-
AML-Efficient-Approximate-Membership
AML EFFICIENT APPROXIMATE MEMBERSHIP LOCALIZATION WITHIN A WEB-BASED JOIN FRAMEWORK ABSTRACT: In this paper, we propose a new type of Dictionary-based Entity Recognition Problem, named Approximate Membership Localization (AML). The popular Approximate Membership Extraction (AME) provides a full coverage to the true matched substrings from a given document, but many redundancies cause a low efficiency of the AME process and deteriorate the performance of real-world applications using the extracted substrings. The AML problem targets at locating non overlapped substrings which is a better approximation to the true matched substrings without generating overlapped redundancies. In order to perform AML efficiently, we propose the optimized algorithm P-Prune that prunes a large part of overlapped redundant matched substrings before generating them.
- 2013-12-15 16:45:53下载
- 积分：1
-
罗宾
素数Robin―Millor测试法的C语言源码-Robin-Millor prime test of the C language source
- 2022-12-02 15:55:03下载
- 积分：1
-
c语言编写可对文件加密
c语言编写可对文件加密-c language of the document can be encrypted
- 2022-05-20 20:11:13下载
- 积分：1
-
Base64Coding源码
Base64Coding源码-Base64Coding FOSS
- 2023-03-08 15:00:03下载
- 积分：1
-
MD5生成器，可以生成字符串或文件的MD5值
MD5生成器，可以生成字符串或文件的MD5值-MD5 Generator can generate the MD5 string or document the value of
- 2023-04-30 22:10:03下载
- 积分：1
-
AES Crypto PIC Code Example
AES Crypto PIC Code Example
- 2022-03-04 05:57:43下载
- 积分：1
-
文件中有说明,本程序主要提供了MD5算法的功能实现!
文件中有说明,本程序主要提供了MD5算法的功能实现!-Document has indicate that this procedure provides the main function of MD5 algorithm realize!
- 2022-03-13 13:36:05下载
- 积分：1
-
RSA的类的源代码，Visual C++ 6.0实现。该算法（不是指代码）虽然有诸多缺点，但目前应用广泛。该代码的思路清晰，容易学习掌握。...
RSA的类的源代码，Visual C++ 6.0实现。该算法（不是指代码）虽然有诸多缺点，但目前应用广泛。该代码的思路清晰，容易学习掌握。-RSA class
- 2022-01-24 15:25:01下载
- 积分：1
-
这是用C编写的命令加密法算法，请大虾们指教。
这是用C编写的命令加密法算法，请大虾们指教。-This is prepared by using C commands encryption algorithm, please advise their shrimp.
- 2022-04-13 02:33:36下载
- 积分：1
-
ecc
ecc加密算法，通过c语言实现。希望有帮助(ecc c++)
- 2020-12-21 19:19:08下载
- 积分：1
