sdkautooff_prog

这是c++的经典例题，一共一百个，大家共同学习共同进步-classical c++ programs

快速排序与归并排序的比较-rapid sequencing and merging sort of comparison

最小覆盖的树型DP,解决最小覆盖的树型DP-Minimum tree cover DP, to solve the smallest tree-covered DP

制作avi文件的程序。-production avi file.

c#Windows窗体设计, Windows窗体基本操作，Windows窗体高级操作，特殊类型的窗体，菜单、工具栏的设计，其它。 -c# Windows Forms Designer, Windows Forms basic operation, Windows Forms advanced operators, special types of forms, menus, toolbars, design, other.

wince 5.0上基于personal java写的的虚拟魔方，可以自动拼图 需安装NSIcom的虑拟机 CrE-ME 之后直接双击jar文件即可。-wince 5.0 based on written personal java virtual Rubik

使用协议的OAuth2.0认证的桌面应用程序VKontakte等perevoditsya。根据用于授权API的文档通过在3个阶段：1.打开浏览器窗口，为用户的授权; 2.获得访问权限的用户的数据; 3.制备键访问令牌与VKontakte等API的工作。后经过这些步骤，并得到access_token可以查询VKontakte等API。为了简单起见，我结合所述第一和第二阶段在一个

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

JPEGXR JPEGXR JPEGXR-JPEGXR

旋转裁剪图片的源文件，实现图片任意角度的旋转裁剪功能-Crop rotation picture header file, to achieve picture at any angle of rotation cropping feature