a XOR b> a,然后a XOR b< b,and both a and b are dependent data
上传时间： 2014-01-26
上传用户：yxgi5
Signal Processing and Linear Systems,B.P. Lathi,Berkeley-Cambridge Press book matlab codes
标签： Berkeley-Cambridge Processing Systems Signal
上传时间： 2017-06-01
上传用户：541657925
~{JGR 8vQ IzWwR5SC5D2V?bD#DbO5M3~} ~{3v?b~} ~{Hk?b~} ~{2iQ/5H9&D\~} ~{?IRTWw@)3d~} ~{TZ~}JDK1.4.2~{OBM(9}~}
上传时间： 2015-02-21
上传用户：ommshaggar
b to b 模式 电子商务系统 ，c# 开发 , B/S结构
上传时间： 2014-01-20
上传用户：hanli8870
樣板 B 樹 ( B - tree ) 規則 : (1) 每個節點內元素個數在 [MIN,2*MIN] 之間, 但根節點元素個數為 [1,2*MIN] (2) 節點內元素由小排到大, 元素不重複 (3) 每個節點內的指標個數為元素個數加一 (4) 第 i 個指標所指向的子節點內的所有元素值皆小於父節點的第 i 個元素 (5) B 樹內的所有末端節點深度一樣
上传时间： 2017-05-13
上传用户：日光微澜
欧几里德算法：辗转求余 原理: gcd(a,b)=gcd(b,a mod b) 当b为0时,两数的最大公约数即为a getchar()会接受前一个scanf的回车符
上传时间： 2014-01-10
上传用户：2467478207
数据结构课程设计 数据结构B+树 B+ tree Library
上传时间： 2013-12-31
上传用户：semi1981
All inputs of the C16x family have Schmitt-Trigger input characteristics. These Schmitt-Triggers are intended to always provide proper internal low and high levels, even if anundefined voltage level (between TTL-VIL and TTL-VIH) is externally applied to the pin.The hysteresis of these inputs, however, is very small, and can not be properly used in anapplication to suppress signal noise, and to shape slow rising/falling input transitions.Thus, it must be taken care that rising/falling input signals pass the undefined area of theTTL-specification between VIL and VIH with a sufficient rise/fall time, as generally usualand specified for TTL components (e.g. 74LS series: gates 1V/us, clock inputs 20V/us).The effect of the implemented Schmitt-Trigger is that even if the input signal remains inthe undefined area, well defined low/high levels are generated internally. Note that allinput signals are evaluated at specific sample points (depending on the input and theperipheral function connected to it), at that signal transitions are detected if twoconsecutive samples show different levels. Thus, only the current level of an input signalat these sample points is relevant, that means, the necessary rise/fall times of the inputsignal is only dependant on the sample rate, that is the distance in time between twoconsecutive evaluation time points. If an input signal, for instance, is sampled throughsoftware every 10us, it is irrelevant, which input level would be seen between thesamples. Thus, it would be allowable for the signal to take 10us to pass through theundefined area. Due to the sample rate of 10us, it is assured that only one sample canoccur while the signal is within the undefined area, and no incorrect transition will bedetected. For inputs which are connected to a peripheral function, e.g. capture inputs, thesample rate is determined by the clock cycle of the peripheral unit. In the case of theCAPCOM unit this means a sample rate of 400ns @ 20MHz CPU clock. This requiresinput signals to pass through the undefined area within these 400ns in order to avoidmultiple capture events.For input signals, which do not provide the required rise/fall times, external circuitry mustbe used to shape the signal transitions.In the attached diagram, the effect of the sample rate is shown. The numbers 1 to 5 in thediagram represent possible sample points. Waveform a) shows the result if the inputsignal transition time through the undefined TTL-level area is less than the time distancebetween the sample points (sampling at 1, 2, 3, and 4). Waveform b) can be the result ifthe sampling is performed more than once within the undefined area (sampling at 1, 2, 5,3, and 4).Sample points:1. Evaluation of the signal clearly results in a low level2. Either a low or a high level can be sampled here. If low is sampled, no transition willbe detected. If the sample results in a high level, a transition is detected, and anappropriate action (e.g. capture) might take place.3. Evaluation here clearly results in a high level. If the previous sample 2) had alreadydetected a high, there is no change. If the previous sample 2) showed a low, atransition from low to high is detected now.
上传时间： 2013-10-23
上传用户：copu
* 高斯列主元素消去法求解矩阵方程AX=B,其中A是N*N的矩阵,B是N*M矩阵 * 输入: n----方阵A的行数 * a----矩阵A * m----矩阵B的列数 * b----矩阵B * 输出: det----矩阵A的行列式值 * a----A消元后的上三角矩阵 * b----矩阵方程的解X
上传时间： 2015-07-25
上传用户：xauthu
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.
标签： represented integers group items
上传时间： 2016-01-16
上传用户：jeffery
