随机生成可被N整除的数的最佳算法

Is there a better agorithm to do the following?

有更好的算法来实现以下目标吗?

I am trying to generate 50 random numbers that are divisible by 7. I then select one of those 50 at random & return that number.

我试着生成50个随机数字可以被7整除。然后我随机选择其中的一个，并返回那个数字。

Is there a more efficient/better way to randomly generate numbers that are divisible by 7? Is there a better way I could code/do this?

有没有更有效/更好的方法来随机生成能被7整除的数?有更好的方法来编码/做这个吗?

    unsigned int generateRandomNumberDivisibleByN( unsigned int n, unsigned int num=10 )
    {
        // Post: Generate many different random numbers that are divisible by n, then randomly select one of
        //       of those numbers to return.

        unsigned int potentialNums[num];

        for (int i=0, j=2; i<num; i++, j=rand()%INT_MAX)
        {
            potentialNums[i] = j*n;
        }

        return potentialNums[ rand()%num ]; // should this be rand()%(num-1) so it never returns an invalid array index?
    }

4 个解决方案

#1

Why can't you just do this?

你为什么不能这么做?

return (rand() % MAX) * 7;

It does almost the same thing.

几乎是一样的。

Where MAX is small enough to avoid overflow during the multiplication by 7. Which you can define as:

当最大值足够小的时候，可以避免在7的乘法运算中溢出。你可以将其定义为:

const MAX = INT_MAX / 7;

Or if you want it to be fast, you can do something like:

或者如果你想要它快一点，你可以这样做:

return (rand() & 0xff) * 7;

#2

The most efficient way to randomly generate a number divisible by 7 is to generate a random number and then multiply it by 7.

随机生成一个能被7整除的数的最有效的方法是生成一个随机数，然后乘以7。

return ( rand() % ( INT_MAX / 7 ) ) * 7

#3

Why generate a bunch of random numbers, then choose one of them at random? You have the core idea already: generate a random number and multiply it by 7. The potentialNums array doesn't add any value.

为什么生成一堆随机数，然后随机选择其中一个呢?您已经有了核心思想:生成一个随机数并将其乘以7。potential alnums数组不添加任何值。

#4

To make this much faster first use the fast random generation function found here this should make the actual process of creating randoms much faster,

为了让这个更快地使用快速随机生成函数这应该会让创建randoms的过程变得更快，

Next, you can multiply the random by 7 and it will always be a multiple of 7, however, if this multiplication causes an overflow of the int (which with randoms is likely) you can always take a mask of the first few bits and multiply them.

接下来，你可以将随机数字乘以7，它将永远是7的倍数，然而，如果这个乘法导致了int(很可能是randoms)的溢出，你总是可以取前几个位的掩码并将它们相乘。

e.g.

如。

randomNumber = (generatedRandom & 0x00FFFFFF) * 7

随机数= (generatedRandom & 0x00FFFFFF) * 7。

#1