- 删除一批数据中的重复数据,删除后的一批数据是互不相同的。
include irvine32.inc
.data
dat word 1,2,4,3,2,1
p word 6 dup(?)
.code
main proc
mov ecx,6
mov edx,0
mov edi,0
again1:
mov esi,0
mov ax,word ptr dat[edi*2]
again2:
cmp ax,word ptr p[esi*2]
jz next
inc esi
cmp esi,edx
jl again2
mov word ptr p[edx*2],ax
inc edx
next:
inc edi
cmp edi,ecx
jnz again1
mov esi,0
pt:
movzx eax,p[esi*2]
call writeint
call crlf
inc esi
cmp esi,edx
jnz pt
exit
main endp
end main
- 有一批互不相同的有符号数,求出其中两数和的绝对值的最大值,要求输出最大值,以及这两个数。
第一种(全正或全负):5, 3, 2, 6, 1,和绝对值最大值是11,两个数是5, 6
第二种(有负有正):-3,2,-2,-4, 1, 5,和绝对值最大值是7,两个数:-3,-4及2,5
此代码有所缺陷,只能输出一组。如果有两组等大的,可以排序后比较头两位和尾两位的和的绝对值,如果等大则两组都输出。(我懒得改了
include irvine32.inc
.data
dat dword 5,3,2,6,1
max dword 0
x dword ?
y dword ?
count dword ?
.code
main proc
mov ecx,lengthof dat
mov count,ecx
mov ecx,0
again1:
mov eax,dat[ecx*4]
inc ecx
cmp ecx,count
jz pt
mov esi,ecx
again2:
mov edx,dat[esi*4]
add edx,eax
cmp edx,0
jg cpare
neg edx
cpare:
cmp edx,max
jle next
mov max,edx
mov edx,dat[esi*4]
mov x,edx
mov y,eax
next:
inc esi
cmp esi,count
jnz again2
cmp esi,count
jz again1
pt:
mov eax,max
call writeint
call crlf
mov eax,x
call writeint
call crlf
mov eax,y
call writeint
exit
main endp
end main
3、从键盘输入两个非0的无符号数,要求用子程序实现求两个数的最大公约数和最小公倍数。
include irvine32.inc
.data
x dd ?
y dd ?
max dd ?
min dd ?
.code
main proc
call readint
mov x,eax
call readint
mov y,eax
push x
push y
push offset max
push offset min
call mpublic
mov eax,max
call writeint
call crlf
mov eax,min
call writeint
exit
main endp
mpublic proc
push ebp
mov ebp,esp
push eax
push ecx
push edx
push ebx
mov eax,dword ptr [ebp+20] ;x的值
mov ecx,dword ptr [ebp+16] ;y的值
mov edx,dword ptr [ebp+12] ;max的地址
mov ebx,dword ptr [ebp+8] ;min的地址
again:
cmp eax,ecx
ja next
xchg eax,ecx
next:
sub eax,ecx
cmp eax,0
jnz again
mov [edx],ecx
mov eax,[ebp+20]
mov ecx,[ebp+16]
mul cl ;x*y
mov ecx,[edx]
div cl ;x*y/max
mov [ebx],eax
pop ebx
pop edx
pop ecx
pop eax
pop ebp
ret 16
mpublic endp
end main
4、从键盘输入字符串,查找是否包含指定子字符串’abc’
include irvine32.inc
.data
dat byte 10 dup(?)
count dword ?
.code
main proc
mov edi,0
lea edx,dat
mov ecx,10
call readstring
mov count,eax
mov esi,0
again:
mov ecx,esi
cmp dat[ecx],61h
jnz next
inc ecx
cmp dat[ecx],62h
jnz next
inc ecx
cmp dat[ecx],63h
jnz next
inc edi
jmp pt
next:
inc esi
cmp esi,count
jnz again
pt:
mov eax,edi
call writeint
exit
main endp
end main