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
ARM2410 step motor controlARM2410 step motor controlARM2410 step motor controlARM2410 step motor controlARM2410 step motor controlARM2410 step motor control
标签： controlARM motor 2410 step
上传时间： 2014-01-09
上传用户：songyue1991
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
常用芯片DIP SOT SOIC QFP电阻电容二极管等3D模型库 3D视图封装库 STEP后缀三维视图（154个）:050-9.STEP0805R.STEP1001-1.STEP1001-2.STEP1001-3.STEP1001-4.STEP1001-5.STEP1001-6.STEP1001-7.STEP1001-8.STEP103_1KV.STEP10X5JT.STEP1206R.STEP13PX2.STEP15PX2.STEP20P插针.STEP25V1000UF.STEP3296W.STEP35V2200UF.STEP3mmLED.STEP3mmLEDH.STEP3X3可调电阻.STEP400V0.1UF.STEP455.STEP630V0.1UF.STEP7805.STEP8P4R.STEPAXIAL-0.2-0.125W.STEPAXIAL-0.4-0.25W.STEPaxial-0.6-2W.STEPB-3528.STEPC-0805.STEPC06x18.STEPCAP-6032.STEPCH3.96 X2.STEPCH3.96-3P.STEPD-PAK.STEPDB25.STEPDC-30.STEPDIP14.STEPDIP16.STEPDIP6.STEPDIP8.STEPDO-214AA.STEPDO-214AB.STEPDO-214AC.STEPDO-41.STEPDO-41Z.STEPFMQ.STEPGNR14D.STEPH9700.STEPILI4981.STEPIN4007.STEPIN5408.STEPJP051-6P6C_02.STEPJQC-3F.STEPJS-1132-10.STEPJS-1132-11.STEPJS-1132-12.STEPJS-1132-13.STEPJS-1132-14.STEPJS-1132-15.STEPJS-1132-2.STEPJS-1132-3.STEPJS-1132-4.STEPJS-1132-5.STEPJS-1132-6.STEPJS-1132-7.STEPJS-1132-8.STEPJS-1132-9.STEPJS-1132R-2.STEPJS-1132R-3.STEPJS-1132R-4.STEPJS-1132R-5.STEPJS-1132R-6.STEPJS-1132R-7.STEPJS-1132R-8.STEPJZC-33F.STEPKBP210.STEPKE2108.STEPKF2510 X8.STEPKF301.STEPKF301x3.STEPKSD-9700.STEPLED5_BLUE.STEPLED5_GRE.STEPLED5_RED.STEPLED5_YEL.STEPLFCSP_WQ.STEPLQFP100.STEPLQFP48.STEPMC-146.STEPmolex-22-27-2021.STEPmolex-22-27-2031.STEPmolex-22-27-2041.STEPmolex-22-27-2051.STEPmolex-22-27-2061.STEPmolex-22-27-2071.STEPmolex-22-27-2081.STEPMSOP10.STEPMSOP8.STEPPA0630NOXOX-HA1.STEPPIN10.STEPPIN24.STEPPIN24A.STEPR 0805.STEPR0402.STEPR0603.STEPR0805.STEPR1206.STEPRA-15.STEPRA-20.STEPRS808.STEPSIP-3-3.96 22-27-2031.STEPSL-B.STEPSL-D.STEPSL-E.STEPSL-G.STEPSL-H.STEPSOD-123.STEPSOD-323.STEPSOD-523.STEPSOD-723.STEPSOD-80.STEPSOIC-8.STEPSOP-4.STEPSOP14.STEPSOP16.STEPSOP18.STEPSOT-89.STEPSOT223.STEPSOT23-3.STEPSOT23-5.STEPSSOP28.STEPTAJ-A.STEPTAJ-B.STEPTAJ-C.STEPTAJ-D.STEPTAJ-E.STEPTAJ-R.STEPTHB6064H.STEPTO-126.STEPTO-126X.STEPTO-220.STEPTO-247.STEPTO-252-3L.STEPTOSHIBA_11-4C1.STEPTSSOP-8.STEPTSSOP14-BOTTON.STEPTSSOP14.STEPTSSOP28.STEPUSB-A.STEPUSB-B.STEPWT.STEP
标签： 芯片 dip sot soic qfp 电阻 电容 二极管 封装
上传时间： 2021-11-21
上传用户：XuVshu
C++ Demystified: A Self-Teaching Guide by Jeff Kent ISBN:0072253703 McGraw-Hill/Osborne © 2004 This hands-on, step-by-step resource will guide you through each phase of C++ programming, providing you with the foundation to discover how computer programs and programming languages work.
标签： Self-Teaching Demystified McGraw-Hill 0072253703
上传时间： 2013-12-13
上传用户：1427796291
* 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： 2015-07-26
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： 2016-01-17
上传用户：jeffery
How the K-mean Cluster work Step 1. Begin with a decision the value of k = number of clusters Step 2. Put any initial partition that classifies the data into k clusters. You may assign the training samples randomly, or systematically as the following: Take the first k training sample as single-element clusters Assign each of the remaining (N-k) training sample to the cluster with the nearest centroid. After each assignment, recomputed the centroid of the gaining cluster. Step 3 . Take each sample in sequence and compute its distance from the centroid of each of the clusters. If a sample is not currently in the cluster with the closest centroid, switch this sample to that cluster and update the centroid of the cluster gaining the new sample and the cluster losing the sample. Step 4 . Repeat step 3 until convergence is achieved, that is until a pass through the training sample causes no new assignments.
标签： the decision clusters Cluster
上传时间： 2013-12-21
上传用户：gxmm
(1) 、用下述两条具体规则和规则形式实现.设大写字母表示魔王语言的词汇 小写字母表示人的语言词汇 希腊字母表示可以用大写字母或小写字母代换的变量.魔王语言可含人的词汇. (2) 、B→tAdA A→sae (3) 、将魔王语言B(ehnxgz)B解释成人的语言.每个字母对应下列的语言.
上传时间： 2013-12-30
上传用户：ayfeixiao
1.有三根杆子A,B,C。A杆上有若干碟子 2.每次移动一块碟子,小的只能叠在大的上面 3.把所有碟子从A杆全部移到C杆上 经过研究发现，汉诺塔的破解很简单，就是按照移动规则向一个方向移动金片： 如3阶汉诺塔的移动：A→C,A→B,C→B,A→C,B→A,B→C,A→C 此外，汉诺塔问题也是程序设计中的经典递归问题
上传时间： 2016-07-25
上传用户：gxrui1991
