如何使随机数独一无二

我正在制作一个随机数生成器。它询问用户希望在数字中包含多少位数字。例如，如果他们输入2，它将生成10到99之间的随机数。我制作了生成器，但我的问题是数字不是唯一的。

这是我的密码。我不知道为什么它不生成唯一的数字。我以为srand（时间（空））能做到。

void TargetGen::randomNumberGen()
{
srand (time(NULL));
if (intLength == 1)
{
    for (int i = 0; i< intQuantity; i++)
    {
        int min = 1;
        int max = 9;
        int number1 = rand();
        if (intQuantity > max)
        {
            intQuantity = max;
        }
        cout << number1 % max + min << "t";
    }
}
else if (intLength == 2)
{
    for (int i = 0; i<intQuantity; i++)
    {
        int min = 10;
        int max = 90;
        int number1 = rand();
        if (intQuantity > max)
        {
            intQuantity = max;
        }
        cout << number1 % max + min << "t";

    }
}
if (intLength == 3)
{
    for (int i = 0; i<intQuantity; i++)
    {
        int min = 100;
        int max = 900;
        int number1 = rand();
        if (intQuantity > max)
        {
            intQuantity = max;
        }
        cout << number1 % max + min << "t";
    }
}
else if (intLength == 4)
{
    for (int i = 0; i<intQuantity; i++)
    {
        int min = 1000;
        int max = 9000;
        int number1 = rand();
        if (intQuantity > max)
        {
            intQuantity = max;
        }
        cout << number1 % max + min << "t";
    }
}
if (intLength == 5)
{
    for (int i = 0; i<intQuantity; i++)
    {
        int min = 10000;
        int max = 90000;
        int number1 = rand();
        if (intQuantity > max)
        {
            intQuantity = max;
        }
        cout << number1 % max + min << "t";
    }
}
else if (intLength == 6)
{
    for (int i = 0; i<intQuantity; i++)
    {
        int min = 100000;
        int max = 900000;
        int number1 = rand();
        if (intQuantity > max)
        {
            intQuantity = max;
        }
        cout << number1 % max + min << "t";

    }
}
if (intLength == 7)
{
    for (int i = 0; i<intQuantity; i++)
    {
        int min = 1000000;
        int max = 9000000;
        int number1 = rand();
        if (intQuantity > max)
        {
            intQuantity = max;
        }
        cout << number1 % max + min << "t";
    }
}
else if (intLength == 8)
{
    for (int i = 0; i <intQuantity; i++)
    {
        int min = 10000000;
        int max = 89999999;
        int number1 = rand();
        if (intQuantity > max)
        {
            intQuantity = max;
        }
        cout << number1 % max + min << "t";
    }
}
if (intLength == 9)
{
    for (int i = 0; i < intQuantity; i++)
    {
        int min = 100000000;
        int max = 900000000;
        int number1 = rand();
        if (intQuantity > max)
        {
            intQuantity = max;
        }
        cout << number1 % max + min << "t";
    }
}
}

好吧，我想我找到了一种不用数组的方法，但在我切换到fisher-yates方法之前，它不起作用。有人能告诉我为什么这不起作用吗？它本质上应该是将随机数放入变量numGen中。然后在变量b=到numgen中。只是为了保持numGen过去的值，所以当循环通过并生成另一个随机数时，它会将其与旧数字进行比较，如果不等于旧数字，它就会输出它。如果它等于旧数字而不是输出它，它会解增量i，这样它就可以在循环中运行，而不会完全跳过这个数字。然而，当我这样做的时候，是无限循环。我不知道为什么。

if（intLength==1）｛

    for (int i = 0; i< intQuantity; ++i)
    {
        int min = 1;
        int max = 9;
        int number1 = rand();
        int numGen = number1 % max + min;

        if (intQuantity > max)
        {
            intQuantity = max;
        }
        for (int k = 0; k < 1; k++)
        {
            cout << numGen << "t";
            int b = numGen;
        }
        int b = numGen;
        if (b != numGen )
        {
            cout << numGen << "t";
        }
        else
        {
            i--;
        }
    }
}