Quick learning method for soft exam questions (personal notes)

1, 19 under 46

2, 19 under 24

Look at the label number, it is now 15, and it is close to 15, 15-12=3, 19-15=4, so it is better to choose 12, that is, ①⑤First, the order is indistinguishable, and then the answer will come out. If you continue Going down, 19 is closer to 12, so next is ②④

This kind of thing is to move the arm and then rotate , remember: arm, to rotate

3, 19 under 4

CFFFFH-A0000H+1=30000, which is 3 * 16^4, then, how many bytes are 3 16 4 = 192

pieces, which is the above 192/64= 3 , 64 is what the title writes 64k

4. Public key and private key

This kind of stuff is always confused. The general topic is that two people exchange information, and then they must be blended by a third party, which is simply the promise of both parties, the third party blending . This third party is called CA. Public key encryption authentication, private key decryption signature . Generally, the CA public key is used to verify the authenticity of the CA signature

5. Lexical, grammar, semantics, intermediate code, optimization, target code

1. Lexical analysis: output the token stream, scan it again, and identify each word
2. Grammatical analysis: A lot of work, anyway, I won’t choose grammatical analysis, there is a high probability that I will miss it. What input token flow, matching, and error checking are all done at this stage, and I have never seen what can be done in the rest of the stage.

6, 19 on 25

At first glance, 1024G, 64-bit, 4M, just add, subtract, multiply and divide these few, the answer is all 4 digits, 1024G is 1024 * 1024M, then the answer is 1024 * 1024/4/64=4096

7, 19 on 49

The question is converted to 0(0|(ε(1ε)+ε))0simplified to 0(0|1+)0, there are 000 and 010 or 011…0
A:, 01*0cannot be converted to 000
B:, 001*0cannot be 000
C:, 0(0|1)1*0equivalent to 0(0|(01+)|1+)0, can be converted to 000, 010 , 01111…0
D:, 01*00cannot be 000
is the question What to convert to, look at the options, and whether the options can all match

8. 25 on 17 (deadlock, mutually exclusive process)

Fast learning method:, 互斥进程*(最大需求-1)+1 > 资源deadlock; if it is, <=it is not deadlock①
: 2*(2-1)+1=3=3, not deadlock②
: 3*(2-1)+1=4>3, deadlock lock
③: 2 * (2-1) + 1 = 3 <5, is not deadlock
④: 3 * (3-1) + 1 = 7> 5, deadlock
⑤: 3 * (3-1) + 1 = 7>6, deadlock
⑥: 4*(2-1)+1=5<6, not deadlock,
so 245 is deadlock

How much resources are added to these conditions is not a deadlock? It is 资源>公式good, 1,2,1 on OK

9. Association, aggregation, combination, dependency

• Association:
  One of the attributes in class A is class B, and the relationship between A and B is 1: 0 or 1: more

• Aggregation
  Special associations, the relationship between A and B is detachable. For example, if you have an assembly machine, you have to buy a new graphics card and install it.

• Combination A
  special association, the relationship between A and B is inextricable. For example, the graphics card of a notebook has no need to change

• 
  Methods that rely on A have classes of B