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[Crack Hack] VC_RSA DL : 0: 一、RSA基本原理对明文分组M和密文分组C，加密与解密过程如下： C = POW (M , e) mod n M = POW(C , d) mod n = POW(POW( M ,e), d) mod n=POW( M,e*d) 其中POW是指数函数，mod是求余数函数。其中收发双方均已知n，发送放已知e，只有接受方已知d，因此公钥加密算法的公钥为 KU={ e , n},私钥为KR={d , n}。该算法要能用做公钥加密，必须满足下列条件： 1. 可以找到e ，d和n，使得对所有M<n ,POW(M ,e*d)=M mod n . 2. 对所有 M<n,计算POW (M , e)和POW(C , d)是比较容易的。 3. 由e 和n确定d是不可行的 -one, the basic tenets of RSA expressly group M and cipher block C, encryption and decryption process is as follows : C = POW (M, e) mod n = M POW (C, d) mod n = POW (POW (M, e), d) mod n = POW (M, e * d), which is an exponential function POW, mod is the pursuit of the remaining functions. Transceivers which both known n, send e Fang known, the only known recipient d, therefore the public key encryption algorithm for public key e KU = (n), private key for KR = (d, n). The algorithm could be used to be a public key encryption, must meet the following conditions : 1. E can be found, and d n, making all the right M
Date : 2008-10-13 Size : 5.39kb User : 烟翔
[Crack Hack] VC_RSA DL : 0: 一、RSA基本原理对明文分组M和密文分组C，加密与解密过程如下： C = POW (M , e) mod n M = POW(C , d) mod n = POW(POW( M ,e), d) mod n=POW( M,e*d) 其中POW是指数函数，mod是求余数函数。其中收发双方均已知n，发送放已知e，只有接受方已知d，因此公钥加密算法的公钥为 KU={ e , n},私钥为KR={d , n}。该算法要能用做公钥加密，必须满足下列条件： 1. 可以找到e ，d和n，使得对所有M<n ,POW(M ,e*d)=M mod n . 2. 对所有 M<n,计算POW (M , e)和POW(C , d)是比较容易的。 3. 由e 和n确定d是不可行的 -one, the basic tenets of RSA expressly group M and cipher block C, encryption and decryption process is as follows : C = POW (M, e) mod n = M POW (C, d) mod n = POW (POW (M, e), d) mod n = POW (M, e* d), which is an exponential function POW, mod is the pursuit of the remaining functions. Transceivers which both known n, send e Fang known, the only known recipient d, therefore the public key encryption algorithm for public key e KU = (n), private key for KR = (d, n). The algorithm could be used to be a public key encryption, must meet the following conditions : 1. E can be found, and d n, making all the right M
Date : 2025-12-27 Size : 1.87mb User :
[Crack Hack] 11 DL : 0: 完成基于公钥证书的非对称密钥分配工作说明：（1）作为通讯的双方A和B，都有一个共同信赖的第三方CA，由CA为每个人生成公钥证书和相应的私钥并分发给A、B。（2）A和B分别将自己的公钥证书提供给对方后，分别在本地验证对方公钥证书是否可靠（假定A、B已经获得CA的公钥），然后A用B的公钥加密一个数据X（整型，可自己定义，比如100）并发送给B，B用自己的私钥解密后得到数据X，然后按协商好的函数F（比如F(X)=2*X）计算得到F(X)，再利用A的公钥加密后发送给A；A在本地用自己的私钥解密后得到F(x)，在同本地计算得到的F(X)进行对比，如果一致，说明公钥分配成功。（3）A用B的公钥对一图象文件（test_pig.bmp)进行加密并发送给B，B收到后解密，对比原图象看是否一致，并记录整个加密/解密的所用时间T1；-Complete the asymmetric public key certificate-based key distribution work Description: (1) as the communication of both A and B, have a common trusted third party CA, the CA generates a public key certificate for each person and the corresponding private key and distributed to A, B. (2) A and B respectively to provide their own public key certificates to each other both in the local verify the reliability of each other s public key certificate (assuming A, B has received CA s public key), then A with B s public key encryption 1 the data X (integer, can be their own definition, such as 100) and send it to B, B with their own private key decrypted by the data X, and then consult a good function by F (eg F (X) = 2* X) calculated F (X), then encrypted using A s public key sent to A A locally with their own private key to decrypt obtained F (x), calculated with the local F (X) are compared, if the same description Successful public-key distribution. (3) A with B s public key on an
Date : 2025-12-27 Size : 70kb User : 熊龙生
[Crack Hack] Practical-digital-signature DL : 0: 实用数字签名的步骤（1）发送方使用MD5算法对通信内容进行计算，获得一个固定长度的信息摘要；（2）发送方用自己的私钥加密生成的信息生成发送方的数字签名，发送方把这个数字签名作为发送信息的附件和明文信息，一同用接收方的公钥进行加密，将加密后的密文一同发送给接收方；（3）接收方首先把接收到的密文用自己的私钥解密，得到明文信息和数字签名，再用发送方的公钥对数字签名进行解密，然后使用相同的单向散列算法来计算解密得到的明文信息，得到信息摘要；对比计算出来的信息摘要和发送方发送过来的信息摘要是否一致思考：（1）请分别选择文件test-1.txt和test_picture.bmp作为通讯内容，分别记录全过程用时，并对比一下；（2）如果所发送的内容数据量非常大，上述方案中的效率不能满足要求，请考虑一下如何解决这个问题。-Practical steps digital signature ( A ) the sender uses the MD5 algorithm to calculate the content of communications , access to a fixed-length message digest ( 2 ) the sender s private key encryption with their own information generated by the sender generates a digital signature , the digital signature of the sender of the message as an attachment and send plaintext , together with the recipient s public key to encrypt the encrypted sent along with the ciphertext to the recipient ( 3 ) First, the recipient receives the ciphertext with his private key to decrypt , get the plaintext and the digital signature , then the sender s public key to decrypt the digital signature , and then using the same one-way hash algorithm to calculate decrypted plaintext information, to receive message digest comparing the calculated message digest and the information sent from the sender summary is consistent Reflections : ( 1 ) Please select the file test-1.txt respectively and test_pictur
Date : 2025-12-27 Size : 1.32mb User : 丁元元