文件名称:rsa
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- 上传时间:2012-11-16
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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.
相关搜索: RSA euclidean
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rsa.cpp
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