RSA算法 :首先, 找出三个数, p, q, r, 其中 p, q 是两个相异的质数, r 是与 (p-1)(q-1) 互质的数...... p, q, r 这三个数便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 这个 m 一定存在, 因为 r 与 (p-1)(q-1) 互质, 用辗转相除法就可以得到了..... 再来, 计算 n = pq....... m, n 这两个数便是 public_key ，编码过程是, 若资料为 a, 将其看成是一个大整数, 假设 a < n.... 如果 a >= n 的话, 就将 a 表成 s 进位 (s <= n, 通常取 s = 2^t), 则每一位数均小于 n, 然後分段编码...... 接下来, 计算 b == a^m mod n, (0 <= b < n), b 就是编码后的资料。解码的过程是, 计算 c == b^r mod pq (0 <= c < pq), 于是, 解码完毕-RSA algorithm : First, find a few 3, p, q, r, p, q is the two different prime number, with the r (p-1) (q-1) coprime several ... p, q, r it person_key number is three, and then find m, making r ^ m == a mod (p-1) (q-1 m ...................................... this must exist, because r (p-1) (q- 1) coprime, with the division algorithm can be a ..... Next, calculate n = pq ....... m, n is the number two public_key, coding process is that if a data and to treat it as a big Integer assuming a <n. ... If a> = n, a table will be rounding into s (s <= n, usually from 2 ^ s = t), each n is less than the median, and then sub-coding ..... . Next, calculate a == b ^ m mod n, (0 <= b <n), b is encoded information. Decoding is the process of calculating c == b ^ r mod pq (0 <= c &