文件名称:rsa
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1) 找出两个相异的大素数P和Q，令N＝P×Q，M＝（P－1）（Q－1）。
2) 找出与M互素的大数E，用欧氏算法计算出大数D，使D×E≡1 MOD M。
3) 丢弃P和Q，公开E，D和N。E和N即加密密钥，D和N即解密密钥。
-1) to identify two different large prime numbers P and Q, so N = P × Q, M = (P-1) (Q-1). 2) to identify and M large numbers coprime E, Euclidean algorithm used to calculate lump sum for the D, so that D × E ≡ 1 MOD M. 3) disposed of P and Q, the open E, D and N. E and N that encryption keys, D and N that the decryption key.
2) 找出与M互素的大数E，用欧氏算法计算出大数D，使D×E≡1 MOD M。
3) 丢弃P和Q，公开E，D和N。E和N即加密密钥，D和N即解密密钥。
-1) to identify two different large prime numbers P and Q, so N = P × Q, M = (P-1) (Q-1). 2) to identify and M large numbers coprime E, Euclidean algorithm used to calculate lump sum for the D, so that D × E ≡ 1 MOD M. 3) disposed of P and Q, the open E, D and N. E and N that encryption keys, D and N that the decryption key.
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rsa.cpp