编写汇编程序段,计算(V-(X*Y Z-100)) X

当然可以啦.另外告诉你,记得我老师曾经说过,解运动学题目时,如果觉得缺个已知条件已知数似的,请想想“平均速度公式”.结果我在高考时真遇上了,所以记得到现在.

assumecs:code,ds:datadatasegmentxdwydwzdwresultdwdataendscodesegmentstart:movax,datamovds,axmovax,xi

#includeusingnamespacestd;doublefun(doublex){\x05if(x>-10&&x-5&&x5&&xx;\x05\x05cout再问：麻烦用C语言写一下再答：#i

PrivateSubCommand1_Click()DimnAsLong,xAsSingle,yAsSingle,y2AsSingle,sumAsSinglex=Val(InputBox("请输入一个

DATASEGMENTBUFDW12,0,-13,45,-23,0,-34,-12,34ZHENGDB0FUDB0COUNTEQU(ZHENG-BUF)/2DATAENDSCODESEGMENTASS

#includeusingnamespacestd;intpow(intx,intn);intf1(intn);intmain(intargc,char*argv[]){intx,n;doubleex

#includevoidmain(){intx,y;//你用的是浮点型,我改成整形的了...你如果真的要用浮点的话...要加上小数点,如2.0等...printf("x=");scanf("%d",&

下面的【deccl】说明每循环一次cl减去1,再下面的【jnzlooper】说明cl为0时结束循环.所以,目测,用于加载循环次数再问：第二句话movcl，[bx]这步执行了以后cl的内容是什么呢？我理

unsignedchari;unsignedintsum;sum=0;for(i=2;i

多绕几圈,可以实现的.>>result=maple('evalf','(Ei(1,y))')result=Ei(1,y)>>y=2y=2>>result=subs(result)result=Ei(1

#includepower(floata,floatb){floatsum1=a*b;returnsum1;}intmain(){floatx,y;floatsum;printf("inputthef

#include#includeintmain(){doubled;scanf("%lf",&d);printf("|y|=%.2lf\n",fabs(d));return0;}

ifx>=0theny=xelsey=-xendifprinty

等会帮你编一个.编好了：datasegmentxdw?ydw?zdw?wdw?dataendsstacksegmentstack\x09db32dup(?)stackendscodesegmentas

% x为一个数或者数组都可以计算function [y]=f(x)[m,n]=size(x);y=zeros(m,n);for i=1:m*ny(i)=(x(i)>

;MOVAX,AANDAX,B;AX=aANDbMOVBX,AXORBX,B;BX=aXORbADDAX,BXADDAX,BX;AX=2*(aXORb)+aANDbADDAX,A;AX=a+2*(aX

;X86汇编_求0-9立方,显示结果;MASM5.0编译后,运行无误codesegmentassumecs:code,ds:codeorg100hstart:pushcspopdsAgain:lead

datasegmentmsgxdb'inputx:$'msgydb'inputy:$'msgzdb'inputz:$'msgvdb'inputv:$'wdbdataendscodesegmentass

CMP[SI],0这句才不对应该把[SI]付给一个寄存器里,再比较.

鸟棉b再问：这家是我看过信用最好的再答：请查看33088.info