输入两个数,输出这两个数,要求先输出其中的小数,再输出大数.

#include"stdio.h"intmain(){inta,b,num;printf("输入一个两位数");scanf("%d",&num);a=num/10;b=num%10;printf("%

C/C++版本,请楼主参考.#include"stdio.h"#include"stdlib.h"intgetMax(inta,intb){if(a>b)returna;elsereturnb;}vo

#includevoidmach(intx,inty){printf("x²+y²=%d\n",(x*x+y*y));printf("x²-y²=%d\n",(

#includeintmain(intargc,char*argv[]){inti,a,b,sum=0,c;scanf("%d%d",&a,&b);if(a>b){c=a;a=b;b=a;}for(i

#include/*最大公约数*/intGreatestCommonDivisor(intm,intn){\x05intr;\x05do\x05{\x05\x05r=m%n;\x05\x05m=n;\

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

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

#includefloatplus(floatm,floatn){floatx;x=m*n;return(x);}voidmain(){floatm,n;printf("请输入求积数据(以空格键隔开)

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

clearinputtoainputtobifa>b?aelse?bendifcancel

求逻辑表达式还是画图?再问：逻辑表达式，不知道怎么抽象再问：麻烦你了再答：再答：列表：A1A0---------Y3Y2Y1Y0000011010100100111111100所以（反斜杠表示“非”）

#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);

PrivateSubCommand1_Click()a=InputBox("请输入数字","提示")b=InputBox("请输入数字","提示")Print"和："&Val(a)+Val(b)Pri

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

//#include"stdafx.h"//vc++6.0加上这一行.#include"stdio.h"voidmain(void){intm,n,k,l;while(1){printf("Typem

#includeintmain(){inta,b,i,c,d,m,n;scanf("%d%d",&m,&n);for(i=m;i>=1;i--){a=m%i;b=n%i;if(a==0&&b==0){

说实话,谁出的题目那么坑爹,没啥技术感,也没算法训练这种说法,就知道猛得叠加任务,实际涉及的技术含量却很低.就是欺负你时间多.下面是用最笨的办法,最直接的办法做的,控制台程序：你看着调用吧!别说调用方

#include#include#includeusingnamespacestd;intmax(inta,intb,intc){returna>b?a>c?a:c:b>c?b:c;}intmain(

#includemain(){inti,j,cnt=0;scanf("%d''"%d",&m,&n);if(m>n){temp=m;m=n;n=temp;}for(i=m;i