输入一个整数,将其各位上的偶数去掉

#includeintGetNumber(intn)//用递归来实现很简单{intsum=0;if(n/10!=0){\x09sum+=GetNumber(n/10);}sum+=n%10;retur

#includevoidmain(){intx,y,a,b,c;coutx;if((x999)){cout

你没要求具体的语言,用VB给你写一个吧.subchkNumdimNumasintegerdimlenNasintegerdimiasintegerdimsumNasintegerdimstrNasst

#include#includeintisPrime(intnum){inti;if(num

将浮点数先转换为整形数,得到整数部分,再将该实数乘以几百倍,该数减去整数部分得到小数.

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

不对,实质上这种情况该用char类型,并判断,因为int可输入多位,例如输入11,12,14你这种算法就不能反向输出

#includeintmain(void){intn,m,j=0,k=0;printf("请输入一个数;\n");scanf("%d",&n);for(;n>0;){m=n%10;if(m%2==0)

varn,s:longint;beginread(n);s:=0;whilen0dobegins:=s+nmod10;n:=ndiv10;end;writeln(s);end.自创!

#includevoidmain(){usignedintinput,output=0;coutinput;for(inti=intput;i>0;i/=10){intx=i%10;if(i%2==0

#include"stdio.h"intmain(){intcnt;inti,j,max,n;ints[100];intbuf[100];printf("pleaseenternumber\n");s

#includeintjudge(intnum){returnnum%2==0?0:1;}main(){intn;while(scanf("%d",&n)!=EOF){if(judge(n))prin

#includeusingnamespacestd;intmain(){intn;intunit,tens,hund;cin>>n;unit=n%10;tens=n/10%10;hund=n/100;

#includeintmain(void){inti;scanf("%d",&i);if(i%2==0)printf("it'saEven");/*输出偶数*/elseprintf("it'saodd

#include <stdio.h>void main(void){ int input[20]; int a=0,b=0;&n

如该数为m则其十位数为a=m/10整个位数为b=m/10取余

程序:#include(iostream.h)voidmain(void){intn,right_digit,newnum=0;coutn;cout

先把整形转化成String在用数组,

#includeusingnamespacestd;intmain(){inta,m;cin>>a;while(a!=0){m=a%10;a=a/10;cout

inti,a,n;int*pl=NULL;printf("inputn:");scanf("%d",&n);if(n>0){pl=(int*)malloc(n*sizeof(int));}for(i=