编写一个方程,要求输入两个正数

#includeintmain(){inta,b;printf("请输入两个整数,以空格隔开（形如12）：\n",a,b);scanf("%d%d",&a,&b);if(b!=0&&a!=0&&a%b

+=functionadd(){varA=document.getElementById("a").value;varB=document.getElementById("b").value;varA

首先明确gbs（最小公倍数）=m*n/gys（最大公约数）；然后求最大公约数用欧几里得辗转相除法；代码很短的.#include"stdio.h"intgys(intm,intn){returnn==0

main(){inta=0,i;scanf("%d",&a);for(i=0;i

#include"stdio.h"//voidmain(void){inta,b,c;printf("请输入两个十进制整数!\na=");scanf("%d",&a);printf("b=")

Reverse

#includeintmain(){intm,n;do{printf("输入两个非零整数:\n");scanf("%d%d",&m,&n);}while(m==0||n==0);printf("%d+

maylab软件挺好编写的用c语言也不难

#includevoidmain(){inta1,a2;floate;scanf("%d%d",&a1,&a2);printf("%d+%d=%d\n",a1,a2,a1+a2);printf("%d

importjava.util.Scanner;publicclassTest{publicstaticvoidmain(String[]args){Scannersc=newScanner(Syst

intHCF(intx,inty)//定义最大公约数函数{inti,change;if(x>y)//保证x是最小数{change=y;x=change;y=x;}for(i=x;i>=1;i--)if

考虑两种情况：（1）f(x)=0只有一根.此时一.若m-2=0,即f(x)为一次函数,此时f(x)=-8x-2有一负根x=-1/4.二.若m-2不为零,方程f(x)=0判别式（4m）^2-4(m-2)

#include#includemain(){\x09longintx;\x09inti=0;printf("input:");\x09scanf("%ld",&x);do{x=x/10;i++;\x

#includevoidmain(){doublea,b;printf("请输入两个数a,和b\n");scanf("%lf%lf",&a,&b);printf("两个数和是:%lf\n",a+b);

#include#include"stdafx.h"#includechare;floatmain(floatx,floaty){voidaction(floatx,floaty);voidactio

staticvoidMain(string[]args)　　{　　stringstr="",math="";　　while(true)　　{　　boolIsOk=true;　　Console.Writ

#includeusingnamespacestd;intmain(){inta,b;cout

#includeintmain(){floata,b,s,m,n,j;scanf("%f",&a,b);s=a+b;m=a-b;n=a*b;j=a/b;printf("%f\n",s,m,n,j);r

#include"stdio.h"main(){inta,b;printf("请输入两个数")；scanf("%d",&a);scanf("%d",&b);printf("a=%d,b=%d,a+b=

#includeusingnamespacestd;intmain(){floata,b,c,d,e,f;cin>>a;cin>>b;if(a!=0&&b!=0){c=a+b;d=a-b;e=a*b;