输入一个四位数怎么将每位输出
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![输入一个四位数怎么将每位输出](/uploads/image/f/7440972-60-2.jpg?t=%E8%BE%93%E5%85%A5%E4%B8%80%E4%B8%AA%E5%9B%9B%E4%BD%8D%E6%95%B0%E6%80%8E%E4%B9%88%E5%B0%86%E6%AF%8F%E4%BD%8D%E8%BE%93%E5%87%BA)
楼主需要什么语言版的再问:C++再答:我只给出反序数你自己琢磨一下大致思路是这样的voidReverse(intm,intlen){if(m/10){cout
//z2.cpp:定义控制台应用程序的入口点.//#include"stdafx.h"#include#includeusingnamespacestd;voidReverse(int&m)//m为4
inta=0;inti=0;intb[10]={0};//存放解析后的数字scanf("%d"&a);for(i=0;;i++){b[i]=a%10;if(abreak;a=a/10;}你把头尾完善一
/*12342103Pressanykeytocontinue*/#include<iostream.h>int main() {\x09int digit1
#includeusingnamespacestd;voidmain(){inta[4];cout再问:还是有好几个错误再答:#includeusingnamespacestd;voidmain(
int类型所能容纳的数字位数不能超过10.我写的这个程序稍微长了点,但不受int类型容量的限制,能够处理很长的整数输入(由buffer数组的大小决定).这程序只处理纯整数输入.有疑问尽管问. 
#includevoidmain(){inti,j,k,m,n;printf("inputanumber:\n");scanf("%d",&n);i=n/1000;j=(n%1000)/100;k=(
#includeusingnamespacestd;intmain(){intn;cin>>n;do{cout
publicstaticvoidmain(String[]arg){intk=0;intm=0;System.out.println("输入一个四位数");Scannerinput=newScanne
#include#includeintmain(){intoldnum;intnewnum=0;inttemp;printf("pleaseinputnumber\n");scanf("%d",&ol
#includevoidmain(){inta[9][9],i,j,n;scanf("%d",&n);for(i=0;i
charsNum[8];intnumber=0;intnCount=0;cin>>number;charstring[25];memset(string,0,25);itoa(number,strin
#includeintmain(){intn,m;scanf("%d",&n);m=0;while(n>0){m=m*10+n%10;n/=10;}printf("%d\n",m);return0;}
#includeusingnamespacestd;intreve_int(intn){intt=0;while(n){t*=10;t+=n%10;n/=10;}returnt;}voidmain()
importjava.util.Scanner;publicclassEncpryt{publicEncpryt(){Scannerscan=newScanner(System.in);System.
用字符数组存储这N个数再用ASCII码将字符类型的数字转换成整数数字再加起来即可;vara:array[1..10000]ofchar;i,s,n:longint;beginfori:=1to4dor
#include"math.h"#includeusingnamespacestd;#defineM4voidmain(){intnum,p[M],i;cout
#include<stdio.h>int main(){int n,s=0;scanf("%d",&n);while(n){s=
intnum=4568;intshiwei=0,baiwei=0,qianwei=0,gewei=0;qianwei=num/1000;baiwei=(num%1000)/100;shiwei=(nu
作为字符串形式读入比用整形读入好处理,而且数的位数不受限制(整型有数值超界问题)#includevoidmain(){chars[80];//最长80位数字inti,L;printf("pleasei