优化目标： 16行393步

见：https://github.com/atesgoral/hrm-solutions/blob/master/solutions/32-Inventory-Report-16.393/16.377-cowboy.asm

想比16行385步方案，通过优化执行顺序，省掉了计数器初始化时减1的操作

snippet.asm

-- HUMAN RESOURCE MACHINE PROGRAM --
 
    JUMP     b
a:
    COPYFROM 19
    OUTBOX  
b:
    INBOX   
    COPYTO   15
    COPYFROM 14
    COPYTO   19
    COPYTO   18
    JUMP     d
sum:
    BUMPUP   19
counter:
    BUMPUP   18	
d:
    COPYFROM [18]
    JUMPZ    a
    SUB      15
    JUMPZ    sum
    JUMP     counter

参考：https://github.com/atesgoral/hrm-solutions/blob/master/solutions/32-Inventory-Report-16.393/15.31.exploit-kyanet.asm

用于INBOX中ABCX数量都是固定的情况（游戏中A=4, B=5, C=2, X=3），

优化掉一个 BUMPUP 18，由于无论找没找到，计数器18都会+1，因此在 COPYFROM [18] 之前就将计数器加一，这样就要求计数器初始化时为 -1。另外 先INBOX在初始化计数器，能省4步

snippet.asm

-- HUMAN RESOURCE MACHINE PROGRAM --
 
    JUMP     b
a:
    COPYFROM 19
    OUTBOX  
b:
    INBOX   
    COPYTO   15
    COPYFROM 14
    COPYTO   19
    BUMPDN   19
    COPYTO   18    
e:
    BUMPUP   19
d:
    BUMPUP   18
    COPYFROM [18]
    JUMPZ    a
    SUB      15
    JUMPZ    e
    JUMP     d

snippet.asm

-- HUMAN RESOURCE MACHINE PROGRAM --
 
    JUMP     b
a:
    COPYFROM 19
    OUTBOX  
b:
    COPYFROM 14
    COPYTO   19
    COPYTO   18
    INBOX   
    COPYTO   15
c:
d:
    COPYFROM [18]
    JUMPZ    a
    SUB      15
    JUMPZ    e
    BUMPUP   18
    JUMP     c
e:
    BUMPUP   19
    BUMPUP   18
    JUMP     d