algorithm - 设计一个算法，输入3个数字，然后输出小人计算中的最高值

Question

我可以对其进行编程以输出 2 个数字中的最高值，但我一直坚持如何输出 3 个输入中的最高值。有人可以帮忙吗？

Answer 1

用一句话来说：

read input into mailboxes M0, M1, M2
if M1 > M2
    store M1 into M2
if M0 > M2
    store M0 into M2
output M2

在小人电脑汇编程序中：

    INP
    STA M0
    INP
    STA M1
    INP
    STA M2

    SUB M1
    BRP J1
    LDA M1
    STA M2
J1  LDA M2
    SUB M0
    BRP J2
    LDA M0
    STA M2
J2  LDA M2
    OUT
    HLT
M0  DAT
M1  DAT
M2  DAT

你可以在这里运行程序：Max of 3 in LMC Emulator

Answer 2

一个小人程序，它接受三个值作为输入并产生三个值中的最大值作为输出。

00  IN  901 Take first value as input 
01  STO 320 Store the first value in 20th mailbox
02  IN  901 Take second value as input 
03  STO 321 Store the second value in 21st mailbox 
04  IN  901 Take third value as input 
05  STO 322 Store the third value in 22nd mailbox
06  SUB 221 Subtract the second value (21st mailbox) from third value 
07  BRP 810 If the difference is positive then, move to the 10th instruction
08  LDA 521 As the second value is greater than third value, load second value
09  STO 322 Store the second value in 22nd mailbox
10  LDA 522 Load the value from 22nd mailbox
11  SUB 220 Subtract the first value (20th mailbox) from loaded value 
12  BRP 815 If the difference is positive then, go to the 15th instruction
13  LDA 520 As the first value is greater than loaded value, load first value
14  STO 322 Store the first value in 22nd mailbox
15  LDA 522 Load the value from 22nd mailbox
16  OUT 902 Output the loaded value
17  HLT 000 Stop

Answer 3

LMC Assembly code Comment

INP

L1 STA hi

L2 LDA co

 SUB two

 BRZ op

 LDA co

 ADD one

 STA co

 INP

 STA wo

 SUB hi

 BRP sw

 BRA L2

sw LDA wo

 BRA L1

op LDA hi

 OUT

 HLT

hi DAT

co DAT

wo DAT

one DAT 1

two DAT 2

Input the first number

L1 Store the number as the highest so far

L2 Load the loop counter

Subtract two from the counter

If we are finished jump to output op

Otherwise load the counter

Add one to the counter

Store the updated counter

Get the next number

Store this as the working number

Subtract the highest number so far

Branch to sw(ap) if we have a new high

Otherwise branch back to L2

sw Load the working number (the new high)

Branch to L1

op Load the highest number again

Send it to output

Halt the program

Data to hold the current high number

Data to hold the loop counter

Data to hold the working number

Data hold the loop increment number

Data the number of times to loop

This can easily be adapted for more numbers by changing the last DAT statement. To do ten numbers, use 'two DAT 9' instead.