The challenge ciphertext provided below is the result of encrypting a short secret ASCII plaintext using the RSA modulus given in the first factorization challenge. The encryption exponent used is e=65537. The ASCII plaintext was encoded using PKCS v1.5 before the RSA function was applied, as described in PKCS. Use the factorization you obtained for this RSA modulus to decrypt this challenge ciphertext and enter the resulting English plaintext in the box below. Recall that the factorization of N enables you to compute φ(N) from which you can obtain the RSA decryption exponent. Challenge ciphertext (as a decimal integer): 22096451867410381776306561134883418017410069787892831071731839143676135600120538004282329650473509424343946219751512256465839967942889460764542040581564748988013734864120452325229320176487916666402997509188729971690526083222067771600019329260870009579993724077458967773697817571267229951148662959627934791540 After you use the decryption exponent to decrypt the challenge ciphertext you will obtain a PKCS1 encoded plaintext. To undo the encoding it is best to write the decrypted value in hex. You will observe that the number starts with a '0x02' followed by many random non-zero digits. Look for the '0x00' separator and the digits following this separator are the ASCII letters of the plaintext. (note: the separator used here is '0x00', not '0xFF')

RSA加密解密字符串

python-xtea.tar.gz_The Chain_block_block chain_encryption_xtea

Python implementation of the XTEA Block Encryption Algorithm. Also includes a CBC decrypt function for decrypting Cipher Block Chain mode encrypted ciphertext.

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plaintext := PowerMod(ciphertext, d, n); printf("Original message: %s\n", message); printf("Encrypted message: %d\n", ciphertext); printf("Decrypted message: %s\n", convert(plaintext, 'string', ...

The private key d is used to decrypt ciphertexts created with the corresponding public key (it's also used to "sign" a message but we'll get to that later). The private key is the secret piece of information or "trapdoor" which allows us to quickly invert the encryption function. If RSA is implemented well, if you do not have the private key the fastest way to decrypt the ciphertext is to first factorise the modulus. In RSA the private key is the modular multiplicative inverse of the exponent e modulo the totient of N. Given the two primes: p = 857504083339712752489993810777 q = 1029224947942998075080348647219 and the exponent: e = 65537 What is the private key d? please givethe code

To find the private key d, we need to calculate the modular multiplicative inverse of e modulo the totient of N: N = p * q totient(N) = (p - 1) * (q - 1) d ≡ e^(-1) (mod totient(N)) Here's the ...

用JAVA编写将一个明文文件plaintext.txt中的内容，按照一定的方法，对每个字符加密后存放到另一个密文文件ciphertext.txt中。可以鼠标右键另存为下载明文文件： plaintext.txt 【输入】文件plaintext.txt。（该文件已经存在，无需自己创建）【输出】生成文件ciphertext.txt，里面存放加密后的信息。不需要在屏幕上显示信息。【输入示例】文件plaintext.txt，其中内容： Welcometo Java! 【输出示例】文件ciphertext.txt，其中内容： Ygneqog"vq"Lcxc#

这道题目需要使用JAVA编写程序，将明文文件plaintext.txt中的内容按照一定的方法进行加密，并将加密后的信息存放到另外一个密文文件ciphertext.txt中。程序需要完成以下几个步骤：步骤1：读取明文文件plaintext....

用JAVA编程将一个明文文件plaintext.txt中的内容，按照一定的方法，对每个字符加密后存放到另一个密文文件ciphertext.txt中。可以鼠标右键另存为下载明文文件： plaintext.txt 【输入】文件plaintext.txt。（该文件已经存在，无需自己创建）【输出】生成文件ciphertext.txt，里面存放加密后的信息。不需要在屏幕上显示信息。【输入示例】文件plaintext.txt，其中内容： Welcometo Java! 【输出示例】文件ciphertext.txt，其中内容： Ygneqog"vq"Lcxc#

其次，需要读取明文文件plaintext.txt中的内容，逐个字符进行加密，同时将加密后的字符存入新的文件ciphertext.txt中。以下是Java代码实现，仅供参考： java import java.io.BufferedReader; import java.io....

将一个明文文件plaintext.txt中的内容，按照一定的方法，对每个字符加密后存放到另一个密文文件ciphertext.txt中。可以鼠标右键另存为下载明文文件： plaintext.txt 【输入】文件plaintext.txt。（该文件已经存在，无需自己创建）【输出】生成文件ciphertext.txt，里面存放加密后的信息。不需要在屏幕上显示信息。【输入示例】文件plaintext.txt，其中内容： Welcometo Java! 【输出示例】文件ciphertext.txt，其中内容： Ygneqog"vq"Lcxc# 【提示】这里采用一种简单的加密方法，将每个字符的编码加2。需使用 Main 作为主类名。

这是一个加密程序，将明文文件plaintext.txt中的内容按照一定的方法进行加密，然后将加密后的结果存储到另一个密文文件ciphertext.txt中。输入是明文文件plaintext.txt，输出是密文文件ciphertext.txt，不需要在屏幕...

from tkinter import * from crypto.Cipher import DES def encrypt(): key = bytes(key_entry.get(), 'utf-8') cipher = DES.new(key, DES.MODE_ECB) plaintext = bytes(plaintext_entry.get(), 'utf-8') ciphertext = cipher.encrypt(plaintext) ciphertext_entry.delete(0, END) ciphertext_entry.insert(0, ciphertext.hex()) def decrypt(): key = bytes(key_entry.get(), 'utf-8') cipher = DES.new(key, DES.MODE_ECB) ciphertext = bytes.fromhex(ciphertext_entry.get()) plaintext = cipher.decrypt(ciphertext) plaintext_entry.delete(0, END) plaintext_entry.insert(0, plaintext.decode('utf-8')) root = Tk() plaintext_label = Label(root, text="Plaintext:") plaintext_label.grid(row=0, column=0) plaintext_entry = Entry(root) plaintext_entry.grid(row=0, column=1) key_label = Label(root, text="Key:") key_label.grid(row=1, column=0) key_entry = Entry(root) key_entry.grid(row=1, column=1) encrypt_button = Button(root, text="Encrypt",command=encrypt) encrypt_button.grid(row=2, column=0) decrypt_button = Button(root, text="Decrypt",command=decrypt) decrypt_button.grid(row=2, column=1) ciphertext_label = Label(root, text="Ciphertext:") ciphertext_label.grid(row=3, column=0) ciphertext_entry = Entry(root) ciphertext_entry.grid(row=3, column=1) root.mainloop()

这段代码存在一个问题：在函数 encrypt() 中，程序尝试使用一个名为 plaintext_entry 的变量，但是该变量在函数外部并没有定义，导致程序无法识别该变量，进而报错。同样的问题也出现在函数 decrypt() 中。 ...

解释代码：def encrypt_file(): file_path = file_path_entry.get() save_file_path = save_file_path_entry.get() key = key_entry.get() if file_path and save_file_path and key: with open(file_path, "r") as file: plaintext = file.read() ciphertext = vigenere_cipher(plaintext, key) with open(save_file_path, "w") as save_file: save_file.write(ciphertext) result_label.config(text="Encryption completed.") else: result_label.config(text="Please fill in all fields.")

这段代码定义了一个名为"encrypt_file"的函数，该函数用于加密一个文件。...如果所有输入框都被填写，则标签"result_label"将会显示"Encryption completed."，否则将会显示"Please fill in all fields."。

ciphertext should be 1021BF0420 key = 'Wiki' plaintext = 'pedia' ciphertext should be 45A01F645FC35B383552544B9BF5 key = 'Secret' plaintext = 'Attack at dawn' .3 根据RC4加密，当key='2022112623007'时，plaintext是多少

RC4加密算法中，plaintext和key进行异或运算生成密钥流，然后将密钥流与plaintext进行异或运算得到ciphertext。首先将key转换为ASCII码： 50 48 50 50 49 49 50 54 50 51 48 48 55 然后将plaintext转换为...

解释一干下，下面的代码 using System; using System.Windows.Forms; namespace VigenereCipher { public partial class MainForm : Form { public MainForm() { InitializeComponent(); } private void encryptButton_Click(object sender, EventArgs e) { string text = inputTextBox.Text; string key = keyTextBox.Text; string cipherText = Encrypt(text, key); outputTextBox.Text = cipherText; } private void decryptButton_Click(object sender, EventArgs e) { string text = inputTextBox.Text; string key = keyTextBox.Text; string plainText = Decrypt(text, key); outputTextBox.Text = plainText; } static string Encrypt(string text, string key) { string cipherText = ""; int keyIndex = 0; for (int i = 0; i < text.Length; i++) { char c = text[i]; int offset = key[keyIndex] - 'A'; char encryptedChar = (char)((c + offset) % 256); cipherText += encryptedChar; keyIndex = (keyIndex + 1) % key.Length; } return cipherText; } static string Decrypt(string text, string key) { string plainText = ""; int keyIndex = 0; for (int i = 0; i < text.Length; i++) { char c = text[i]; int offset = key[keyIndex] - 'A'; char decryptedChar = (char)((c - offset + 256) % 256); plainText += decryptedChar; keyIndex = (keyIndex + 1) % key.Length; } return plainText; } } }

Encrypt 和 Decrypt 方法都是静态方法，它们接受两个字符串参数：text 和 key，分别是要加密或解密的文本和密钥。在方法的实现中，程序会使用 Vigenere 密码算法对文本进行加密或解密，并返回结果。

解释一干下，下面的代码加密和解密算法 using System; using System.Windows.Forms; namespace VigenereCipher { public partial class MainForm : Form { public MainForm() { InitializeComponent(); } private void encryptButton_Click(object sender, EventArgs e) { string text = inputTextBox.Text; string key = keyTextBox.Text; string cipherText = Encrypt(text, key); outputTextBox.Text = cipherText; } private void decryptButton_Click(object sender, EventArgs e) { string text = inputTextBox.Text; string key = keyTextBox.Text; string plainText = Decrypt(text, key); outputTextBox.Text = plainText; } static string Encrypt(string text, string key) { string cipherText = ""; int keyIndex = 0; for (int i = 0; i < text.Length; i++) { char c = text[i]; int offset = key[keyIndex] - 'A'; char encryptedChar = (char)((c + offset) % 256); cipherText += encryptedChar; keyIndex = (keyIndex + 1) % key.Length; } return cipherText; } static string Decrypt(string text, string key) { string plainText = ""; int keyIndex = 0; for (int i = 0; i < text.Length; i++) { char c = text[i]; int offset = key[keyIndex] - 'A'; char decryptedChar = (char)((c - offset + 256) % 256); plainText += decryptedChar; keyIndex = (keyIndex + 1) % key.Length; } return plainText; } } }

具体地，程序会将密钥中当前字符的 ASCII 值减去字符 'A' 的 ASCII 值，得到一个偏移量 offset，然后将文本中当前字符的 ASCII 值加上这个偏移量，并对 256 取模，得到一个加密后的字符。最后将加密后的字符追加到 ...

为什么下面的代码解密按钮后没有输出到文本框 using System; using System.Windows.Forms; namespace VigenereCipher { public partial class MainForm : Form { public MainForm() { InitializeComponent(); } private void encryptButton_Click(object sender, EventArgs e) { string text = inputTextBox.Text; string key = keyTextBox.Text; string cipherText = Encrypt(text, key); outputTextBox.Text = cipherText; } private void decryptButton_Click(object sender, EventArgs e) { string text = inputTextBox.Text; string key = keyTextBox.Text; string plainText = Decrypt(text, key); outputTextBox.Text = plainText; } static string Encrypt(string text, string key) { string cipherText = ""; int keyIndex = 0; for (int i = 0; i < text.Length; i++) { char c = text[i]; int offset = key[keyIndex] - 'A'; char encryptedChar = (char)((c + offset) % 256); cipherText += encryptedChar; keyIndex = (keyIndex + 1) % key.Length; } return cipherText; } static string Decrypt(string text, string key) { string plainText = ""; int keyIndex = 0; for (int i = 0; i < text.Length; i++) { char c = text[i]; int offset = key[keyIndex] - 'A'; char decryptedChar = (char)((c - offset + 256) % 256); plainText += decryptedChar; keyIndex = (keyIndex + 1) % key.Length; } return plainText; } } }

这段代码实现了维吉尼亚密码的加密和解密功能，其中Encrypt函数实现了加密，Decrypt函数实现了解密。该程序界面中有三个文本框，分别是输入文本框、密钥文本框和输出文本框。当用户在输入文本框中输入明文并在密钥...

Suppose you are a cyptanalyst and you have intercepted a ciphertext c = 4 that has been sent to Alice. You find Alice’s RSA public key in a key directory: (e, n) = (5, 247). You assume this key has been used to produce the ciphertext. Find Alice’s private key. Show all your working.

To find Alice's private key, we need to use the RSA encryption and decryption formula: Encryption: c = m^e mod n Decryption: m = c^d mod n We know the ciphertext c = 4 and the public key (e, n) = (5...

from Crypto.Cipher import DES from binascii import unhexlify def pkcs7_decrypt(ciphertext, key): # 将密文解码为二进制数据 ciphertext = unhexlify(ciphertext) # 创建 DES 密码器，使用 ECB 模式 cipher = DES.new(key.encode(), DES.MODE_ECB) # 解密密文 plaintext = cipher.decrypt(ciphertext) # 去除 PKCS7 填充 padding_len = plaintext[-1] plaintext = plaintext[:-padding_len] # 将明文返回 return plaintext.decode() # 测试 ciphertext = "6d1e39d9ccaecc7a" key = "xuae5618" plaintext = pkcs7_decrypt(ciphertext, key) print(plaintext)

具体来说，该函数 pkcs7_decrypt 接受两个参数：密文 ciphertext 和密钥 key。它首先将密文从十六进制字符串解码为二进制数据，然后使用 ECB 模式创建一个 DES 密码器，并使用密钥对密文进行解密。接下来，...

修复下面代码：private void btnEncrypt_Click(object sender, EventArgs e) { try { // 获取密钥和向量 string key = txtKey.Text.Trim(); string iv = txtIV.Text.Trim(); byte[] keyBytes = Encoding.UTF8.GetBytes(key); byte[] ivBytes = Encoding.UTF8.GetBytes(iv); // 获取明文 string plaintext = txtPlaintext.Text.Trim(); byte[] plaintextBytes = Encoding.UTF8.GetBytes(plaintext); // 创建AES加密器 using (Aes aes = Aes.Create()) { aes.Key = keyBytes; aes.IV = ivBytes; // 创建加密流 using (MemoryStream msEncrypt = new MemoryStream()) { using (CryptoStream csEncrypt = new CryptoStream(msEncrypt, aes.CreateEncryptor(), CryptoStreamMode.Write)) { // 将明文写入加密流 csEncrypt.Write(plaintextBytes, 0, plaintextBytes.Length); csEncrypt.FlushFinalBlock(); // 获取加密结果 byte[] ciphertextBytes = msEncrypt.ToArray(); string ciphertext = Convert.ToBase64String(ciphertextBytes); // 显示加密结果 txtCiphertext.Text = ciphertext; } } } } catch (Exception ex) { MessageBox.Show(ex.Message); } } private void btnDecrypt_Click(object sender, EventArgs e) { try { // 获取密钥和向量 string key = txtKey.Text.Trim(); string iv = txtIV.Text.Trim(); byte[] keyBytes = Encoding.UTF8.GetBytes(key); byte[] ivBytes = Encoding.UTF8.GetBytes(iv); // 获取密文 string ciphertext = txtCiphertext.Text.Trim(); byte[] ciphertextBytes = Convert.FromBase64String(ciphertext); // 创建AES解密器 using (Aes aes = Aes.Create()) { aes.Key = keyBytes; aes.IV = ivBytes; // 创建解密流 using (MemoryStream msDecrypt = new MemoryStream(ciphertextBytes)) { using (CryptoStream csDecrypt = new CryptoStream(msDecrypt, aes.CreateDecryptor(), CryptoStreamMode.Read)) { // 读取解密结果 byte[] plaintextBytes = new byte[ciphertextBytes.Length]; int bytesRead = csDecrypt.Read(plaintextBytes, 0, plaintextBytes.Length); // 显示解密结果 string plaintext = Encoding.UTF8.GetString(plaintextBytes, 0, bytesRead); txtPlaintext.Text = plaintext; } } } } catch (Exception ex) { MessageBox.Show(ex.Message); } }

string plaintext = txtPlaintext.Text.Trim(); byte[] plaintextBytes = Encoding.UTF8.GetBytes(plaintext); // 创建AES加密器 using (Aes aes = Aes.Create()) { aes.Key = keyBytes; aes.IV = ivBytes; ...

详细讲解以下代码import java.util.Scanner; public class Encrypt { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); System.out.print("请输入要加密的字符串："); String plaintext = scanner.nextLine(); StringBuilder ciphertext = new StringBuilder(); for (int i = 0; i < plaintext.length(); i++) { char c = plaintext.charAt(i); if (c >= 'a' && c < 'z' || c >= 'A' && c < 'Z') { c++; } else if (c == 'z') { c = 'a'; } else if (c == 'Z') { c = 'A'; } ciphertext.append(c); } System.out.println("加密后的字符串为：" + ciphertext); } }

12. ciphertext.append(c); 这行代码将加密后的字符c添加到ciphertext中。 13. System.out.println("加密后的字符串为：" + ciphertext); 这行代码输出加密后的字符串。总的来说，这个程序将输入的字符串...

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ciphertext should be 1021BF0420 key = 'Wiki' plaintext = 'pedia' ciphertext should be 45A01F645FC35B383552544B9BF5 key = 'Secret' plaintext = 'Attack at dawn' .3 根据RC4加密，当key='2022112623007'时，plaintext是多少

Suppose you are a cyptanalyst and you have intercepted a ciphertext c = 4 that has been sent to Alice. You find Alice’s RSA public key in a key directory: (e, n) = (5, 247). You assume this key has been used to produce the ciphertext. Find Alice’s private key. Show all your working.

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