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Paillier

于 2021-03-02 发布文件大小:4563KB

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下载积分: 1 下载次数: 29

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代码说明：

Paillier加密算法的实现。产生512位大数，进行大数运算。1、产生密钥2、加密、3、解密。(Paillier encryption algorithm implementation. Produce 512 Tarsus, Tarsus operation performed. 1, 2 generates a key, encrypts, 3, decryption.)

文件列表：

Paillier
........\Paillier
........\........\2800素数.h,18792,2015-06-15
........\........\BigNumber.h,13337,2015-06-15
........\........\Debug
........\........\.....\Pailliar.Build.CppClean.log,912,2015-06-15
........\........\.....\Pailliar.exe,1090048,2015-06-15
........\........\.....\Pailliar.ilk,2776540,2015-06-15
........\........\.....\Pailliar.log,1394,2015-06-15
........\........\.....\Pailliar.obj,486623,2015-06-15
........\........\.....\Pailliar.pdb,4926464,2015-06-15
........\........\.....\Pailliar.tlog
........\........\.....\.............\CL.read.1.tlog,6014,2015-06-15
........\........\.....\.............\CL.write.1.tlog,380,2015-06-15
........\........\.....\.............\link.read.1.tlog,2272,2015-06-15
........\........\.....\.............\link.write.1.tlog,404,2015-06-15
........\........\.....\.............\Pailliar.lastbuildstate,175,2015-06-15
........\........\.....\vc120.idb,273408,2015-06-15
........\........\.....\vc120.pdb,339968,2015-06-15
........\........\Pailliar.cpp,2294,2015-06-15
........\........\Pailliar.dsp,4432,2013-10-09
........\........\Pailliar.dsw,522,2013-10-09
........\........\Pailliar.ncb,58368,2015-03-17
........\........\Pailliar.opt,53760,2015-03-17
........\........\Pailliar.plg,1485,2015-03-17
........\........\Pailliar.sdf,7667712,2015-06-15
........\........\Pailliar.sln,961,2015-05-17
........\........\Pailliar.vcxproj,6394,2015-06-04
........\........\Pailliar.vcxproj.filters,1124,2015-06-04
........\README.txt,27,2015-06-15

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VMPC加密解密算法。已对x86优化，其他平台也可用。(VMPC encryption and decryption algorithms. Has been optimized for x86, other platforms are also available.)

2013-07-15 18:51:17下载

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2012-05-02 11:47:41下载

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2005-04-30 11:59:45下载

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39.3
说明：编制程序，将输入的一行字符以加密的形式输出，然后将其解密，解密的字符串序列与输入的正文相比较，吻合时输出解密的正文，否则解密失败。加密时，将每个字符的ascii码依次反复加上“4962873”中的数字，并在32（‘’）~122（‘z’）之间作模运算。解密与加密的顺序相反。例如输入正文“the result of 3 and 2 is not 8” 结果为： (Programming, the input line character output in an encrypted form, and then decrypted, the decrypted text sequence of characters compared with the input, output to decrypt the body of the agreement, otherwise the decryption failed. Encryption, ascii code of each character in turn repeatedly with &quot 4962873&quot in the figures, and 32 (' ' )~ 122 (' z' ) as a model between the operations. Decryption and encryption in reverse order. For example, enter the text &quot the result of 3 and 2 is not 8&quot results:)

2011-04-05 08:43:48下载

积分：1

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