输入一个数 判断其为奇数还是偶数,如果是奇数则进一步判断它是否为5的倍数

voidmain(){inta;scanf("%d",&a);if(a%2==0){printf("偶数");}else{printf("奇数");}}programJO;vara:integer;b

#includeintmain(){intn;printf("请输入一个整数:");scanf("%d",&n);if(n>0){if(n%2)printf("正奇数\n");elseprintf("

#include <stdio.h>void main(){    int n;   &

#includevoidmain(){inta;printf("Pleaseinputanumber:");scanf("%d",&a);if(a==0)printf("%disaevennumber

main(){intn;printf("pleaseinputanumber\n");scanf("d%",&n);if(n%2==0)printf("thenumberisoushu");elsep

DimXAsInteger这句出的毛病.你直接把它定义成整数型变量,输入3.5直接四舍五入转换为4.把它换成DimXAsSingle

个位是2的倍数:偶数个位不是2的倍数:奇数数一下奇数有多少个,如是偶数个,则是偶数如是奇数个,则是奇数奇数：1,3,5,7……偶数:2,4,6,8……再问：数一下奇数有多少个,如是偶数个,则是偶数如是

不知道楼主的编译器是不是和我的一样,不支持longdouble类型,我的改成这样就可以了：#include <stdio.h>int main(void){\x05do

楼上是C语言的.我补充VB的.设你的数是num（整型）.a=nummod2若a=1则num为奇数,若a=0则num为偶数.

能被2整除的整数是偶数,不能被2整除的整数是奇数.核心部分用%就可以了if(x%2==0)输出“YES”else输出“NO”

供参考……#include"stdio.h"//#include"math.h"//voidmain(void){intn,i,cx;printf("请输入一个整数:\n");scanf("%d

#includeintmain(void){longi;printf("请输入要判断的数\n");scanf("%ld",&i);if(i%2==0){printf("您输入的%ld是偶数\n",i)

#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(){    int num;    

方法一:movax,1234horax,ax;或运算jnpn1;奇转jpen2;偶转...方法二:movax,1234htestax,1;与测试jzt1;奇转jnzt2;偶转...

package zhidao;import java.util.InputMismatchException;import java.util.Scanner;publi

#include"stdio.h"main(){intx;printf("请输入一个整数");scanf("%d",&x);if(x%2==0)printf("%d是偶数\n",x);elseprin

importjava.util.*;publicclassMath{publicstaticvoidmain(Stringargs[]){inta,b;Scannersc=newScanner(Sys

#include"stdio.h"intmain(){      intn;     &n

#includemain(){intnum,yushu;printf("Pleaseenterthenumber\n");scanf("%d",&num);yushu=num%2;if(yushu==