逐个输入整数,直到输入0为止

//prob.cpp:Definestheentrypointfortheconsoleapplication.//#include"stdafx.h"#include//#includeusingn

#include/**希望你自己能够看得懂..利用了递归的思想*//**@author:banxi1988*@date:2010-12-12**/intcanuse(intx,intk){if(x

PrivateSubCommand1_Click()Dimaa=InputBox("请输入一个数")PrintStrReverse(a)EndSu

PrivateSubCommand1_Click()Doa=Val(InputBox("输入整数,-1结束:"))IfaMod2=0Thens=s+1LoopUntila=-1Print"共有";s;

inti,sum=0;do{scanf("%d",&i);sum=sum+i;}while(i!=0);再问：intsum=0,n=0;scanf("%d",&n);while(n!=0){sum=s

#includevoidmain(){inta=0;intb;intc=0;do{\x05scanf("%d",&b);\x05if(b!=0)\x05{\x05\x05if(b%2==0)\x05\

inta,index;inti;while(1){cout

#include“stdio.h”voidmain(){intn;//输入的整数longsum=0;所有输入数的和,定义为long是为了防止溢出intflag=1;作为一个标记数while(flag)

#includeintmain(){intx,y;while(scanf("%d%d",&x,&y)==2&&(x!=0||y!=0)){\x09printf("%d\n",x+y);}return0

你那是lindo语法不要跟lingo语法混用min=100*x1+100*x2+40*y1+40*y2+40*y3+40*y4+40*y5;x1+x2+y1>=4;x1+x2+y1+y2>=3;x1+

已经调试,请采纳,如需注释,请追问另,负数都返回0,如需返回负数整数个数,使用fabs,或将字符串指针挪到符号之后.#include <stdio.h>#include 

同学你是女的吧,我亲自来指导你~

importjava.util.Scanner;publicclassMain{publicstaticvoidmain(Stringargs[]){Scannerinput=newScanner(S

intmain(void){\x09inti=0,j=0,n;\x09printf("\ninputnumbers:");\x09scanf("%d",&n);\x09//\x09\x09scanf(

.你把做为结束符的0也算进去了吧.在循环里判定一下,如果是0不做MIN==NUMif（NUM!=0）MIN=MIN

#include <stdio.h>#include <string.h>main(){\x05int n=0;\x05int s=0;

#include#includeintmain(){inta,sum=0;do{scanf("%d",&a);if((a/100)%10==3)sum+=a;}while(a!=0);printf("

先看它一共有几位假设有54321位,该数字除以10000取模,就是第一位数字5,然后减去50000,得4321,继续

#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){

-1-8*9=-73绝对值小于100-73-8*9=-145绝对值大于100返回最后输出的结果是-145