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[Console] clean DL : 1: 一个清除系统垃圾的bat程序，可以清楚大量的系统垃圾，特别是IE文件夹的垃圾和临时文件，这里面往往有病毒。
Date : 2008-10-13 Size : 1009byte User : zhaochundong
[Console] hist DL : 0: 计算如下图像bread.jpg的直方图，并将图像进行均衡化，比较均衡化前后的图像及直方图。直方图是统计在每一个灰度级上像素的个数，直方图均衡化是通过点运算使一幅输入图像转换为在每一灰度级上都有相同的象素点数的输出图像（即输出的直方图是平的）。公式如下：连续情况：离散的情况：其中表示输入图像的直方图，表示灰度等级，表示输入图像的面积。 -Calculated as follows bread.jpg image histogram, and a balanced image, and compare images before and after equalization and histogram. Histogram is a statistical gray level in each of the number of pixels, histogram equalization is to make a point of computing the input image is converted to gray level in each pixel have the same points of the output image (ie output histogram is flat). Formula is as follows: continuous case: discrete case: which states that the histogram of the input image, said gray levels, said the size of the input image.
Date : 2026-01-10 Size : 2.21mb User : outuba
[Console] clean DL : 0: 一个清除系统垃圾的bat程序，可以清楚大量的系统垃圾，特别是IE文件夹的垃圾和临时文件，这里面往往有病毒。-A clear system of garbage bat procedures, a large number of systems can clearly rubbish, especially IE folder garbage and temporary files, which tend to have the virus inside.
Date : 2026-01-10 Size : 1kb User : zhaochundong
[Console] DisableIEView DL : 0: 禁用IE[查看]菜单下[工具栏]中的子菜单\禁用IE[查看]菜单下[工具栏]中的子菜单-Disable IE [View] menu [Toolbar] in the sub-menus \ Disable IE [View] menu [Tool Bar] in the sub-menu
Date : 2026-01-10 Size : 2kb User : bobo
[Console] ferry DL : 0: 某汽车轮渡口，有n辆车要过河。n辆车只有两种要么是客车，要么是货车。已知过江渡船每次能载10辆车，从0分开始每10分钟来一次（即0分一辆，10分一辆，以此类推）。又知上渡船要遵守下述规定：若x分来了一辆渡船所有到大渡口时间<=x分且还没过河的车辆才可能上船，且客车先于货车上船，每上4辆客车必上1辆货车，这时若没有货车就上客车，若没有客车就上货车。试编写一个程序，模拟渡口的管理，统计客车和货车的平均等待时间。输入格式：第一行一个正整数n（1000 <= n <= 1000 000），表示要过江的车辆总数。第二行到n+1行每行两个非负整数type，time（0 <= time <= 1000 000）,表示time分到达渡口的车是type类型，type为0表示客车，为1表示货车，时间从0开始计，且按到达时间先后给出数据。输出格式：一行两个保留两位的小数（四舍五入）carWaitTime，trunkWaitTime以空格分开，carWaitTime表示客车的平均等待时间，trunkWaitTime表示货车的平均等待时间。-A car ferry port, there are n vehicles to cross the river. n is either the car only two buses, either truck. Known across the river ferry can carry 10 cars each, starting from 0 to once every 10 minutes (ie, 0 a, 10 a, and so on). They know on the ferry to comply with the following requirements: if x points to a car ferry to the large ferry all the time <= x min and it could not cross the river on board the vehicle, and passenger vehicles on board prior to each on the 4 buses will be on a truck, then if there is no vehicle on the bus, if there is no passenger on the truck. Try to write a program to simulate the management of the ferry, bus and truck statistical average waiting time. Input formats: The first line of a positive integer n (1000 <= n <= 1000 000), said the total number of vehicles to cross the river. The second line to the n 1 line per line two non-negative integer type, time (0 <= time <= 1000 000), said the time points to reach the ferry car is of ty
Date : 2026-01-10 Size : 2kb User : lijinping
[Console] cmobliphone DL : 0: C语言实现设计一个简易的手机通讯录管理系统。该题目要求在熟练掌握C语言的基本知识：数据类型（整形、实型、字符型、指针、数组、结构等）；运算类型（算术运算、逻辑运算、自增自减运算、赋值运算等）；程序结构（顺序结构、判断选择结构、循环结构）；大程序的功能分解方法（即函数的使用）等。进一步掌握各种函数的应用，包括时间函数、、绘图函数，以及文件的读写操作等。-Design a simple mobile phone address book management system. The subject is required to master the basic knowledge of C language: data types (plastic, real, character, pointers, arrays, structures, etc.) operator type (arithmetic, logical operators, increment decrement assignment operator) program structure (structure of the order, judgment to select a structure, loop structure) functional decomposition of large programs (ie, the function used). To further understand the application of a variety of functions, including a function of time, drawing functions, as well as file read and write operation.
Date : 2026-01-10 Size : 63kb User : alex
[Console] HuffmanTree DL : 0: 哈夫曼编/译码器利用哈夫曼编码进行信道通信可以大大提高信道利用率，缩短信息传输时间，降低传输成本。但是，这要求在发送端通过一个编码系统系统对待传数据预先编码，在接收端将传来的数据进行译码(复原)。对于双工信道(即可以双向传输信息的信道)，每端都需要一个完整的编/译码系统。试为这样的信息收发站写一个哈夫曼码的编/译码系统。要求：一个完整的系统应具有以下功能： (1)初始化(Initialization)。从终端读入字符集大小n，以及n个字符和权值, 建立哈夫曼树，并将它存于文件hfmTree中。 (2)编码(Encoding)。利用已建好的哈夫曼树(如不在内存，则从文件hfmTree中读入)对文件ToBeTrans中的正文进行编码，然后将结果存入文件CodeFile中。 (3)译码(Decoding)。利用已建好的哈夫曼树将文件CodeFile中的代码进行译码，结果存入文件TextFile中。- Huffman encoder/decoder The use of Huffman coding channel communication can greatly improve the channel utilization, time-to-information transmission, and reduce transmission costs. However, this requires the sending end through a coding system pre-treatment data encoding, decoding at the receiving end of the data (recovery). For duplex channel (ie two-way transmission channel), each side needs a complete encoder/decoder system. Write a Huffman encoder/decoder system test for such information transceiver station. Requirements: a complete system should have the following features: (1) initialization (Initialization). Read from the terminal character set size n, and n characters and weights, Create a Huffman tree, and save it in a file hfmTree. (2) encoding (encoding). Huffman tree has been built (if not in memory, read from the file hfmTree Into) the file ToBeTrans the text encoding, and the result is stored in the file CodeFile. (3) decoding
Date : 2026-01-10 Size : 248kb User : 丁洁琼
[Console] shiyan3 DL : 0: 海萍夫妇为了彻底告别“蜗居”生活，痛下决心贷款60万元购买了一套三居室。若贷款月利率为0.5 ，还款期限为120个月，还款方式为等额本金还款法（即贷款期限内每期以相等的额度偿还贷款本金，贷款利息随本金逐期递减）。试求出每个月还款的本金、每个月的利息以及总利息分别是多少元。若还款方式采用等额本息还款法（即贷款期限内每期以相等的额度偿还贷款本息，贷款利息随本金逐期递减）。试求出每个月还款的本金、每个月的利息以及总利息分别是多少元。-Haiping couple in order to bid farewell to " dwelling" life, faced up to 600,000 yuan loans purchased a three-bedroom. If the lending interest rate of 0.5 in January, the repayment period of 120 months, repayment is equal principal repayment method (ie, within the term of the loan in equal installments of principal repayment of the loan amount, loan interest of decreasing gradually with the principal ). Find the monthly repayment of principal, interest and total interest each month how many dollars respectively. If the repayment of principal and interest repayment method using equal (ie, within the term of the loan in equal installments of principal and interest repayment of the loan amount, loan interest with the principal by period descending). Find the monthly repayment of principal, interest and total interest each month how many dollars respectively.
Date : 2026-01-10 Size : 1kb User : 许之上
[Console] VC-find-all-ie-web-element DL : 0: VC++下的，实现遍历当前打开IE网页的所有元素，前在控制台上输出显示-vc.6 platform develop
Date : 2026-01-10 Size : 62kb User : xiaoliu
[Console] list DL : 0: 1）编一程序把A表B表重复的节点提取出来，形成一条新的单向递增链表D。 2）将A表和B表归并成一个新的递增有序不重复的有环单向链表C（即重复的数字只出现一次），要求合并后链表C的最后一个节点的下一个节点指向链表C中间一个节点, 形成平卧的 6 字形链表。连接结点为最接近链表C的结点数值平均值的那个结点。 3）判断一个链表是否存在环。输入：DataSource.txt 输出：链表C各个结点和链表D各个结点，并输出显示结点个数，平均值，是否有环，有环的情况下要输出交叉的节点-1) compiling a program to extract A Table B Table duplicate node out of the formation of a new one-way increments list D. 2) Table A and Table B will merge into a new ascending order not to repeat the list are unidirectional ring C (ie duplicate numbers appear only once), The next node in the list after the requirements of the merger last node C of C in the middle of a node point list form supine 6 shape list. Connecting the node to the node that node C is the value closest to the average of the list. 3) to determine whether there is a chain ring. Input: DataSource.txt Output: list of each node C and D list each node, and an output node number is displayed point average, Output node to the next whether there is cross ring, there is the case of the ring
Date : 2026-01-10 Size : 4.88mb User : 陈雨霞
[Console] Class-matrix DL : 0: 设计一个矩阵类CMatrix，该矩阵的行列数可根据需要设定（即不指定固定的行列数），并完成基本的计算功能。-Designing a matrix class CMatrix, the number of ranks of the matrix may need to be set (ie do not specify a fixed number of ranks), and perform basic computing functions.
Date : 2026-01-10 Size : 171kb User : HiJune
[Console] XMSAALib_demo DL : 0: 实时获取应用控件属性值，浏览器当前网页URL，只支持IE，谷歌，firefox- U5B9E u65F6 u83B7 u53D6 u5E94 u7528 u63A7 u4EF6 u5C5E u6027 u503C uFF0C u6D4F u89C8 u5668 u5F53 u524D u7F51 u9875URL uFF0C u53EA u652F u6301IE uFF0C u8C37 u6B4C uFF0Cfirefox
Date : 2026-01-10 Size : 2.89mb User : luo