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...

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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