輸入兩個數(shù)，使用輾轉(zhuǎn)相除法求最大公約數(shù)

程序運行：

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代碼：

datas segmentM_max_len db 0ffhM_len db 0M_string db 100 dup(?)M dw 0 N_max_len db 0ffhN_len db 0N_string db 100 dup(?)N dw 0inputM db 'input M=$'inputN db 0dh,0ah,'input N=$'output db 0dh,0ah,'R=$'error_number db 0dh,0ah,'error number$' datas endsstacks segment stackdb 100h dup(?)stacks endscodes segmentassume cs:codes,ds:datas,ss:stacks main proc far start:push dsmov ax,0hpush axmov ax,datas ;初始化dsmov ds,ax;輸入M提示lea dx,inputMmov ah,9int 21h;輸入Mlea dx,M_max_lenmov ah,10int 21h;M轉(zhuǎn)成十進制數(shù)lea si,M_lencall translate_to_numbermov M,ax;輸入N提示lea dx,inputNmov ah,9int 21h;輸入Nlea dx,N_max_lenmov ah,10int 21h;N轉(zhuǎn)成十進制數(shù)lea si,N_lencall translate_to_numbermov N,ax ;輾轉(zhuǎn)相除法mov ax,Mmov bx,Ncmp ax,0jle no_pluscmp bx,0 jle no_pluss:xor dx,dx div bx cmp dx,0 jz breakmov ax, bxmov bx,dxjmp s;輸出R結(jié)果break: lea dx,outputmov ah,9int 21hmov ax,bx call decimaljmp exit;錯誤數(shù)據(jù)no_plus:lea dx,error_numbermov ah,9int 21hexit:retmain endp;字符串轉(zhuǎn)換為十進制數(shù) translate_to_number proc near ;si:lenght firstpush cx push dx push bx push si push di mov di,10mov ax,0mov cl,[si]mov ch,0 cmp cx,0 jz errinc sitran:mov bl,[si]inc sicmp bl,'0'jb errcmp bl,'9'ja err sub bl,30hxor bh,bh mul di add ax,bx loop tranjmp exiterr:lea dx,error_numbermov ah,9int 21h mov ax,4c00hint 21hexit:pop di pop si pop bx pop dxpop cx ret translate_to_number endp;輸出十進制數(shù) decimal proc near push axpush cxpush dxpush bxmov bx,ax cmp ax,0jge plusmov dl,'-'mov ah,2int 21hneg bx mov ax,bx plus:mov cx,0mov bx,10 de:xor dx,dx div bx push dx inc cx cmp ax,0hjnz de de1:pop dx add dl,30hmov ah,2 int 21hloop de1 pop bx pop dx pop cx pop ax retdecimal endpcodes endsend main

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