汇编判断一元二次方程是否有根

.model small
.stack
.data
_a db 0h
_b db 0h
_c db 0h
tag db ?
string1 db 255
        db 0
        db 255 dup(0)
        
string2 db 255
        db 0
        db 255 dup(0)
        
string3 db 255
        db 0
        db 255 dup(0)
        
Prompta db 'please input a  :',0dh,0ah,'$'
Promptb db 'please input b :',0dh,0ah,'$'
Promptc db 'please input c :',0dh,0ah,'$'

Newline	db 0dh, 0ah, "$"	;回车换行

.code


      ALdisp proc
        push ax
        push cx
        push dx
        push ax
        mov dl,al
        mov cl,4
        shr dl,cl
        or dl,30h
        cmp dl,39h
        jbe aldisp1
        add dl,7
aldisp1:mov ah, 2
        int 21h
        pop dx
        and dl,0fh
        or dl,30h
        cmp dl,39h
        jbe aldisp2
        add dl,7
aldisp2: mov ah,2
        int 21h
        pop dx
        pop cx
        pop ax
        ret
ALdisp  endp

.startup
    ;输入a
    mov dx,offset Prompta
    mov ah,09h
    int 21h
	mov	ax, seg string1	;取数据段存入ds中
	mov	ds, ax
 
	mov dx,offset string1
	mov	ah, 0ah
	int	21h
    
    
	lea	dx, Newline	;输出回车换行
	mov	ah, 9h
	int	21h

	
	;输入b
    mov dx,offset Promptb
    mov ah,09h
    int 21h
	mov	ax, seg string2	;取数据段存入ds中
	mov	ds, ax
 
	mov dx,offset string2
	mov	ah, 0ah
	int	21h
 
	lea	dx, Newline	;输出回车换行
	mov	ah, 9h
	int	21h
	
	
	;输入c
    mov dx,offset Promptc
    mov ah,09h
    int 21h
	mov	ax, seg string3	;取数据段存入ds中
	mov	ds, ax
 
	mov dx,offset string3
	mov	ah, 0ah
	int	21h
 
	lea	dx, Newline	;输出回车换行
	mov	ah, 9h
	int	21h
	not_use macro
	mov ax,offset string3+2h
	push ax
	mov cl,string3+1h
	mov ch,0
	push cx
	call GetC
	endm
	
	;求a
	mov bx,offset string1+2h
      mov cl,string1+1h
	  mov ch,0
again:mov al,[bx]
      sub al,30h
      cmp cx,2
      ja next2
      cmp cx,1
      ja next1
      add _a,al
      jmp done
next2:mov dl,64h
      mul dl
      add _a,al
      jmp done
next1:mov dl,0ah
      mul dl
      add _a,al
done: inc bx
      loop again

      
      
      ;求b
      mov bx,offset string2+2h
      mov cl,string2+1h
	  mov ch,0
againb:mov al,[bx]
      sub al,30h
      cmp cx,2
      ja next2b
      cmp cx,1
      ja next1b
      add _b,al
      jmp doneb
next2b:mov dl,64h
      mul dl
      add _b,al
      jmp doneb
next1b:mov dl,0ah
      mul dl
      add _b,al
doneb: inc bx
      loop againb
      
	
	;求c
	mov bx,offset string3+2h
      mov cl,string3+1h
	  mov ch,0
againc:mov al,[bx]
      sub al,30h
      cmp cx,2
      ja next2c
      cmp cx,1
      ja next1c
      add _c,al
      jmp donec
next2c:mov dl,64h
      mul dl
      add _c,al
      jmp donec
next1c:mov dl,0ah
      mul dl
      add _c,al
donec: inc bx
      loop againc
      
      ;求是否有根
      mov al,_b
      imul al
      mov bx,ax
      mov al,_a
      imul _c
      mov cx,4
      imul cx
      cmp bx,ax
      jge yes
      mov tag,'N'
      jmp donew
yes:  mov tag,'Y'
donew:mov ah,02h
      mov dl,tag
      int 21h
	.exit 0
	end