C 输入一个4位的整数,然后分别输出组成改4位数的各位数字.

doublea,b;scanf("%lf",&a);b=a;longintp;p=(longint)a;printf("整数部分:%d\n",p);printf("小数部分:%lf",(b-p));再

这设计到计算机的数值表示问题,一般32位的计算机用unsignedlonglong最大能表示的也只到18446744073709551615,所以超过20位做运算时就因为计算机不能表示而出错了.对于这

#include"stdio.h"intmain(){inta[10][10]={0};intn,i,j,sum1=0,sum2=0;scanf("%d",&n);for(i=0;i

#includevoidmain(){intx,a,b,c;scanf("%d",&x);a=x/100;b=x/10%10;c=x%10;x=c*100+b*10+a;printf("%d",x);

/>#include <stdio.h>int main(){\x09int n, cnt = 11;\x09scanf(&qu

#includeintmain(){floatn,n2;intn1;printf("输入一个数：\n");scanf("%f",&n);n1=int(n);n2=n-n1;printf("%d%f",

太简单了吧#includevoidmain(){intA,bai,shi,ge;printf("请输入一个三位整数A：");scanf("%d",&A);bai=A/100;shi=A%100/10;

main(){intx;scanf("%d",&x);x=(x%10)*100+(x/10%10)*10+(x/100)printf("%d\n",x);}

请输入:100357111317192329313741434753596167717379838997Pressanykeytocontinue#include#includeintmain(){

方法一：//用数学函数#include#includevoidmain(){inta;scanf("%d",&a);printf("%d\n",abs(a));}方法二：//判断#includevoi

#include#includeintmain()//这里{\x05intx;doubley,z,num;//这里\x05printf("inputanumber:");\x05scanf("%d",

intiIsRunAround(intnNum){intnCount=0;//位数intnStep=0;intnCur=nNum;intarrNumber[10];//各位数,个位到最高位intarr

voidmain(){inta,b;inty;floatx;printf("请输入2个数字：");scanf("%d%d",&a,&b);x=a/b;y=a%b;printf("商：%0.3f",x)

#includemain(){inta,b;scanf("%d",&a);for(;a>=1;){b=a%10;printf("%d",b);a/=10;}printf("\n");}

#include <stdio.h>void main(){ int n,a[100],i,j,temp;\x09scanf("%d&qu

import java.util.*;public class Demo{    public static 

Subtest()Dimarr(1To3)AsIntegerDimiAsIntegerDimaAsInteger,bAsInteger,cAsIntegerDimxAsStringFori=1To3a

#include#includevoidconv(char*s){inti;for(i=strlen(s)-1;i>=0;i--)putchar(s[i]);printf("\n");}in

#includeintmain(){intn=0;inti=0;intd[4]={0};printf("Inputanum(-99999999):");scanf("%5d",&n);n%=10000

intb[4];intsum=0;b[0]=a/1000;//千位b[1]=a%10;//个位b[2]=(a-b[1])%100;//十位b[3]=(a-b[0]*1000)/100;//百位for(