-
this program is very good
数据结构中单链表的源代码,该程序可以实现单链表的插入,删除,查找等基本功能-this program is very good
- 2022-06-28 22:43:43下载
- 积分:1
-
本程序是在turbo c 下运行,实现多项式的相加,相减,相乘.由4个函数组成...
本程序是在turbo c 下运行,实现多项式的相加,相减,相乘.由4个函数组成-This procedure is in turbo c run, realize the sum of polynomials, subtract, multiply. 4 function by the composition of
- 2022-07-10 03:54:35下载
- 积分:1
-
student select course system,powerbuilder做的,很容易实现 ,不会有人上载了吧...
student select course system,powerbuilder做的,很容易实现 ,不会有人上载了吧-student select course system, powerbuilder done, it is easy to realize that no one on the right is contained
- 2022-02-27 05:46:13下载
- 积分:1
-
假设一个文件中出现了8种符号S0,SQ,S2,S3,S4,S5,S6,S7,那么每种符号要编码,至少需要3bit。假设编码成000,001, 010,011,1...
假设一个文件中出现了8种符号S0,SQ,S2,S3,S4,S5,S6,S7,那么每种符号要编码,至少需要3bit。假设编码成000,001, 010,011,100,101,110,111。那么符号序列S0S1S7S0S1S6S2S2S3S4S5S0S0S1编码后变成 000001111000001110010010011100101000000001,共用了42bit。我们发现S0,S1,S2这3个符号出现的频率比较大,其它符号出现的频率比较小,我们采用这样的编码方案:S0到S7的码辽分别01,11,101,0000,0001,0010,0011, 100,那么上述符号序列变成011110001110011101101000000010010010111,共用了39bit。尽管有些码字如 S3,S4,S5,S6变长了(由3位变成4位),但使用频繁的几个码字如S0,S1变短了,所以实现了压缩。对于上述的编码可能导致解码出现非单值性:比如说,如果S0的码字为01,S2的码字为011,那么当序列中出现011时,你不知道是S0的码字后面跟了个1,还是完整的一个S2的码字。因此,编码必须保证较短的编码决不能是较长编码的前缀。符合这种要求的编码称之为前缀编码。要构造符合这样的二进制编码体系,可以通过二叉树来实现。-Suppose a file appears in eight kinds of symbols S0, SQ, S2, S3, S4, S5, S6, S7, then each symbol to be encoded, at least 3bit. Suppose encoding 000,001, 010,011,100,101,110,111. Then the symbolic sequence S0S1S7S0S1S6S2S2S3S4S5S0S0S1 encoded into 000001111000001110010010011100101000000001, sharing a 42bit. We found that S0, S1, S2 these three symbols the frequency of relatively large, the other symbols the frequency is relatively smal
- 2022-04-27 21:17:34下载
- 积分:1
-
using Windows API, including how to control the menu display and hidden menu ite...
使用windowsapi,包括如何控制菜单的显示和隐藏菜单项
- 2022-09-10 09:40:02下载
- 积分:1
-
旅行商问题, 旅行商问题,旅行商问题。
旅行商问题, 旅行商问题,旅行商问题。-Traveling salesman problem, traveling salesman problem, traveling salesman problem.
- 2022-03-23 03:13:03下载
- 积分:1
-
问题描述:假设一个商店,有一个架子和一个仓库,当第…
问题描述: 假设一个商店,它有一个货架和一个仓库,当货架上的商品数量少于一定的数目时,从仓库运一定数量的商品摆到货架上,当仓库里的商品的数量少于一定的数目时,购买商品把仓库填满,商品的出售要按照商品的生产日期来,快要过期的商品要先出售。 解决问题的方法: 用了两个栈和 一个队列,把队列当作仓库,用一个栈作为货架,把另一个栈当作临时的存储箱。当要往货架上添商品时,先把作为货架的栈中的元素全都压到作为存储箱的栈中,再把仓库中的元素压到存储箱中,然后再把存储箱中的所有元素都压到货架上,这样,就能保证快要过期的商品先被出售。-Problem description : Suppose a shop, which has a shelf and a warehouse, when the merchandise on the shelves less than a certain quantity of the number, a certain number from the warehouses of goods on the shelves, when the warehouse volume of goods is less than certain number, purchase goods warehouse filled, the goods according to the sale of commodity production to date, is about to expire first sale of goods. The solution : a two stack and a queue, queue as a warehouse, used as a stack shelves, and another stack as a temporary storage bins. When to go to the shelves Tim commodity, first as a stack shelves of all elements of the pressure storage tank as the stack, then the warehouse down to the storage element box, and then s
- 2022-07-13 01:56:00下载
- 积分:1
-
模拟分页式存储管理中硬件的地址转换和用先进先出调度算法(FIFO)处理缺页中断...
模拟分页式存储管理中硬件的地址转换和用先进先出调度算法(FIFO)处理缺页中断-simulation tabbed storage management hardware address translation, and the use of FIFO scheduling algorithm (FIFO) interrupt handling missing pages
- 2022-01-26 06:59:50下载
- 积分:1
-
迷宫的又一解法。该算法简单、实用、可读性强,是学习递归和回溯很好的例子。
迷宫的又一解法。该算法简单、实用、可读性强,是学习递归和回溯很好的例子。-maze of another solution. The algorithm is simple, practical and readable study is recursive and backtracking good example.
- 2022-03-05 14:22:00下载
- 积分:1
-
对N皇后问题进行优化,可以在较短时间内计算出皇后排列方法的个数...
对N皇后问题进行优化,可以在较短时间内计算出皇后排列方法的个数-Optimized for N queens problem, you can calculate the Queen in a relatively short period of time means the number of ordered
- 2022-03-16 23:06:24下载
- 积分:1