(60%)Assume+that+the+following+sequences+of+instructions+are+exe-cuted+on+a+five-stage+pipelined+dat
时间: 2023-12-29 14:26:31 浏览: 23
以下是一个用于找到方程ax^2 + bx + c = 0的实根的伪代码算法:
```python
1. 输入a, b, c
2. 计算delta = b^2 - 4ac
3. 如果delta < 0,则方程没有实根,结束
4. 如果delta = 0,则方程有一个实根,计算x = -b / (2a),结束
5. 如果delta > 0,则方程有两个实根,计算x1 = (-b + sqrt(delta)) / (2a) 和 x2 = (-b - sqrt(delta)) / (2a),结束
```
请注意,这只是一个伪代码算法,具体的实现可能会根据编程语言和环境的不同而有所不同。
相关问题
Consider+a+circuit-switched+network.+Suppose+that+all+links+in+the+network+use+TDM+with+10+slots+and
A circuit-switched network is a type of communication network where a dedicated communication path is established between two nodes for the duration of a communication session. In this network, the available bandwidth is divided into fixed-size time slots, and each slot can be used to transmit data or voice signals.
In the given scenario, let's assume that all links in the network use Time Division Multiplexing (TDM) with 10 slots. TDM is a technique that allows multiple signals to be transmitted simultaneously over a single communication channel by dividing the channel into time slots.
With 10 slots in TDM, each link in the circuit-switched network can accommodate up to 10 different communication sessions simultaneously. Each session is allocated a dedicated time slot for transmitting data or voice signals. The time slots are allocated in a cyclic manner, ensuring fair access to all sessions.
用汇编语言计算3X-2|Y|+|Z|
根据提供的汇编代码,我们可以使用类似的思路来计算3X-2|Y|+|Z|。下面是汇编代码:
```
DATA SEGMENT
X DW 10 ;存放X的值
Y DW -14 ;存放Y的值
Z DW 8 ;存放Z的值
RESULT DW ? ;存放最后的结果
DATA ENDS
STACK SEGMENT
STACK DB 100 DUP(?)
STACK ENDS
CODE SEGMENT
ASSUME CS:CODE,DS:DATA,SS:STACK
STACK:
MOV AX,DATA
MOV DS,AX
MOV AX,X ;将X的值赋值给AX
MOV BX,3 ;将3赋值给BX
MUL BX ;AX = AX * BX,即AX = 3X
MOV AX,Y ;将Y的值赋值给AX
CMP AX,0 ;比较AX和0的大小
JGE POSITIVE ;若AX >= 0,跳转到POSITIVE标签处
NEG AX ;否则,AX < 0,求补运算,即取绝对值
POSITIVE:
MOV BX,2 ;将2赋值给BX
MUL BX ;AX = AX * BX,即AX = 2|Y|
SUB AX,BX ;AX = AX - BX,即AX = 2|Y| - 2
MOV BX,Z ;将Z的值赋值给BX
CMP BX,0 ;比较BX和0的大小
JGE POSITIVE2 ;若BX >= 0,跳转到POSITIVE2标签处
NEG BX ;否则,BX < 0,求补运算,即取绝对值
POSITIVE2:
ADD AX,BX ;AX = AX + BX,即AX = 3X - 2|Y| + |Z|
MOV RESULT,AX ;将最后的结果赋值给RESULT单元
MOV AX,4CH ;调用4CH号功能,返回DOS操作系统
INT 21H ;终止当前程序的运行,并返回DOS系统
CODE ENDS
END START
```