mov ah,09h int 21h lea dx,newline mov ah,09h int 21h jmp s1 s2:
mov cx,ax lea dx,newline mov ah,09h int 21h mov ax,cx pop dx pop cx
pop bx ret
input endp 输入： (ax)=年份 judge proc near push bx
；现场恢复 9
；判断一个年份是否闰年 push cx push dx
push ax ；保存ax;输出要判断的年份 mov cx,4 lea bx,inbuff add bx,2 lop2:
mov dl,byte ptr [bx] mov ah,02h int 21h inc bx
loop lop2 ; 循环输出年份正确时结果中的年份 lop3: pop ax mov dx,0
mov cx,ax ；保存ax到cx中 mov bx,4 ；判断是否能被4整除 div bx cmp dx,0 jne not leap mov ax,cx
10
mov bx,100 ；判断是否能被100整除 div bx cmp dx,0 jne is leap mov ax,cx
mov bx,400 div bx cmp dx,0 jne notleap isleap: lea dx,yes mov ah,09h int 21h jmp s3 notleap: lea dx,no mov ah,09h int 21h s3:
lea dx,newline mov ah,09h
；判断是否能被400整除 ；换行 11
int 21h ；输出年份正确时结果的剩余部分 pop dx pop cx pop bx ret judge endp code ends end start
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