good morning,my dear teachers,my dear professors.i am very glad to be here for your interview.my name is song yonghao,i am 22 years old .i come from luoyang,a very beautiful aicent city.my undergratuade period will be accomplished in changan university in july ,2004 and now,i am trying my best for o
標簽： dear professors interview my
上傳時間： 2015-05-24
上傳用戶：shus521
這是很好用的利用JAVA語言編寫的程序，本來我也不懂 是朋友給我的，我看到了，很好 再就是我真的需要拿個程序 所以謝謝！
上傳時間： 2015-09-15
上傳用戶：王者A
I think this the first time every one can look at a PE crypter source in top level language such VC++. So as I promised ... if some one sent me one nice compress source I would publish my source. I dedicate this source to all people who involve in this field. I hope it helps someone. Have good days ashkbiz Check: yodap.cjb.net
標簽： language crypter source think
上傳時間： 2013-12-29
上傳用戶：dianxin61
#include <reg51.h> void delay_ms(unsigned short ms) { unsigned short i unsigned char j for(i=0 i<ms i++) { for(j=0 j<200 j++) for(j=0 j<102 j++) } }
標簽： unsigned short delay_ms include
上傳時間： 2016-03-30
上傳用戶：cuibaigao
Ex3-23 親兄弟問題 « 問題描述： 給定n 個整數0 1 1 , , , n- a a a 組成的序列。序列中元素i a 的親兄弟元素k a 定義為： min{ | } k i j n j j i a = a a ³ a < < 。 親兄弟問題要求給定序列中每個元素的親兄弟元素的位置。元素i a 的親兄弟元素為k a 時，稱k 為元素i a 的親兄弟元素的位置。當元素i a 沒有親兄弟元素時，約定其親兄弟元素 的位置為-1。 例如，當n=10，整數序列為6，1，4，3，6，2，4，7，3，5 時，相應的親兄弟元素位 置序列為：4，2，4，4，7，6，7，-1，9，-1。 « 編程任務： 對于給定的n個整數0 1 1 , , , n- a a a 組成的序列，試用抽象數據類型棧，設計一個O(n) 時間算法，計算相應的親兄弟元素位置序列。 « 數據輸入： 由文件input.txt提供輸入數據。文件的第1 行有1 個正整數n，表示給定給n個整數。 第2 行是0 1 1 , , , n- a a a 。 « 結果輸出: 程序運行結束時，將計算出的與給定序列相應的親兄弟元素位置序列輸出到output.txt 中。 輸入文件示例 輸出文件示例 input.txt 10 4 2 4 4 7 6 7 -1 9 -1 output.txt 6 1 4 3 6 2 4 7 3 5
上傳時間： 2013-12-17
上傳用戶：shizhanincc
好東西，圖像處理的 關于分割的還有識別的 全市好東西i全市看看吧
標簽：
上傳時間： 2013-12-21
上傳用戶：dongbaobao
This Handbook, published under the auspices of the Senior NATO Logisticians’ Conference (SNLC), i s i ntended as a simple guide to logistics i n NATO. It does not attempt to examine current i ssues or provide answers to the problems that logisticians will face, but i t rather aims at i ntroducing them to some of the basic principles, policies, concepts and organisations with which they will work.
標簽： Logisticians Conference published the
上傳時間： 2014-12-07
上傳用戶：www240697738
I often need a simple function generator. Just to generate a certain frequency. After all the years I ve worked with electronics, I still haven t got me one. Even though I need it now and then, I just couldn t seem to justify the cost of one. So, standard solution - build one yourself. I designed a simple sinewave generator based on a Analog Devices AD9832 chip. It will generate a sinewave from 0.005 to 12 MHz in 0.005 Hz steps. That s pretty good, and definitely good enough for me ! But while waiting for the AD9832 chip to arrive, I came up with a very simple version of the DDS synth, using just the 2313 and a resistor network.
標簽： frequency generator function generate
上傳時間： 2013-12-17
上傳用戶：thesk123
Good morning, dear teachers. I am very glad to be here for your interview. my name is xx.I am 21 years old. I come from Dafang, a small town of Guizhou province. My undergraduate period will be accomplished in East China Jiaotong University. I major in electrical engineering and automation. I am interesting in computer, especially in program design. I am a hard study student, especially in the things which I interesting in. I am a person with great perseverance. During the days I preparing for the postgraduate examination, I insist on study for more than 10 hours every day. Just owing to this, I could pass the first examination finally. I am also a person with great ambition.
標簽： interview teachers morning Good
上傳時間： 2014-01-11
上傳用戶：釣鰲牧馬
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
