编程:用if语句判断一个数的奇偶性。 编写一个程序，判断任意输入的整数的奇偶性。
\u6c42\u7528if\u8bed\u53e5\u6c42\u4e00\u4e2a\u6570\u7684\u5947\u5076\u6027\u7684javascript\u4ee3\u7801\u8bf7\u8f93\u5165\u4e00\u4e2a\u6570\u5b57:function check(){ var value = document.getElementById('text'); if(!value.match(/^-?\d+$/)){ alert('\u8f93\u5165\u6709\u8bef'); return; } if(value%2){ alert('\u5947\u6570'); }else{ alert('\u5076\u6570'); }}
\u6574\u6570\u7684\u5947\u5076\u6027\uff1f\u662f\u5224\u65ad\u6570\u5b57\u662f\u5947\u6570\u8fd8\u662f\u5076\u6570\u5417\uff1f
\u53ea\u8981\u9664\u4ee52\u4f59\u6570\u4e3a0\u7684\u5c31\u662f\u5076\u6570\uff0c\u4e0d\u4e3a0\u5c31\u662f\u5947\u6570\u3002\u4e5f\u4e0d\u77e5\u9053\u4f60\u662f\u8981\u54ea\u79cd\u8bed\u8a00\u7f16\u7a0b\uff0c\u7b80\u5355\u5199\u4e0b\uff1a
t= x mod 2 //x\u662f\u8f93\u5165\u7684\u6574\u6570\uff0ct\u7528\u6765\u5b58\u653e\u4f59\u6570 \u8fd9\u6761\u4e5f\u53ef\u4ee5\u76f4\u63a5\u5199\u5230if\u7684\u5224\u65ad\u6761\u4ef6\u91cc\u53bb\uff0c\u5c31\u7701\u4e00\u884c\u4ee3\u7801\u4e86
if(t) \u8f93\u51fa\u201c\u8fd9\u662f\u4e00\u4e2a\u5947\u6570\u201d
else \u8f93\u51fa\u201c\u8fd9\u662f\u4e00\u4e2a\u5076\u6570\u201d
判断奇偶性的方法如下：
定义一个变量的num，初始化为0
键盘输入数字，并赋予num
一个数能被2整除，则为偶数。
一个数不能被2整除，则为奇数
输出判断结果
#include <stdio.h>
int main() {
int num = 0;
scanf("%d",&num);
if (0 == num % 2)
{
printf("%d is an even number
",num);
} else{
printf("%d is an odd number
",num);
}
return 0;
}
执行结果：
5
5是奇数
double num = 10;
if (num % 2 == 0)
{
System.out.pirnt(num+"是偶数")
}else{
System.out.print(num+"是奇数")
}
if(shu%2)
{ } //奇数
else
{ } //偶数
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