#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,
上传时间： 2013-10-21
上传用户：13788529953
Keil C251设计软件V3.11(4k代码限制） keil c251 v3.11 demo版的安装说明：安装方法是先将V3.11安装程序用winzip解压缩到某个目录下。（不要直接点击setup.exe安装），比如解压缩到c:\c251然后执行c:\c251\setup\setup.exe 安装程序，这个程序会让你选择安装Eval Version版还是Full Version版，选择Eval Version版进行安装 安装好之后就可以使用，但有代码大小的限制。需要无限制的版本，请购买正版软件。 C251是intel公司的16位单片机，指令兼容51的单片机。8位单片机风靡了20年，下一个20年应该是16位单片机的世界。
上传时间： 2014-01-23
上传用户：yanqie
单片机原理及应用实验报告:实验1 WAVE软件的学习应用*.1实验2下载软件的学习应用*.2实验3运算指令的应用编程.3实验4 P0口输入、输出实验.4实验5数码管显示5实验6按键的识别*6实验7计算器设计实验*7实验8中断实验8实验9定时器/计数器实验9实验10串行口通讯实验.10实验11直流电机调速实验*.12实验12 IC卡读写实验*13实验13 TLC2543的应用实验*14实验14温度测量及控制实验*.15附录A 伟福仿真器系统概述16附录B STC-ISP-V3.1 界面23附录C 实验板.24
上传时间： 2013-11-13
上传用户：王小奇
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，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
/*** *** *** *** *** *** *** *** *** *** *** *** **/ //**此映射表用来映射LED模块不译码时，显示的字符和必须输入的数据的关系 //**每段和对应比特位的关系见示意图 // g // --- --- // b | a |f | | <---显示0时点亮的段为gfedcb // --- // c | |e | | 那么写入数据为0x7e // --- --- // d // bit: 7 6 5 4 3 2 1 0 // 段位： g f e d c b a
上传时间： 2013-11-25
上传用户：
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
LCD1602在ARM7上调试程序，开发环境为KEIL V3.0
上传时间： 2016-12-23
上传用户：luke5347
ARM7上音乐播放程序，开发环境为KEIL V3.0
上传时间： 2016-12-23
上传用户：thuyenvinh
ARM7串口程序，芯片为LPC2138，开发环境为KEIL V3.0
上传时间： 2016-12-23
上传用户：jhksyghr
