文件名称:New-folder-(4)
- 所属分类：
- 加密解密
- 资源属性：
- [Windows] [Visual C] [Basic/ASP] [源码]
- 上传时间：
- 2015-04-29
- 文件大小：
- 9kb
- 下载次数：
- 0次
- 提 供 者：
- eng.yass********
- 相关连接：
- 无
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We begin with choosing two random large distinct primes p and q. We
also pick e, a random integer that is relatively prime to (p-1)*(q-1). The
random integer e is the encryption exponent. Let n = p*q. Using Euclid s
greatest common divisor algorithm, one can compute d, the decryption
exponent, such that:
e*d = 1 (mod (p-1)*(q-1))
Both plaintext m and ciphertext c should be in the set of nonnegative
integers. Furthermore, before encrypting a plaintext message m, we need
to make sure that 0 <= m < n. If m is greater than the modulus n, the result
c of the encryption will not be a unique one-to-one mapping m to c.
From one of the theorems of Euler s, we know that for all integers m,
med = m (mod n) -We begin with choosing two random large distinct primes p and q. We
also pick e, a random integer that is relatively prime to (p-1)*(q-1). The
random integer e is the encryption exponent. Let n = p*q. Using Euclid s
greatest common divisor algorithm, one can compute d, the decryption
exponent, such that:
e*d = 1 (mod (p-1)*(q-1))
Both plaintext m and ciphertext c should be in the set of nonnegative
integers. Furthermore, before encrypting a plaintext message m, we need
to make sure that 0 <= m < n. If m is greater than the modulus n, the result
c of the encryption will not be a unique one-to-one mapping m to c.
From one of the theorems of Euler s, we know that for all integers m,
med = m (mod n)
also pick e, a random integer that is relatively prime to (p-1)*(q-1). The
random integer e is the encryption exponent. Let n = p*q. Using Euclid s
greatest common divisor algorithm, one can compute d, the decryption
exponent, such that:
e*d = 1 (mod (p-1)*(q-1))
Both plaintext m and ciphertext c should be in the set of nonnegative
integers. Furthermore, before encrypting a plaintext message m, we need
to make sure that 0 <= m < n. If m is greater than the modulus n, the result
c of the encryption will not be a unique one-to-one mapping m to c.
From one of the theorems of Euler s, we know that for all integers m,
med = m (mod n) -We begin with choosing two random large distinct primes p and q. We
also pick e, a random integer that is relatively prime to (p-1)*(q-1). The
random integer e is the encryption exponent. Let n = p*q. Using Euclid s
greatest common divisor algorithm, one can compute d, the decryption
exponent, such that:
e*d = 1 (mod (p-1)*(q-1))
Both plaintext m and ciphertext c should be in the set of nonnegative
integers. Furthermore, before encrypting a plaintext message m, we need
to make sure that 0 <= m < n. If m is greater than the modulus n, the result
c of the encryption will not be a unique one-to-one mapping m to c.
From one of the theorems of Euler s, we know that for all integers m,
med = m (mod n)
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下载文件列表
RSA.frm
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RSA.bas