求个汇编语言流程图

Posted 2023-04-18

DATA1 SEGMENT
DATA DW 10
DW 2130,-43,31,-321,-1234,345,0,3213,0,5477
POSITIVE DW 0
NEGATIVE DW 0
ZERO DW 0
DATA1 ENDS
CODE SEGMENT
ASSUME CS:CODE,DS:DATA1
START:MOV AX, DATA1
MOV DS, AX
MOV AX, 0
MOV BX, 0
MOV DX, 0
MOV CX, DATA
JCXZ SAVE
LEA SI, DATA+2
AGAIN:CMP WORD PTR [SI], 0
JL LOWER
JE EQUAL
INC AX
JMP LOOP1
LOWER:INC BX
JMP LOOP1
EQUAL:INC DX
LOOP1:ADD SI, 2
LOOP AGAIN
SAVE:MOV POSITIVE, AX
MOV NEGATIVE, BX
MOV ZERO, DX
MOV AH, 4CH
INT 21H
CODE ENDS
END START

参考技术A stack segment para stack 'stack'
db 256 dup(0)
top label word
stack ends
data segment para public 'data'
buffer db 16h dup(0)
bufpt1 dw 0
bufpt2 dw 0
kbflag db 0
prompt db ' *please practise typing*',0dh,0ah,'$'
scantab db 0,0,'1234567890-=',8,0
db 'qwertyuiop[]',0dh,0
db 'asdfghjkl;',0,0,0,0
db 'zxcvbnm,./',0,0,0
db ' ',0,0,0,0,0,0,0,0,0,0,0,0,0
db '789-456+1230.'
even
oldcs9 dw ?
oldip9 dw ?
str1 db 'abcd efgh ijkl mnop qrst uvwx yz.'
db 0dh,0ah,'$'
str2 db 'there are some newspapers on the table.'
db 0dh,0ah,'$'
str3 db 'there are some clouds in the sky.'
db 0dh,0ah,'$'
str4 db 'she always eats her lunch at noon.'
db 0dh,0ah,'$'
str5 db 'i do not like autumn and winter.'
crlf db 0dh,0ah,'$'
colon db ':','$'
even
saddr dw str1,str2,str3,str4,str5
count dw 0
sec dw 0
min dw 0
hours dw 0
save_lc dw 2 dup(?)
data ends
code segment
assume cs:code,ds:data,es:data,ss:stack
main proc far
start:
mov ax,stack
mov ss,ax
mov sp,offset top

push ds
sub ax,ax
push ax
mov ax,data
mov ds,ax
mov es,ax

; mov ah,0
; mov al,4
; int 21h
; mov ah,0bh
; mov bh,0
; mov bl,4
; int 11h
mov ah,35h
mov al,09h
int 21h
mov oldcs9,es
mov oldip9,bx

push ds
mov dx,seg kbint
mov ds,dx
mov dx,offset kbint
mov al,09h
mov ah,25h
int 21h
pop ds

mov ah,35h
mov al,1ch
int 21h
mov save_lc,bx
mov save_lc+2,es

push ds
mov dx,seg clint
mov ds,dx
mov dx,offset clint
mov al,1ch
mov ah,25h
int 21h
pop ds

in al,21h
and al,11111100b
out 21h,al

first: mov ah,0
mov al,3
int 10h

mov dx,offset prompt
mov ah,9
int 21h

mov si,0
next: mov dx,saddr[si]
mov ah,09h
int 21h

mov count,0
mov sec,0
mov min,0
mov hours,0

sti
forever:
call kbget
test kbflag,80h
jnz endint
push ax
call dispchar
pop ax
cmp al,0dh
jnz forever
mov al,0ah
call dispchar

call disptime
lea dx,crlf
mov ah,09h
int 21h

add si,2
cmp si,5*2
jne next
jmp first
endint: cli
push ds
mov dx,save_lc
mov ax,save_lc+2
mov ds,ax
mov al,1ch
mov ah,25h
int 21h
pop ds

push ds
mov dx,oldip9
mov ax,oldcs9
mov ds,ax
mov al,09h
mov ah,25h
int 21h
pop ds

sti
ret
main endp
clint proc near
push ds
mov bx,data
mov ds,bx

lea bx,count
inc word ptr[bx]
cmp word ptr[bx],18
jne return
call inct
adj:
cmp hours,12
jle return
sub hours,12
return:
pop ds
sti
iret
clint endp
inct proc near
mov word ptr[bx],0
add bx,2
inc word ptr[bx]
cmp word ptr[bx],60
jne exit
call inct
exit: ret
inct endp
disptime proc near
mov ax,min
call bindec

mov bx,0
mov al,':'
mov ah,0eh
int 10h
mov ax,sec
call bindec

mov bx,0
mov al,':'
mov ah,0eh
int 10h

mov bx,count
mov al,55d
mul bl
call bindec

ret
disptime endp
bindec proc near
mov cx,100d
call decdiv
mov cx,10d
call decdiv
mov cx,1
call decdiv
ret
bindec endp
decdiv proc near
mov dx,0
div cx

mov bx,0
add al,30h
mov ah,0eh
int 10h

mov ax,dx
ret
decdiv endp
kbget proc near
push bx
cli
mov bx,bufpt1
cmp bx,bufpt2
jnz kbget2
cmp kbflag,0
jnz kbget3
sti
pop bx
jmp kbget
kbget2:
mov al,[buffer+bx]
inc bx
cmp bx,16h
jc kbget3
mov bx,0
kbget3:
mov bufpt1,bx
pop bx
ret
kbget endp
kbint proc near
push bx
push ax

in al,60h
push ax
in al,61h
or al,80h
out 61h,al
and al,7fh
out 61h,al

pop ax
test al,80h
jnz kbint2
mov bx,offset scantab
xlat scantab
cmp al,0
jnz kbint4
mov kbflag,80h
jmp kbint2
kbint4:
mov bx,bufpt2
mov [buffer+bx],al
inc bx
cmp bx,16h
jc kbint3
mov bx,0
kbint3:
cmp bx,bufpt1
jz kbint2
mov bufpt2,bx
kbint2: cli
mov al,20h
out 20h,al
pop ax
pop bx
sti
iret
kbint endp
dispchar proc near
push bx
mov bx,0
mov ah,0eh
int 10h
pop bx
ret
dispchar endp
code ends
end start 参考技术B

这是一个在一串数据中分别统计大于0、小于0、等于0的数据个数的程序。

你的程序中似乎有个小问题，流程图应该如图所示：

本回答被提问者采纳 参考技术C

如图！

Problem A: 求个最大值

 

Problem A: 求个最大值

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 1635  Solved: 1339
[Submit][Status][Web Board]

Description

定义MaxValue类，用于求一系列非零整数的最大值。其中：

1. 数据成员elements用于存储所有输入的非零整数。

2. void append(int)用于向elements中添加一个新数据。

3. int getMax()用于求出elements中的最大值。

 

Input

输入若干个整数，以输入0表示输入结束。

 

Output

所有输入的非零整数中的最大值。

 

Sample Input

321
496
553
338
837
463
158
154
929
537
0

　　

Sample Output

929

HINT

 

使用vector更为容易实现。

 

 

Append Code

append.cc, 

int main()
{
    int a;
    MaxValue test;
    cin>>a;
    while (a != 0)
    {
        test.append(a);
        cin>>a;
    }
    cout<<test.getMax()<<endl;
    return 0;
}

　　

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class MaxValue
{
public :
    vector<int> elements;
    void append(int num)
    {
        elements.push_back(num);
    }
     int getMax()
     {
         sort(elements.begin(), elements.end());//排序
         return elements.back();//返回最后一个元素
     }
};
int main()
{
    int a;
    MaxValue test;
    cin>>a;
    while (a != 0)
    {
        test.append(a);
        cin>>a;
    }
    cout<<test.getMax()<<endl;
    return 0;
}