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rsa加密过程中需要大数的幂运算后求模,本程序即计算a的b次方的结果对c求模,其中a,b,c均为大自然数,编写rsa加密的朋友可以作为参考(其实就是rsa的精华)-rsa encryption need computing power of large numbers after the seek mode, the procedure to calculate a result of the b-th power of c for mold, which a, b, c are the number of nature, the preparation of rsa encryption friends can serve as a reference ( in fact the essence of rsa)
Date : 2026-01-08 Size : 15kb User : showerxu1979

本代码能实现密码学领域中的大数幂模运算,并且运算效率较高,对进一步实现RSA加密、解密运算的实现有很大的帮助。-This code can achieve in the field of cryptography module computing power of large numbers, and operations more efficient, to further the realization of RSA encryption, decryption operations of great help to realize.
Date : 2026-01-08 Size : 189kb User : zhang

文件包括rsa算法原程序及详细注释。可以实现使用1024位以上大素数进行加解密。其中包括大整数的加、减、乘、除、模幂运算,求逆元运算,以及大素数的判定等算法。稍做扩展即可在实际中应用。-Rsa algorithm for the original documents, including procedures and detailed notes. 1024 can be achieved over the use of large prime numbers for encryption and decryption. This includes integer add, subtract, multiply, divide, modulus computing power, and inverse operations, as well as the large prime numbers, such as algorithms to determine. Slightly extended to the application in practice.
Date : 2026-01-08 Size : 149kb User : 易天行

利用大数库写的一个RSA模幂运算的小软件,该软件能够方便的进行RSA模幂运算,并且可以让读者学习大数库的使用方法。-Use of large numbers of a library to write a small RSA Modular Exponentiation software, the software can easily be RSA modular power, and large numbers can help readers learn to use the library.
Date : 2026-01-08 Size : 328kb User : 稀有品种

基于大数因子不可再分的RSA加密系统,因为这里只是用长整型来存储加密数据,如果素数都比较大,在乘方运算时肯定会造成内存溢出,因此这里仅仅只是验证RSA算法的思想的正确性-Can not be divided based on factors of large numbers of the RSA encryption system, because there is a long integer used to store encrypted data, if the prime was relatively large, the power of computing time will definitely lead to memory overflow, so here just thinking of authentication RSA algorithm correctness
Date : 2026-01-08 Size : 1.5mb User : 刘志伟
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