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A母<b>usb2</b>.0

  • by Randal L. Schwartz and Tom Phoenix ISBN 0-596-00132-0 Third Edition, published July 2001. (See

    by Randal L. Schwartz and Tom Phoenix ISBN 0-596-00132-0 Third Edition, published July 2001. (See the catalog page for this book.) the text of Learning Perl, 3rd Edition. Table of Contents Copyright Page Preface Chapter 1: Introduction Chapter 2: Scalar Data Chapter 3: Lists and Arrays Chapter 4: Subroutines Chapter 5: Hashes Chapter 6: I/O Basics Chapter 7: Concepts of Regular Expressions Chapter 8: More About Regular Expressions Chapter 9: Using Regular Expressions Chapter 10: More Control Structures Chapter 11: Filehandles and File Tests Chapter 12: Directory Operations Chapter 13: Manipulating Files and Directories Chapter 14: Process Management Chapter 15: Strings and Sorting Chapter 16: Simple Databases Chapter 17: Some Advanced Perl Techniques Appendix A: Exercise Answers Appendix B: Beyond the Llama Index Colophon

    标签: L. published Schwartz Edition

    上传时间: 2014-11-29

    上传用户:kr770906

  • by Randal L. Schwartz and Tom Phoenix ISBN 0-596-00132-0 Third Edition, published July 2001. (See

    by Randal L. Schwartz and Tom Phoenix ISBN 0-596-00132-0 Third Edition, published July 2001. (See the catalog page for this book.) Learning Perl, 3rd Edition. Table of Contents Copyright Page Preface Chapter 1: Introduction Chapter 2: Scalar Data Chapter 3: Lists and Arrays Chapter 4: Subroutines Chapter 5: Hashes Chapter 6: I/O Basics Chapter 7: Concepts of Regular Expressions Chapter 8: More About Regular Expressions Chapter 9: Using Regular Expressions Chapter 10: More Control Structures Chapter 11: Filehandles and File Tests Chapter 12: Directory Operations Chapter 13: Manipulating Files and Directories Chapter 14: Process Management Chapter 15: Strings and Sorting Chapter 16: Simple Databases Chapter 17: Some Advanced Perl Techniques Appendix A: Exercise Answers Appendix B: Beyond the Llama Index Colophon

    标签: L. published Schwartz Edition

    上传时间: 2015-09-03

    上传用户:lifangyuan12

  • We have a group of N items (represented by integers from 1 to N), and we know that there is some tot

    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

  • Implement the following integer methods: a) Method celsius returns the Celsius equivalent of a Fahr

    Implement the following integer methods: a) Method celsius returns the Celsius equivalent of a Fahrenheit calculation celsius = 5.0 / 9.0 * ( fahrenheit - 32 ) b) Method fahrenheit returns the Fahrenheit equivalent of a Celsius the calculation fahrenheit = 9.0 / 5.0 * celsius + 32 c) Use the methods from parts (a) and (b) to write an application either to enter a Fahrenheit temperature and display the Celsius or to enter a Celsius temperature and display the Fahrenheit equivalent.

    标签: equivalent Implement the following

    上传时间: 2014-01-19

    上传用户:jackgao

  • Thinking in Java, 3rd ed. Revision 4.0 Preface Introduction 1: Introduction to Objects 2

    Thinking in Java, 3rd ed. Revision 4.0 Preface Introduction 1: Introduction to Objects 2: Everything is an Object 3: Controlling Program Flow 4: Initialization & Cleanup 5: Hiding the Implementation 6: Reusing Classes 7: Polymorphism 8: Interfaces & Inner Classes 9: Error Handling with Exceptions 10: Detecting Types 11: Collections of Objects 12: The Java I/O System 13: Concurrency 14: Creating Windows & Applets 15: Discovering Problems 16: Analysis and Design A: Passing & Returning Objects B: Java Programming Guidelines C: Supplements D: Resources Index

    标签: Introduction Thinking Revision Preface

    上传时间: 2014-07-13

    上传用户:netwolf

  • 利用随机数产生1000个0~100之间的考试分数

    利用随机数产生1000个0~100之间的考试分数,将其存入一个文本文件中,然后程序从这个文件中读取这1000个考试分数,统计其中各分数段(A[90,100],B[80,90), C[70,80), D[60,70), F[0,60))的人数 、所占百分比,平均分。

    标签: 1000 100 随机数 分数

    上传时间: 2014-01-19

    上传用户:问题问题

  • AVR单片机数码管秒表显示

    #include<iom16v.h> #include<macros.h> #define uint unsigned int #define uchar unsigned char uint a,b,c,d=0; void delay(c) { for for(a=0;a<c;a++) for(b=0;b<12;b++); }; uchar tab[]={ 0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90,

    标签: AVR 单片机 数码管

    上传时间: 2013-10-21

    上传用户:13788529953

  • 可编程外围接口82C55A

    82C55A是高性能,工业标准,并行I/O的LSI外围芯片;提供24条I/O脚线。     在三种主要的操作方式下分组进行程序设计82C88A的几个特点:(1)与所有Intel系列微处理器兼容;(2)有较高的操作速度;(3)24条可编程I/O脚线;(4)底功耗的CHMOS;(5)与TTL兼容;(6)拥有控制字读回功能;(7)拥有直接置位/复位功能;(8)在所有I/O输出端口有2.5mA  DC驱动能力;(9)适应性强。方式0操作称为简单I/O操作,是指端口的信号线可工作在电平敏感输入方式或锁存输出。所以,须将控制寄存器设计为:控制寄存器中:D7=1; D6 D5=00;  D2=0。D7位为1代表一个有效的方式。通过对D4 D3 D1和D0的置位/复位来实现端口A及端口B是输入或输出。P56表2-1列出了操作方式0端口管脚功能。

    标签: 82C55A 可编程 外围接口

    上传时间: 2013-10-26

    上传用户:brilliantchen

  • RSA算法 :首先, 找出三个数, p, q, r, 其中 p, q 是两个相异的质数, r 是与 (p-1)(q-1) 互质的数...... p, q, r 这三个数便是 person_key

    RSA算法 :首先, 找出三个数, p, q, r, 其中 p, q 是两个相异的质数, r 是与 (p-1)(q-1) 互质的数...... p, q, r 这三个数便是 person_key,接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 这个 m 一定存在, 因为 r 与 (p-1)(q-1) 互质, 用辗转相除法就可以得到了..... 再来, 计算 n = pq....... m, n 这两个数便是 public_key ,编码过程是, 若资料为 a, 将其看成是一个大整数, 假设 a < n.... 如果 a >= n 的话, 就将 a 表成 s 进位 (s

    标签: person_key RSA 算法

    上传时间: 2013-12-14

    上传用户:zhuyibin

  • 数字运算

    数字运算,判断一个数是否接近素数 A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in base b is a Niven number if the sum of its digits divides its value. Given b (2 <= b <= 10) and a number in base b, determine whether it is a Niven number or not. Input Each line of input contains the base b, followed by a string of digits representing a positive integer in that base. There are no leading zeroes. The input is terminated by a line consisting of 0 alone. Output For each case, print "yes" on a line if the given number is a Niven number, and "no" otherwise. Sample Input 10 111 2 110 10 123 6 1000 8 2314 0 Sample Output yes yes no yes no

    标签: 数字 运算

    上传时间: 2015-05-21

    上传用户:daguda