本人写的浙大acm题目的源码，贴出来供大家分享，希望对大家的编程水平能有所帮助...

存储过程sql： create or replace procedure job_proc is begin --免费票状态修改 需修改ID update O_2413 set attr_24131004="作废" where attr_24131004="已售"; --普通票状态修改 需修改ID update O_2405 set attr_24051004="作废" where attr_24051004="已售"; --团体票状态修改 需修改ID update O_2412 set attr_24121004="作废" where attr_24121004="已售"; end; job的sql： declare   job number;   begin   dbms_job.submit(job, "job_proc;", sysdate, "TRUNC(SYSDATE + 1)");   end;

科研管理系统是用PoerBuilder编写-research management system is used to prepare PoerBuilder

一个出勤的数据库系统，很实用的，欢迎大家下在-An attendance database system, it is practical, welcomed everyone in

宾馆客房管理系统的源码-Hotels Rooms source code management system

一个很好的员工信息管理系统-a good staff information management system

1.[问题描述] 编写递归算法，计算二叉树中叶子结点的数目 [输入] 按照先序序列的顺序输入该结点的内容。其输入abd eh cf i g . [输出] 按中序序列输出，输出的结果为；dbheaficg并计算出二叉树中叶子结点的数目为4 [存储结构] 采用二叉表存储 [算法的基本思想] 采用递归方法建立和遍历二叉树。首先建立二叉树的根结点，然后建立其左右子树，直到空子树为止，中序遍历二叉树时，先遍厉左子树，后遍厉右子树，最后访问根结点。根据左右子树的最后一个结点计算出二叉树中叶子结点的数目。 程序如下： #include #include #include"stdlib.h" -1. [Description of the issue] to prepare recursive algorithm, Binary calculation leaves the number of nodes [imported] in accordance with the first order input sequence in the order of the node contents. Input abd eh i g cf. [Output] by the order sequence output, the results of the output; dbheaficg calculated Binary leaf node to the number four [storage structure] Table 2 forks storage [ The basic idea algorithm] recursive method and traverse binary tree. First established binary tree root node, and then to build their son around the tree, the tree until the loopholes, which preorder binary tree, Li first times the left sub-tree, right af

学生管理系统功能的信息量大，数据安全性和保密性要求最高。本程序可实现这些功能。1)输出学生成绩单2)按姓名查询3)增加学生4)删除学生5)按班级修改学生成绩6)显示不及格学生成绩7)按平均成绩排序并输出成绩

这是数据结构中模拟栈的一个精典例子利用火车进站出站模拟出栈和入栈-This is the data structure of a simulated stack classic examples of using the train station stop simulated stack and pushed

在数据结构中实现两个一元多项式的相加、相减和相乘-In the data structure of the polynomial realize one of the two together, subtract and multiply

约瑟夫问题有原始的约瑟夫和发展了的约瑟夫两种。 原始的约瑟夫的描述： 古代某法官要判决N个犯人的死刑，他有一条荒唐的法律，将犯人站成一个圆圈，从第S个人开始数起，每数到第D个犯人，就拉出来处决，然后再数D个，数到的人再处决―――直到剩下的最后一个可赦免。 发展的约瑟夫的描述： 古代某法官要判决N个犯人的死刑，但这N个人每人持有一个密码，他有一条荒唐的法律，将犯人站成一个圆圈，法官先给出一个密码M，从第S个人开始数起，每数到第M个犯人，就拉出来处决，再根据这个人所持有的密码F，然后再数F个，数到的人再处决，以此类推―――直到剩下的最后一个可赦免。-Joseph problems have original development of the Joseph and Joseph two. The original Joseph Description : Ancient a judge to sentence N inmates of the death penalty, he is a ridiculous law, prisoners station into a circle, from S began a few individuals, every few months to D prisoners, pull out executions, then D, the number of people to be re-executed--- until the last one left to be pardoned. The development of the Joseph Description : Ancient a judge to sentence N inmates of the death penalty, but each individual N holders a password, he is a ridiculous law, prisoners station into a circle, the judges first gave a password M, S individuals from the beginning of several, each of the first few inmates M , pull hi