题目：利用条件运算符的嵌套来完成此题：学习成绩>=90分的同学用A表示，60-89分之间的用B表示，60分以下的用C表示。 1.程序分析：(a>b)?a:b这是条件运算符的基本例子。
上传时间： 2015-01-08
上传用户：lifangyuan12
英文书籍，Introduction.to.Mobile.Telephone.Systems.1G.2G.2.5G.and.3G.Wireless.Technologies.and.Services
上传时间： 2015-02-25
上传用户：hoperingcong
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
Overview In this chapter I introduce Borland C++Builder (BCB) and explain what it is about. I also devote considerable time to explaining the purpose of this book and the philosophy behind my approach to technical writing. Technical subjects covered in this chapter include Creating a simple Multimedia RAD program that plays movies, WAV files, and MIDI files. Shutting down the BCB RAD programming tools and writing raw Windows API code instead. Creating components dynamically on the heap at runtime. Setting up event handlers (closures) dynamically at runtime. A brief introduction to using exceptions. This topic is covered in more depth in Chapter 5, "Exceptions." A brief introduction to ANSI strings. This subject is covered in more depth in Chapter 3, "C++Builder and the VCL." Using the online help. Greping through the include and source files that come with the product and with this book.
标签： introduce Overview Borland Builder
上传时间： 2014-01-04
上传用户：小鹏
aPLib is a 32-bit compression library based on the compression algorithm used in aPACK (my executable packer). aPLib is an easy-to-use alternative to many of the heavy-weight compression libraries available.
标签： compression algorithm executa library
上传时间： 2013-11-30
上传用户：一诺88
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
上传时间： 2015-05-21
上传用户：daguda
Chapter 1. Creational Patterns Chapter 2. Behavioral Patterns Chapter 3. Structural Patterns Chapter 4. System Patterns Chapter 5. Introduction to Java Programming Language Patterns Chapter 6. Java Core APIs Chapter 7. Distributed Technologies Chapter 8. Jini and J2EE Architectures
标签： Patterns Chapter Behavioral Creational
上传时间： 2014-02-02
上传用户：bakdesec
数字水印的嵌入和检测，dct， 作者Mao Li Purpose:an algorithm to achieve the embeding watermarking to a image by dct bloking and hvs,then detect the watermarking by comparability.
上传时间： 2015-06-06
上传用户：hullow
源代码\用动态规划算法计算序列关系个数 用关系"<"和"="将3个数a，b，c依次序排列时，有13种不同的序列关系： a=b=c,a=b<c,a<b=v,a<b<c,a<c<b a=c<b,b<a=c,b<a<c,b<c<a,b=c<a c<a=b,c<a<b,c<b<a 若要将n个数依序列，设计一个动态规划算法，计算出有多少种不同的序列关系， 要求算法只占用O(n),只耗时O(n*n).
上传时间： 2013-12-26
上传用户：siguazgb
电力系统在台稳定计算式电力系统不正常运行方式的一种计算。它的任务是已知电力系统某一正常运行状态和受到某种扰动，计算电力系统所有发电机能否同步运行 1运行说明： 请输入初始功率S0，形如a+bi 请输入无限大系统母线电压V0 请输入系统等值电抗矩阵B 矩阵B有以下元素组成的行矩阵 1正常运行时的系统直轴等值电抗Xd 2故障运行时的系统直轴等值电抗X d 3故障切除后的系统直轴等值电抗 请输入惯性时间常数Tj 请输入时段数N 请输入哪个时段发生故障Ni 请输入每时段间隔的时间dt
上传时间： 2015-06-13
上传用户：it男一枚
