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带灯<b>rj45</b>规格书

  • MDQ80-12产品规格书

    本文由电子工程师林工精心起草并负责解释

    标签: MDQ 80 12 产品规格书

    上传时间: 2015-01-03

    上传用户:哈哈haha

  • MDQ100A产品规格书(中文)

    本文由电子工程师林工精心起草并负责解释

    标签: 100A MDQ 100 产品规格书

    上传时间: 2013-11-16

    上传用户:aa7821634

  • TM1628规格书

    单片机LED按键接口扩展集成电路。减少单片机IO口

    标签: 1628 TM 规格书

    上传时间: 2015-01-03

    上传用户:dljwq

  • THC63LVDM83D THine LVDS芯片datasheet

    THC63LVDM83D THine LVDS芯片datasheet 规格书

    标签: datasheet THine LVDM LVDS

    上传时间: 2015-01-03

    上传用户:xiaojie

  • 智能SD卡座规范

    中国银联主导智能SD卡座接口规范,卡座规格书。智能SD卡又名SIMT卡或NFC TF卡、NFC Micro SD卡,支持NFC近场通信,可应用于一卡通手机、一卡通手机挂件,是国家银联主导推行的便捷支付标准。

    标签: SD卡

    上传时间: 2013-10-18

    上传用户:dbs012280

  • 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

    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。

    标签: gt 90 运算符 嵌套

    上传时间: 2015-01-08

    上传用户:lifangyuan12

  • 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

  • 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

    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).

    标签: lt 源代码 动态规划 序列

    上传时间: 2013-12-26

    上传用户:siguazgb

  • The government of a small but important country has decided that the alphabet needs to be streamline

    The government of a small but important country has decided that the alphabet needs to be streamlined and reordered. Uppercase letters will be eliminated. They will issue a royal decree in the form of a String of B and A characters. The first character in the decree specifies whether a must come ( B )Before b in the new alphabet or ( A )After b . The second character determines the relative placement of b and c , etc. So, for example, "BAA" means that a must come Before b , b must come After c , and c must come After d . Any letters beyond these requirements are to be excluded, so if the decree specifies k comparisons then the new alphabet will contain the first k+1 lowercase letters of the current alphabet. Create a class Alphabet that contains the method choices that takes the decree as input and returns the number of possible new alphabets that conform to the decree. If more than 1,000,000,000 are possible, return -1. Definition

    标签: government streamline important alphabet

    上传时间: 2015-06-09

    上传用户:weixiao99