此這里我們可以直接寫上 phi[i] = i-1 2. 接著我們會看prime[j]是否是i的約數 如果是，那么根據上述推導，我們有：phi[ i * prime[j] ] = phi[i] * prime[j] 否則 phi[ i * prime[j] ] = phi[i] * (prime[j]-1) (其實這里prime[j]-1就是phi[prime[j]]，利用了歐拉函數的積性) 經過以上改良，在篩完素數后，我們就計算出了phi[]的所有值。 我們求出phi[]的前綴和 */
上傳時間： 2016-12-31
上傳用戶：gyq
（一） 求a~b 之間各個數的約數個數之和。(其中包括a和b在內) ans = sigma(f(i)) , (a <= i <= b) , 其中f(i)表示i的約數的個數
上傳時間： 2016-12-31
上傳用戶：daoxiang126
