第1题只能输入数字的程序。.doc

第1题：只能输入数字的程序。;程序如下：data segment??????? msg1 db Input Number:0~9,0ah,0dh,$??????? msg2 db Error!,0ah,0dh,$data ends code segment??????? assume cs:codestart:??????? mov ax,data??????? mov ds,ax??????? ??????? lea dx,msg1??????? mov ah,9h??????? int 21h??????????????? ;显示第一个字符串??????? lp:???? mov ah,01h??????? int 21h??????????????? ;接收字符??????? ??????? cmp al,0??????? jb again??????? ;ASCII码小于0则显示跳到标号again处??????? cmp al,9??????? ja again??????? ;ASCII码大于9则显示跳到标号again处??????? jmp short quit? ;在0~9之间则跳出循环again:? lea dx,msg2??????? mov ah,9h??????? int 21h???????????? ;显示第二个字符串Error!??????? jmp short lp??????? ;跳到标号lp处 quit:?? mov ax,4c00h??????? int 21hcode endsend start ============================================第2题：输入一个字符串(最大长度为10)，统计出现字母A的个数。;程序如下：data segment??????? str db 10,13,input a? string,10,13,$??????? strb db 10,13,the number of A is ,10,13,$??????? buf db 10??????????? db ???????????? db 10 dup(?) data ends code segment? assume cs:code,ds:data?? ?? start:mov ax,data???????? mov ds,ax????????? lea dx,str???????? mov ah,9???????? int 21h???????? ? lea dx,buf???????? mov ah,0ah???????? int 21h???????? ? xor cx,cx???????? xor si,si?;存放A的个数[不能使用al进行]???????? mov cl,buf+1?;改为cl,否则操作数类型不一致???????? lea bx,buf+2???????? ;mov dl,A???? lp: cmp byte ptr [bx],A???????? jne next? inc si?next: ?inc bx??????? loop lp ?lea dx,strb??????? mov ah,9??????? int 21h??????? ?xor dx,dx?mov dx,si?add dl,30h?;使dl成为可显示的阿拉伯数字?mov ah,2??????? int 21h??????? ?mov ah,4ch??????? int 21hcode ends??? end start ==========================================第3题：计算12+22+32+42+……，计算前N项的和，使和大于1000。编写程序求解N。;程序如下：assume cs:code,ds:datadata segment??????? count db 0 ;计数器??????? sum dw 0?? ;sum中存放和data ends code segmentmain:??????? mov ax,data??????? mov ds,ax??????? and ax,0??????? mov al,12??? ;将al的值初始为12??????? lea bx,sum s:????? add byte ptr [bx],al??????? adc byte ptr [bx+1],ah??????????? ;以上两步做累加和??????? a