英文:
Random even numbers on array from 1 to 100
问题
我需要在一个数组上写入20个从1到100的随机偶数。由于某种原因，有些数组值上会得到0。对此有解决方案吗？```javafor (int i = 0; i < 20; i++) {Random random = new Random();int randomNumber = random.nextInt(100) + 1;if (randomNumber % 2 == 0) {arregloAleatorio[i] = randomNumber;}}
打印结果，例如：
0 26 0 4 0 14 0 78 0 16 0 0 86 10 0 0 0 72 34 70
但我不希望数组中出现0。
<details><summary>英文:</summary>I need to write on an array 20 random numbers from 1 to 100, but just even numbers.. For some reason i'm getting 0 on some array value, any solution for this?
for (int i = 0; i < 20; i++) {
Random random = new Random();
int randomNumber = random.nextInt(100) + 1;
if (randomNumber % 2 == 0) {
arregloAleatorio[i] = randomNumber;
}
The print is, for example:
0 26 0 4 0 14 0 78 0 16 0 0 86 10 0 0 0 72 34 70
But I don't want 0 in the array.</details># 答案1**得分**: 1我建议先考虑约束条件。我们希望生成的数字是偶数，并且介于2和100之间。这组数字的大小为50。因此，我们可以生成一个介于0和49（包括在内）之间的随机数，然后加1并乘以2，从而得到所需的数字。代码：```javaint[] arr = new int[20];for (int i = 0; i < 20; i++) {Random random = new Random();int randomNumber = random.nextInt(50) + 1;arr[i] = randomNumber * 2;}
英文:
I would suggest to think about the constraints first. We want to the generated numbers to be the even numbers, between 2 and 100. This group of numbers, is of the size of 50. Therefore, we can generate a random number between 0-49 (Inclusive), and then add 1 and multiply by 2, and get the required number. The code:
int[] arr = new int[20];for (int i = 0; i < 20; i++) {Random random = new Random();int randomNumber = random.nextInt(50) + 1arr[i] = randomNumber * 2;}
答案2
得分: 0
尝试以下内容：
int[] arregloAleatorio = new int[20];for (int i = 0; i < 20; i++) {Random random = new Random();int randomNumber = random.nextInt(101);while (randomNumber % 2 != 0 || randomNumber == 0){randomNumber = random.nextInt(101);}arregloAleatorio[i] = randomNumber;}
如果数字是奇数，则不将其添加到数组中，因此默认为零。
此解决方案生成随机数，直到它不为0或奇数。
英文:
Try the following:
int[] arregloAleatorio = new int[20];for (int i = 0; i < 20; i++) {Random random = new Random();int randomNumber = random.nextInt(101);while (randomNumber % 2 != 0 || randomNumber == 0){randomNumber = random.nextInt(101);}arregloAleatorio[i] = randomNumber;}
You are not adding a number to the array if it is odd, therefore by default it was zero.
This solution generates a random number until it is not 0 or odd.
答案3
得分: 0
因为你的迭代在数字为奇数时仍然继续进行，所以会出现这种情况；当你输出数组时，由于该索引处没有数字，它看起来像是 0！快速的解决方法就是，如果数字是奇数，只需减少 i 的值。我只添加了一行代码。我没有尝试过，但应该能够解决问题。
for (int i = 0; i < 20; i++) {Random random = new Random();int randomNumber = random.nextInt(100) + 1;if (randomNumber % 2 == 0) {arregloAleatorio[i] = randomNumber;} else i--;}
英文:
This is happening because your iteration continues even if the number is odd, and when you are outputting the array, it appears to be a 0 because there is no number at that index! The quick fix would be to just decrement i if the number is odd. I am adding only 1 line of code. I didn't try it but it should work.
for (int i = 0; i < 20; i++) {Random random = new Random();int randomNumber = random.nextInt(100) + 1;if (randomNumber % 2 == 0) {arregloAleatorio[i] = randomNumber;}else i--;
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