汇编上机实验C-最大公约数

1、实验c最大公因数实验目的：从键盘输入两个不超过65535的十进制整数a、b,输出a、b的最大公因数 gcd(a,b)o实验要求：键盘输入两个十进制整数a、b (其对应的二进制至多16位)，以回车结束 一个十进制数的输入。接着在下面的一行上输出这两个整数的最大公约数。如无 最大公因数，则输出“nogcd”。注：为方便起见，最大公因数前允许有多余的前导0。实验结果：please input two numbers and press enter key when you want to stop a number!9600003do you want to continue(y/n)?ypleas

2、e input two numbers and press enter key when you want to stop a number!301200006do you want to continue(y/n)?yplease input two numbers and press enter key when you want to stop a number!50000050do you want to continue(y/n)?yplease input two numbers and press enter key when you want to stop a number!

3、00no gcddo you want to continue(y/n)?n实验报告要求：1、分析要点及调试后的正确程序。datas segmentusedb 10dup(0)bcnum dw?dcnum dw ?detail db 'please in put two numbers and press en ter key when you want to stop a number!'/$'continue db 'do you want to continue(y/n)?','$'nodbnogcd','$,tmp

4、db?datas endsexdatas segmentflg db 0；标记是否为第一次计算，0是，1否exdatas endsstacks segmentdw 100 dup(o)stacks endscodes segmentassume cs:codes,ds:datas,ss:stacks,es:exdatasstart:mov ax,datasmov ds,axmov ax,exdatasmov es,axshowque:cmp flg,0；判断是否为第一次计算jztwo；如果是第一次计算则跳过换行call changelinetwo:lea dx,detail ;显示提示mov

5、ah,9int21hsub bx,bxsub cx,cxmov cx, 10mov si,10ready:call changelinesub ax,axsub bx,bxsub dx,dxinput:mov ahz01h ;输入int21hcmpal,odhjz decsitemovtmp,alsub ax,axsub dx,dxmov ax,bxmul cxmov bx,axsubtmp,30h;转换add bx,word ptrtmpmov si,bx保存sub ax,axjmp inputdecsite:mov dx,bxadd si,2cmp si,14jz nextjnz read

6、ynext:sub ax,axsub bx,bxsubcx,cxsub dx,dxmov ah,02hmov dx,odhint21hmov ah,02hmov dx,oahint21hmyready:sub ax,axmov dx,0mov ax,bcnummov cx,dcnumcmp cx,0jz result2div cxcmp dx,0jz resultjnzsub1sub1:mov bcnum,cxmov dcnum,dxjmpmyreadyresult2:cmp ax,0jz noticemov dx,axjmp mycallresult:mov dx,dx mycall: ca

7、ll transfermovah,02hint21hmovah,02hmov dx,odhint21hmovah,02hmov dx,oahint21hsub ax,axshow:movflg,1lea dx,continue movah,09hint21hsub ax,bxmovah,01hint21hcmpal/y* jzshowquecmp al,v jzshowque jnz exit；输出转换transfer proc near mov si,offset use add si,4sub bx,cxmov bl,$mov sl+l,blmov ax,dxmov cx,5ag1:mov

8、 dx,0mov bx,1odiv bxpush dxadd dl,30hmov si,dl dec si popdx loopag1 mov dx,si incdx m0vah,9int21hret transfer endp；换行changeline proc near movah,02h mov dx,0dh int21h movah,02h mov dx,0ah int21hretchangeline endpnotice:lea dx,nomovah,09hint21hmovah,02h mov dx,odh int21hmovah,02h movsi,oah int21h jmp showexit:m0vah,4chint21hcodes endsend start2、实验体会。这次实验并没有什么技术方面的难度，但是它难在输入数据长度的处理。如 果输入了五位数那么一个寄存器无法放下，必须增加一个寄存器,同时在进行 乘法除法时也要考虑这一点。同时有些难到我的是用户友好性的方面,为了去除一个开头显示会换行的问 题,我增加了一个flg标记