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单片机<b>语言</b>

  • 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

  • 超星格式的电子书

    超星格式的电子书,找了很久,不错,对AVR感兴趣的看看吧中文书名《嵌入式参考书\AVR单片机BASIC语言编程及开发_0》

    标签: 超星 电子书

    上传时间: 2013-12-09

    上传用户:zhaiye

  • 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

    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

    标签: 电力系统 计算 运行

    上传时间: 2015-06-13

    上传用户:it男一枚

  • st7565p液晶驱动的液晶模块

    st7565p液晶驱动的液晶模块,lm6038d,单片机c语言开发串口驱动例子,方便移植!

    标签: 7565p 7565 st 液晶驱动

    上传时间: 2015-09-09

    上传用户:familiarsmile

  • 能配合“变压器测试组件”实现LF2000系列各个型号变压器组件与互感器的测试任务要抗干扰能力强

    能配合“变压器测试组件”实现LF2000系列各个型号变压器组件与互感器的测试任务要抗干扰能力强,操作方便,软件升级维护方便可复位保护功能,pic单片机c语言程序

    标签: 2000 LF 变压器测试 变压器

    上传时间: 2014-01-06

    上传用户:lindor

  • 为了进一步理解上述方法如何在编程中得以实现

    为了进一步理解上述方法如何在编程中得以实现,在此提供了1个用C51单片机编程语言编制的8个按键的键处理程序,以供参考。该程序在KEIL C51 V6.02/uVsion2 demo编译环境下编译通过

    标签: 编程

    上传时间: 2013-12-14

    上传用户:jing911003

  • 简单的LCD多级菜单显示

    简单的LCD多级菜单显示,单片机C语言,在KEIL下运行。

    标签: LCD 多级 菜单

    上传时间: 2014-12-03

    上传用户:weixiao99

  • 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