在Java中生成以N的倍数递增的随机数

Question

I need to write a short randomizer which generates a random number between 1 to N where the random number is increasing by multiples of M.

For example: generate numbers between 1 and (N=30) increasing by multiples of (M=5). The only possible generated numbers can be then: 1,5,10,15,20,25 and 30. Hope you get what I mean :)

Normally if you use new Random().nextInt(30)+1 , you get numbers increasing by multiples of 1 (1,2,3,4,5,6,7,etc.). That is not what I want.

Any help, links, or directions are very appreciated?

EDIT :

That the sequence of generated random numbers could include 1 not zero is one of the requirements of the method. Precisely, the sequence always starts with the lower bound (min). In the example the lower bound is 1 and therefore there can't be zero in the sequence. It's odd I know, but those are the requirements I have to follow ;)

Answer 1

You could multiply the random value by 5. Since you want to have 1 instead of 0 just look for that value and alter it specifically:

static final Random random = new Random(System.currentTimeMillis());

public int random(int range, int multiple) {
    int value = random.nextInt(range / multiple) * multiple;
    return value == 0 ? 1 : value;
}

Answer 2

I think that your example is a little bit flawed: the sequence should be (0, 5, 10, 15, 20, 25, 30)

If I'm correct, you can then use:

new Random().nextInt(7) * 5; // will generate a number between 0 and 7 not included then multiply by 5 which should give what you want

If you want to start by 1, then just add it to the previous statement

Answer 3

Try this..

public static void main(String[] args){
    int N = 50;
    int M = 5;
    System.out.println(getRandomNumber(N, M));

}

static int getRandomNumber(int n, int m){
    Random r = new Random();
    int rnd = r.nextInt(n);
    return rnd%m == 0?rnd:getRandomNumber(n, m);
}

Answer 4

int n=Random.range(1,10);
n=n*100;