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yuyundonghui
参加运动会的 个学校编号为 。比赛分成 个男子项目和 个女子项目,项目编号分别为 和 。由于各项目参加人数差别较大,有些项目取前五名,得分顺序为7,5,3,2,1;还有些项目只取前三名,得分顺序为5,3,2。写一个统计程序产生各种成绩单和得分报表。
2、要求产生各学校的成绩单,内容包括各校所取得的每项成绩的项目号、名次(成绩)、姓名和得分;产生团体总分报表,内容包括校号、男子团体总分、女子团体总分和团体总分。
3、测试数据:对于 , , ,编号为奇数的项目取前五名,编号为偶数的项目取前三名,设计一组实例数据。
(The school serial number participating in Games is. Competition divides into men s event and women s event , the project serial number parts for the sum. The difference is bigger since every project participates in a number , some projects choose the first five , score order is 7 , 5 , 3 , 2 , 1 Still have some of projects taking the first three places , score only being 5 , 3 , 2 in proper order. Write a form for report counting procedure producing the various school report card and score. 2, demands the school report card producing every school , content includes every achievement project number , position in a name list (achievement) , full name and score per got by school Produce the group total points form for report , content including the school number , male person group total points , woman group total points and group total points. 3, testing data: Be that the odd number project chooses the first five to the serial number, the serial number is that the even number project c)
- 2009-04-07 22:40:37下载
- 积分:1
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LinkedStack
链栈 对于堆栈stack的链式实现 应用指针(linked stack)
- 2012-10-03 17:29:55下载
- 积分:1
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tudebianli
图的遍历问题 数据结构 C C++ 源码 课程设计
严蔚敏版的书(yrteryryer)
- 2010-05-08 09:24:06下载
- 积分:1
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yinhangmoni
这是一个简单的银行模拟系统
1、 客户的到达时间随机
2、 客户需要处理的业务随机(不同业务处理的平均长度不同,例如:取款时间较短,开户和销户时间较长)
3、 处理客户业务所需时间随机(在一定范围内)
4、 使用文本文件记录每个客户到达时间、业务处理时间、业务结束时间
5、 动态显示(刷新时间可调)目前每个窗口累计处理客户数量、等待人数,每个窗口的平均业务处理时间,不同业务类型业务的累计办理量。
(This is a simple bank simulation system
1 customer arrival time random
2, customers need to deal with the business of random (the average length of the different business processes, such as: the withdrawal shorter, longer to open an account and cancel the account)
3, the time required to deal with customer service random (within certain limits)
4, using a text file to record each customer arrival time, business processing time, the business end of time
5 dynamically display (refresh time is adjustable) each window is accumulated to handle the number of customers waiting for the number the average business processing time of each window, different type of business operations total for the amount.)
- 2012-11-22 00:08:18下载
- 积分:1
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pinfanfangwen
设有一头指针为L的带有表头结点的非循环双向链表,其每个结点中除有prev(前驱指针),data(数据)和next(后继指针)域外,还有一个访问频度域freq。在链表被起用前,其值均初始化为零。每当在链表中进行一次Locate(L,x)运算时,令元素值为x的结点中freq域的值增1,并使此链表中结点保持按访问频度非增(递减)的顺序排列,同时最近访问的结点排在频度相同的结点的最后,以便使频繁访问的结点总是靠近表头。试编写符合上述要求的Locate(L,x)运算的算法,该运算为函数过程,返回找到结点的地址,类型为指针型。(Has a head pointer is non-circular doubly linked list L with a header node, each node in addition to prev (precursor pointer), data (data) and next (successor pointers) outside, there is an access frequency degree domain freq. Before the list was hired, their values are initialized to zero. Whenever carried out in the list once Locate (L, x) when the operation, so that the element node x is the value in freq domain by one, and keep this list in the node-access frequency of non-increasing (decreasing) the order, while a recent visit to the nodes in the same row of the frequency of the last node, so that the nodes are frequently accessed tables are always near the head. Locate (L, x) algorithm is prepared in accordance with the above requirements of the test operation, the operation as a function of the process, return to find the address of the node of type pointer.)
- 2015-01-15 22:31:59下载
- 积分:1
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Monkeys-and-peach
Monkeys and peach,Monkeys and peach(Monkeys and peach)
- 2013-09-01 22:06:42下载
- 积分:1
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Calender
日历显示,查询1900年后的日历。每一屏显示一年的日历祥情。(Calender dispaling. Acess the calender detail of years after 1900.)
- 2011-12-18 14:57:15下载
- 积分:1
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suanfa
匈牙利算法以及指派问题介绍,kM算法和匈牙利算法的程序代码(Hungarian algorithm and the assignment problem introduced, kM algorithm and the Hungarian algorithm code)
- 2011-09-10 09:42:39下载
- 积分:1
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LR(0)
简单LR语法分析程序自动生成器的实现,可用作数据结构课程设计(LR (0) grammar analysis)
- 2013-09-22 09:26:25下载
- 积分:1
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1
说明: 可以实现判断二叉树是否为完全二叉树、判断两棵二叉树是否镜像对称、求一棵二叉树的一个最长枝等功能(Can determine whether the binary tree is complete binary tree to determine whether two binary mirror symmetry, one of the longest seeking a binary tree branches and other functions)
- 2013-11-13 21:16:53下载
- 积分:1