功能說明： 次組件提供的功能有二： 1、解析出數學表達式中的參數 2、計算數學表達式 此組件不但可以解析和計算數學表達式，還提供了一系列的數學基本函數，可應用于表達式中一塊處理。提供的數學基本函數有：將值舍入到最接近的整數 rount(x)、向下取整函數 int(x)、向上取整 ceiling(x)、求兩個數中最大數的函數 max(a,b)、求兩個數中最小數的函數 min(a,b)、冪函數 pow(x,n)、求平方根的函數 sqrt(x)。 下載地址：http://www.qiusuo365.com/qiusuo365/viewthread.php?tid=2422&extra=page％3D1
上傳時間： 2014-01-18
上傳用戶：李夢晗
一個很好用的 lcd 時鐘程序 C語言 #include<reg51.h> #include<stdio.h> //定義計時器0 的重裝值 #define RELOAD_HIGH 0x3C #define RELOAD_LOW 0xD2 //定義按鍵彈跳時間 #define DB_VAL //定義設置模式的最大時間間隔 #define TIMEOUT 200 //定義游標位置常數 #define HOME 0 #define HOUR 1 #define MIN 2 #define SEC 3
標簽： include define RELOAD stdio
上傳時間： 2014-12-19
上傳用戶：zukfu
Distribution generator Here is a simple generator which can build some distributions with given properties. Distributions generator (compile with -lm) Typical use might be: ./distributions -u -m 1 -M 10 -n 100 -s 500 Generates a distribution of 100 uniform random numbers between 1 and 10, such that the sum of numbers is 500. ./distributions -p -2.2 -m 1 -M 100 -n 200 -s 500 Idem with 200 numbers between 1 and 100 following a power law with exponent -
標簽： generator distributions Distribution simple
上傳時間： 2014-01-27
上傳用戶：sammi
C# BigInteger class. BigInteger.cs is a csharp program. It is the BIgInteger class. It has methods: abs() , FermatLittleTest(int confidence) ,gcd(BigInteger bi) , genCoPrime(int bits, Random rand) , genPseudoPrime(int bits, int confidence, Random rand) , genRandomBits(int bits, Random rand) , isProbablePrime(int confidence) , isProbablePrime() , Jacobi(BigInteger a, BigInteger b) , LucasSequence(BigInteger P, BigInteger Q, BigInteger k, BigInteger n) ,max(BigInteger bi) , min(BigInteger bi) , modInverse(BigInteger modulus) , RabinMillerTest(int confidence) ,
標簽： BigInteger class BIgInteger program
上傳時間： 2013-12-23
上傳用戶：ynzfm
Implement the step 2 of two-level logic minimization. Our goal is to find the minimum (exact minimum) sum-of-products expression for a given function.
標簽： minimization Implement the two-level
上傳時間： 2014-01-09
上傳用戶：無聊來刷下
We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.
標簽： represented integers group items
上傳時間： 2016-01-17
上傳用戶：jeffery
To date tests are still the best solution mankind has found to deliver working software. This book is the sum of four years of research and practice in the testing field. The practice comes from my IT consulting background, first at Octo Technology and then at Pivolis the research comes from my involvement with open source development at night and on weekends.
標簽： solution software mankind deliver
上傳時間： 2014-01-11
上傳用戶：sclyutian
分治法解決最近對問題 畫一條垂直線x=c，把這些給定點分為兩個包含n/2個點的子集S1和S2，使得n/2個點位于直線的左側或直線上，另外n/2個點位于直線的右側或直線上；遵循分治法的思想，遞歸地求出左子集S1和右子集S2中的最近對，分別為d1與d2；之后d=min{d1,d2}。合并過程：在以垂線x=c為對稱軸，2d為寬度的區域內求最近兩個點的距離，記為d3；求D=min{d,d3}；
上傳時間： 2013-12-26
上傳用戶：源碼3
MATLAB對11中調度算法進行比較，包括min-min、max-min、遺傳算法等等。
上傳時間： 2014-01-06
上傳用戶：lnnn30
假設在一個ad hoc網絡中，移動節點的發射功率PTx總是恒定的。要發送數據的移動節點總是先監聽信道，測量接收到的信號功率X，其中X= I + N, I為接收到的干擾，N是噪聲。移動節點只有在X＜INThre時，才可以發射。式中，INThre為背景噪聲門限。 在仿真中，我們規定每個移動節點的發射功率是常數，PTx = 1W；接收節點接收機的靈敏度Smin = -80 dBm；信號質量 min = 2 dB；系統的背景噪聲門限INThre = 1.2e-10。
上傳時間： 2016-03-16
上傳用戶：sevenbestfei
