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#2
有容就大2012-10-06 02:02
我晕 真是昏头了 我要发到汇编版的 怎么发到C版了
 麻烦版主们看到后速度移贴 我没有来C版显摆的意思啊 脑袋被搞昏了  |
下面是C代码和效果图
程序代码：#include<stdio.h>
#define M 12
int main(void)
{
int a[M][M], i, j;
printf("\n\n\n");
for (i = 0; i < M; i++)
{
printf(" ");
for (j = 0; j <= i; j++)
{
if (i == j || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i - 1][j - 1];
printf("%-4d", a[i][j]);
if (i == j)
printf("\n");
}
}
return 0;
}
---#define M 12
int main(void)
{
int a[M][M], i, j;
printf("\n\n\n");
for (i = 0; i < M; i++)
{
printf(" ");
for (j = 0; j <= i; j++)
{
if (i == j || j == 0)
a[i][j] = 1;
else
a[i][j] = a[i - 1][j] + a[i - 1][j - 1];
printf("%-4d", a[i][j]);
if (i == j)
printf("\n");
}
}
return 0;
}
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现在 汇编来实现这个目的
不过我做这个有点限制 就是限制在12行 即最大数能到百位 要搞更多行可以对程序做出适当的修改。
上代码和效果图， 有兴趣的看看， 那里可以改进下。
=====
程序代码：;#Mode=DOS
;MASMPlus 单文件代码模板 - 纯 DOS 程序
;--------------------------------------------------------------------
;--------------------------------------------------------------------
; program name: Yang Hui triangle
; producer: yrjd
; program function: Show the Yang Hui triangle as Left angle
; produce data: 2012-10-5
;--------------------------------------------------------------------
assume cs:code, ds:data, ss:stack
stack segment
db 128 dup(?)
stack ends
data segment
prompt db '12 lines of Left angle Yang Hui triangle: ', 0dh, 0ah, '$'
endStr db 0dh, 0ah, 'Press any key to continue', '$'
last db 20 dup (0)
next db 20 dup (0)
data ends
code segment
start: mov ax, stack
mov ss, ax
mov sp, 128
mov ax, data
mov ds, ax
lea dx, prompt
mov ah, 09h
int 21h
mov dh, 4
mov dl, 8
mov ax, 160
mul dh
mov bx, ax
mov ax, 2
mul dl
add bx, ax
mov ax, 0b800h
mov es, ax
mov cx, 11
mov bp, 1
mov last[0], 1
mov es:[bx + 6], 31h
mov es:[bx + 6 + 1], 2
add bx, 160
Set: push cx
mov dx, 1
mov next[0], 1
mov es:[bx + 6], 31h
mov es:[bx + 6 + 1], 2
dis: cmp dx, bp
je SetOne
mov di, dx
mov al, BYTE ptr last[di]
dec di
add al, last[di]
inc di
mov next[di], al
mov al, next[di]
call TurnToBCD
inc dx
jmp dis
SetOne: mov si, dx
mov next[si], 1
mov al, next[si]
call TurnToBCD
push di
mov cx, dx
mov di, 1
SetOK: mov ah, BYTE ptr next[di]
mov BYTE ptr last[di], ah
inc di
loop SetOK
pop di
pop cx
inc bp
add bx, 160
loop Set
mov ah, 2
mov bh, 0
mov dh, 17
mov dl, 0
int 10h
lea dx, endStr
mov ah, 09h
int 21h
mov ah, 01h
int 21h
mov ah, 4ch
int 21h
TurnToBCD:
push bx
push cx
push dx
push si
push di
push ax
push ax
mov ax, dx
mov dl, 8
mul dl
mov di, ax
pop ax
mov ah, 0
mov cl, 10
div cl
add ah, 30h
mov dl, ah
mov ah, 0
mov cl, 10
div cl
add ah, 30h
mov dh, ah
mov ah, 0
mov cl, 10
div cl
add ah, 30h
mov ch, ah
cmp ch, 30h
jz ok1
mov es:[bx + di + 2], ch
mov es:[bx + di + 3], 2
ok1: cmp dh, 30h
jz ok2
mov es:[bx + di + 4], dh
mov es:[bx + di + 5], 2
ok2:
mov es:[bx + di + 6], dl
mov es:[bx + di + 7], 2
OK: pop ax
pop di
pop si
pop dx
pop cx
pop bx
ret
code ends
end start
----;MASMPlus 单文件代码模板 - 纯 DOS 程序
;--------------------------------------------------------------------
;--------------------------------------------------------------------
; program name: Yang Hui triangle
; producer: yrjd
; program function: Show the Yang Hui triangle as Left angle
; produce data: 2012-10-5
;--------------------------------------------------------------------
assume cs:code, ds:data, ss:stack
stack segment
db 128 dup(?)
stack ends
data segment
prompt db '12 lines of Left angle Yang Hui triangle: ', 0dh, 0ah, '$'
endStr db 0dh, 0ah, 'Press any key to continue', '$'
last db 20 dup (0)
next db 20 dup (0)
data ends
code segment
start: mov ax, stack
mov ss, ax
mov sp, 128
mov ax, data
mov ds, ax
lea dx, prompt
mov ah, 09h
int 21h
mov dh, 4
mov dl, 8
mov ax, 160
mul dh
mov bx, ax
mov ax, 2
mul dl
add bx, ax
mov ax, 0b800h
mov es, ax
mov cx, 11
mov bp, 1
mov last[0], 1
mov es:[bx + 6], 31h
mov es:[bx + 6 + 1], 2
add bx, 160
Set: push cx
mov dx, 1
mov next[0], 1
mov es:[bx + 6], 31h
mov es:[bx + 6 + 1], 2
dis: cmp dx, bp
je SetOne
mov di, dx
mov al, BYTE ptr last[di]
dec di
add al, last[di]
inc di
mov next[di], al
mov al, next[di]
call TurnToBCD
inc dx
jmp dis
SetOne: mov si, dx
mov next[si], 1
mov al, next[si]
call TurnToBCD
push di
mov cx, dx
mov di, 1
SetOK: mov ah, BYTE ptr next[di]
mov BYTE ptr last[di], ah
inc di
loop SetOK
pop di
pop cx
inc bp
add bx, 160
loop Set
mov ah, 2
mov bh, 0
mov dh, 17
mov dl, 0
int 10h
lea dx, endStr
mov ah, 09h
int 21h
mov ah, 01h
int 21h
mov ah, 4ch
int 21h
TurnToBCD:
push bx
push cx
push dx
push si
push di
push ax
push ax
mov ax, dx
mov dl, 8
mul dl
mov di, ax
pop ax
mov ah, 0
mov cl, 10
div cl
add ah, 30h
mov dl, ah
mov ah, 0
mov cl, 10
div cl
add ah, 30h
mov dh, ah
mov ah, 0
mov cl, 10
div cl
add ah, 30h
mov ch, ah
cmp ch, 30h
jz ok1
mov es:[bx + di + 2], ch
mov es:[bx + di + 3], 2
ok1: cmp dh, 30h
jz ok2
mov es:[bx + di + 4], dh
mov es:[bx + di + 5], 2
ok2:
mov es:[bx + di + 6], dl
mov es:[bx + di + 7], 2
OK: pop ax
pop di
pop si
pop dx
pop cx
pop bx
ret
code ends
end start
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对贴上去的排版无语 以后有时间修正下 稍微加点注释。。。
眼冒金星了 睡觉去。
