7.18 - daily question - 408

One sentence per day ： There is always a time in life , Full of uneasiness , But in addition to face up to , We have no choice .

data structure
1. Set the height to h The degree of a binary tree is only 0 The sum is 2 The node of , Then the number of nodes contained in this kind of binary tree is at least ______（ Anhui University 2014 year ）
A.h+1
B.2h-1
C.2h
D.2h+1
answer ：B
analysis ： Do it by exclusion , From the properties of binary tree , If the degree of a binary tree is only 0 The sum is 2 The node of , Then the summary point is an odd number , Can be ruled out A and C. When h = 2 when , A binary tree has at most nodes 2^2-1 = 3 individual , and D Item equal to 5, So don't choose .
expand ：
The nodes of a full binary tree are either leaf nodes , Degree is 0, Or the degree is 2 The node of , The nonexistence degree is 1 The node of .
The full binary tree satisfies the following properties .
1、 One layer number is k The total number of full binary trees is ： (2^k) -1. So the number of nodes in a full binary tree must be odd .
2、 The first i The number of nodes on the layer is 2^(i-1)：
3、 One layer number is k The number of leaf nodes of a full binary tree （ That's the last layer ）：2^(k-1)

computer network
2. Let's say the rate of transmission of the signal on the media is 2.3×10^8 m/s, When the media length is 10cm, Data rate is 1M/s when , The number of bits being transmitted in the media is _________.（ Ocean University of China 2017 year ）

A.4.35 x 10^(-4)b
B.4.35 x 10^(-3)b
C.4.35 x 10^(-5)b
D.4.35 x 10^(-2)b

answer ：A
analysis ： That is, find the propagation delay bandwidth product . Propagation delay = Channel length / The propagation rate of electromagnetic wave on the channel . Propagation delay bandwidth product = Propagation delay × data rate . When the media length 0.1m, Data rate is 1Mb/s when , The number of bits in the media is calculated as follows ： Propagation delay =0.1m/2.3×10^8 m/s = 4.35×10^(-10) s Number of bits in the media = 4.35×10^(-10) s ×10^6 b/s=4.35 x 10^(-4)b.
expand ：
data rate 1Mb/s It means megabits per second , The number of bits transmitted per second .
there MB/s yes 10^6B/s , instead of 1024 * 1024 B/s
The unit of data transmission rate is generally MB/s or Mbit/s, Especially in the internal data transmission rate, official data are more used Mbit/s In units of .
MB/s It means megabytes per second ,Mbit/s It means megabits per second , The former refers to the number of bytes transmitted per second , The latter refers to the number of bits transmitted per second .1MB/s be equal to 8Mbit/s.
Detailed description , Please refer to http://www.mancos-co.com/article-72-238696-0.html

operating system
3. Hypothetical job 1、2、3、4 Arrive at the same time , The running time is 2、5、 8、3, The priority numbers of assignments are 4、9、1、8. When using high priority number first scheduling algorithm , The average turnaround time of the operation is ________ Hours .( Beijing Jiaotong University 2014 year )
A.4.5
B.10.5
C.4.75
D.10.25

answer ：D
analysis ：

picture

The average turnaround time is （10+5+18+8）/4 = 10.25, The higher the priority number, the first to execute .

The principle of computer organization
4． In the complement addition operation , Overflow occurs when _______.（ Guangdong University of Technology 2018 year ）

Ｉ ． The sign bits of the two operands are the same , Single sign bit is used in operation , The sign bit of the result is the same as the operand
Ⅱ． The sign bits of the two operands are the same , Single sign bit is used in operation , The sign bit of the result is different from the operand
Ⅲ． Single sign bit is used in operation , If the sign bit of the result is different from the highest digit, carry will be generated
Ⅳ． Single sign bit is used in operation , The sign bit and the highest digit of the result produce a carry at the same time
Ｖ． Double sign bits are used in operation , The two sign bits of the operation result are the same
Ⅵ． Double sign bits are used in operation , The two sign bits of the operation result are different
Ａ.Ｉ Ⅲ Ｖ　　
Ｂ.Ⅱ Ⅳ Ⅵ
Ｃ.Ⅱ Ⅲ Ⅵ　　
Ｄ.Ｉ Ⅲ Ⅵ

answer ：C
analysis ： There are three commonly used overflow judgment methods ： Take a sign bit 、 Use carry bits and double sign bits .
The overflow condition with one sign bit is , The sign bit of the result is different from that of the operand .
The overflow condition using carry bits is ： If the sign bit of the result is different from the highest digit, carry will be generated .
The overflow condition with double sign bits is ： The two sign bits of the operation result are different .

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