Assembly language to implement printing Yang Hui triangle

CODE SEGMENT 
 ASSUME CS:CODE,DS:CODE 
 org 100h
 START: jmp begin
message db 13,10,9,'Input N(N<=10): $'
error db 13,10,9,'Data out of range!$'
 begin:
  push cs
  pop ds
  mov dx,offset message
  mov ah,9
  int 21h

  call shur
  cmp bp,10
  jbe goon
  mov dx,offset error
  mov ah,9
  int 21h
  jmp exit
 goon:
  mov ax,0e0dh
  int 10h
  mov al,0ah
  int 10h
 push bp
  call yhsj
 exit:
  mov ah,0
  int 16h
  mov ah,4ch
  int 21h
 
 shur proc
 push cx
 push bx
 xor bp,bp
 mov bx,10
 mov cx,2
 input:
 mov ah,0; keyboard input data
 int 16h
 cmp al,0dh; end input with Enter
 jz ok
 cmp al,'0' ;Only 0~9 is allowed
 jb input
 cmp al,'9'
ja input
mov ah,0eh ;Show valid input
int 10h
sub al,30h ;changeASCIIforHEX
cbw ;字节扩展for字
xchg ax,bp
mul bx ;expand10Double
add bp,ax ;Add one
loop input
ok:nop ;The numerical result isBPmiddle
;Recover the registers used
pop bx
pop cx
ret
shur endp


; Functions that output Yang Hui triangle，Accept parameters on a stackN
; OutputNYang Hui Triangle
yhsj:
 mov bp, sp
 mov ax, [bp+2]  ; saveNarriveax
 shr ax, 1  ; N = N / 2
 push ax
 mov ax, [bp+2]  ; saveNarriveax
 push ax
 call C; C(N, N/2)获取最后一行middle间的那个值，That is, the maximum value
 call getdigit   ; Calculate the length of this maximum value，like252Then return3
 mov cx, ax ; saveMaximum lengtharrivecx，Used for post-event format
 xor di, di ; Outer loop countdi，外层cycleOutput每一行
 jmp cp1
up1:
 inc di; renewdi
cp1:
 cmp di, [bp+2]  ; Test cycle conditions，cycleNSecond-rate
 jg done1
 mov ax, [bp+2]  ; the following3Sentence calculates the number of spaces before line = (N-i)*cl，clIt is the maximum length
 sub ax, di
 mul cl
 call showspace  ; Output行前空格
 xor si, si ; 内存cycle计数si，内层cycleOutput一行middle的每Number of
 jmp cp2
up2:
 inc si; renewdi
cp2:
 cmp si, di ; Test cycle conditions，cyclediSecond-rate
 jg done2
 push si
 push di
 call C; Get the line's locationsiNumber of combinations of positions，CallC(di, si)
 push ax   ; save该组合数
 call show  ; Output该数
 mov ax, cx ;┒the following3句Output数字间间隔空格，Number of = N - 1
 sub ax, 1  ;┃
 call showspace  ;┚
 pop ax;┒
 call getdigit   ;┃Get the length of the combination
 mov bx, ax ;┃
 mov ax, cx ;┃
 sub ax, bx ;┃Calculate the number of spaces to be filled = Maximum length - The length of this number + 1
 add ax, 1  ;┃本来应该先填充再Output数字间空格，顺序反过来是for了左对齐
 call showspace  ;┚The above brackets2The segments are in the normal order in turn
 jmp up2   ; renew内层cycle
done2: ; 内层cycle结束
 mov ah, 2  ; the following5Sentences implement line breaks
 mov dl, 13
 int 21h
 mov dl, 10
 int 21h
 jmp up1   ; renew外层cycle
done1: ; 外层cycle结束
 ret 2 ; Free the stack space used by function parameters

; Recursive function to find combination numbers，Accept the stack2Parametersn, m(n > m)
; returnC(n, m)，Right nownselectm的Number of
; The algorithm is:
; { C(n, m) = 1 (n < m or m = 0)
; { C(n, m) = C(n-1, m-1) + C(n-1, m) (n > m)
; Right now某位置组合数等于上一行左右两数之和
C:
 push bp
 mov bp, sp
 sub sp, 2  ; Reserve a storage location
 mov bx, [bp+6]  ; savemarrivebx
 cmp bx, [bp+4]  ; like果m > n return1
 jz L1
 cmp bx, 0  ; like果m = 0 return1
 jz L1
 mov ax, [bp+4]  ; savenarriveax
 dec ax; ax = ax - 1
 dec bx; bx = bx - 1
 push bx
 push ax
 call C; return上一行左边的那Number of
 mov [bp-2], ax  ; save左肩膀上的数
 mov ax, [bp+4]  ; the following5Same sentence，return上一行右肩膀上的数
 dec ax
 push [bp+6]
 push ax
 call C
 add ax, [bp-2]  ; Add the numbers on the left shoulder to get the combined number
 jmp L2
L1:
 mov ax, 1
L2: 
 mov sp, bp
 pop bp
 ret 4 ; axreturn组合数

; Recursively10进制Outputax
; The method is very simple，Just find the remainder，Thenax = ax / 10
; ax = 0Exit when，开始逆序Output求出的各位余数
show:
 mov bx, 10
 cmp ax, 0
 jz ok1
 div bl
 push ax
 and ax, 00ffh
 call show
 pop dx
 mov dl, dh
 or dl, 30h
 mov ah, 2
 int 21h
ok1:
 ret

; 获取一Number of的长度，axforparameter，like果ax = 252Then return3
; ax里是return值
getdigit:
 mov bx, 10
 xor dx, dx
next:
 cmp ax, 0
 jle ok2
 div bl
 and ax, 0ffh
 inc dx
 jmp next
ok2:
 mov ax, dx
 ret

; OutputaxA space，parameterax，无return值
showspace:
 mov bx, ax
 mov ah, 2
 mov dl, ' '
nexts:
 cmp bx, 0
 jle dones
 int 21h
 dec bx
 jmp nexts
dones:
 ret

 CODE ENDS 
 END START