取两个数组:
int[] arr2 = new int[5]; int[] arr3 = new int[5];
现在,如果数组元素除以2得到的余数为0,则为偶数。获取这些元素并添加另一个数组。这遍历数组的长度:
if (arr1[i] % 2 == 0) {
arr2[j] = arr1[i];
}在else条件下,您将获得奇数元素。将它们添加到单独的数组中,并分别显示它们,如以下示例所示:
using System;
namespace Demo {
public class Program {
public static void Main(string[] args) {
int[] arr1 = new int[] {
77,
34,
59,
42,
99
};
int[] arr2 = new int[5];
int[] arr3 = new int[5];
int i, j = 0, k = 0;
for (i = 0; i < 5; i++) {
if (arr1[i] % 2 == 0) {
arr2[j] = arr1[i];
j++;
} else {
arr3[k] = arr1[i];
k++;
}
}
Console.WriteLine("Even numbers...");
for (i = 0; i < j; i++) {
Console.WriteLine(arr2[i]);
}
Console.WriteLine("Odd numbers...");
for (i = 0; i < k; i++) {
Console.WriteLine(arr3[i]);
}
}
}
}输出结果
Even numbers... 34 42 Odd numbers... 77 59 99