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使用vc6.0开发的学生管理系统,包括对成绩的增删改查,并有管理员权限以及带柱型图和饼图的个人成绩统计和班级成绩统计(最高成绩、最低成绩、平均分等)。
管理...
使用vc6.0开发的学生管理系统,包括对成绩的增删改查,并有管理员权限以及带柱型图和饼图的个人成绩统计和班级成绩统计(最高成绩、最低成绩、平均分等)。
管理员用户名为space,密码为19871025。
演示程序时可打开文件夹中的050704new,里面有测试数据。-Vc6.0 students to develop the use of management systems, including additions and deletions to search results and have administrator privileges, as well as columnar with pie charts and statistics and personal accomplishments achieved statistical classes (maximum performance, minimum performance, with an average grade) . Administrator user name space, password is 19871025. Demo program can open the folder 050704new, there are test data.
- 2023-03-24 11:40:04下载
- 积分:1
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双向探针1.1 双向探针1.1
双向探针1.1 双向探针1.1 -two-way probe 1.1 1.1 bidirectional two-way probe probe 1.1-1.1 two-way probe to probe 1.1
- 2022-03-15 08:18:21下载
- 积分:1
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有关计算机机房监控的
有关计算机机房监控的-the computer room monitoring
- 2022-07-18 22:23:46下载
- 积分:1
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Simple Binary Tree Class
Simple Binary Tree Class
- 2022-04-07 15:13:54下载
- 积分:1
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程序目的:两个多项式相加
程序目的:两个多项式相加 -procedures Objective : add two polynomials
- 2022-05-16 09:23:28下载
- 积分:1
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使用VC6进行ADO编程
使用VC6进行ADO编程-use ADO programming for VC6
- 2022-07-11 11:43:43下载
- 积分:1
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使用vs2005编程
设计一种基于优先级和安全度的可抢占调度算法,其中安全度参数优于优先级参数,并由此判断给出任务组是否可调度。...
使用vs2005编程
设计一种基于优先级和安全度的可抢占调度算法,其中安全度参数优于优先级参数,并由此判断给出任务组是否可调度。-VS2005 Programming using a priority-based and safety may seize the scheduling algorithm, which is superior to the security parameter priority parameters, and thus given the task group to determine whether scheduling.
- 2023-03-09 18:50:03下载
- 积分:1
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在WINCE下TCP/IP网络开发程序源码
在WINCE下TCP/IP网络开发程序源码-WINCE under the TCP/IP network development process FOSS
- 2022-04-22 04:56:00下载
- 积分:1
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文本形式的字节流转binary格式的字节流
文本形式的字节流转binary格式的字节流-text byte transfer binary format byte stream
- 2022-02-15 01:08:43下载
- 积分:1
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there is a defect in only know that the secret and public key
还有一个缺陷就是在只知道密文 x 及公钥(n,e)的情况下,只要将 (x^e) mod n 所得余数 s 再不断地循环操作 s = s^e mod n,此运算不断地循环 e 次之后,很多情况下都可以循环出原文,只是计算量过余多一些罢了。不过有不少情况下,根本都无须循环 e 次,不过对于1024位的 n 级别来说,e 也是一个相当大的数值,所以循环密文的余数以解得原文是有些不现实。 以上内容仅供参考,如有不实,请予更正-there is a defect in only know that the secret and public key-x (n, e) the circumstances, as long as (x ^ e) mod n from the remaining s to continuously cycle operation s = s ^ e mod n, this constant cycle of Operational e occasion, the very many circumstances can be recycled from the original, but I calculated the volume more than just. There are, however, many instances, simply do not need e cycle times, but for 1024 the level n, e is a very large figure, so secret circle the remainder of the text was obtained in the original is a bit unrealistic. The above is for reference only, if not true, I corrected
- 2022-03-22 01:18:28下载
- 积分:1