mcq 队列_MCQ | 密码学中作为IDEA，DES，AES，RSA的块密码

mcq 队列

1) In the AES-128 algorithm there are mainly __________ similar rounds and _________ round is different from other round.

1)在AES-128算法中，主要有__________个相似的回合，而_________个回合与其他回合不同。

5 similar rounds having 2 pair ; every alternate
5个类似的回合有2对; 每隔一个
9 ; the last
9；最后
8 ; the first and last
8；第一个和最后一个
10 ; no
10；没有

Answer: b. 9; the last

答案： b。 9; 最后

Explanation:

说明：

Basically, in the AES-128 there are 9 rounds and the last round is different from other rounds.

基本上，在AES-128中有9个回合，最后一个回合与其他回合不同。

2) Which of the following modes of operation in DES is used for operating?

2)DES中以下哪种操作模式用于操作？

Cipher Feedback Mode (CFB)
密码反馈模式(CFB)
Cipher Block chaining (CBC)
密码块链接(CBC)
Electronic code book (ECB)
电子密码书(ECB)
Output Feedback Modes (OFB)
输出反馈模式(OFB)

Answer: c. Electronic code book (ECB)

答： c。电子密码书(ECB)

Explanation:

说明：

Mainly, the Electronic codebook mode is used for operating on short data as the same key is used for each block in this algorithm. Thus convert in Plain Text lead to repetitions in Cipher Text in cryptography.

主要是，电子密码本模式用于对短数据进行操作，因为此算法中的每个块都使用相同的键。因此，以纯文本格式转换会导致密码学中的密文重复。

3) Using Rivest, Shamir, Adleman cryptosystem with p=7 and q=9. Encrypt M=24 to find ciphertext. The Ciphertext is:

3)使用p = 7和q = 9的Rivest，Shamir，Adleman密码系统。 加密M = 24以找到密文。 密文是：

42
42
93
93
114
114
103
103

Answer: c. 114

答： c。 114

Explanation:

说明：

    Calculate n = p × q = 63
Phi(n)=(p-1) x (q-1)=48
Let e=5 , where gcd(e,phi(n))=1
Plaintext M = 24
Ciphertext = C ≡ M^e (mod n)
= 24^5 mod 63 = 7962624 mod 161
Ciphertext transmitted = 114.

4) International Data Encryption Algorithm (IDEA) was developed by

4)国际数据加密算法(IDEA)由