C#用两个随机数组合,但它必须等于用户输入

Код*_*145 1 c# random math sum

我想知道是否有可能从用户输入int得到一个总和但每次总和必须是radom.

这是我制作的代码:

    public static string genSum(int askedNumber)
    {
        string outputStr = null;

        bool switchOneTwo = false;
        int neededSum = 0;

        int intOne = 0;
        int intTwo = 0;

        while (askedNumber != neededSum)
        {

            if (switchOneTwo == true)
            {
                intOne += 1;
                switchOneTwo = false;
            }
            else
            {
                intTwo += 1;
                switchOneTwo = true;
            }

            neededSum = intOne + intTwo;

            if (neededSum == askedNumber)
            {
                if (neededSum >= 4)
                {
                    Random randomInt = new Random();

                    int tmpIntOne = intOne;
                    int tmpIntTwo = intTwo;
                    int method;

                    //For now only one method
                    method = randomInt.Next(1, 1);

                    if (method == 1)
                    {
                        tmpIntOne = tmpIntOne / 2;
                        tmpIntTwo = (tmpIntTwo * 2) - tmpIntOne;
                    }

                    intOne = tmpIntOne;
                    intTwo = tmpIntTwo;                
                }

                outputStr = "(" + intOne.ToString() + "+" + intTwo.ToString() + ")";

                return outputStr;
            }
        }
        return outputStr;
    }
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所以我想要的是,如果一个用户输入一个例如20的数字,它将会得到一笔:

用户输入20并按GO!:

结果:

10 + 10 = 20

12 + 8 = 20

5 + 15 = 20

1 + 19 = 20

3 + 17 = 20

用户输入500并按GO!:

结果:

9 + 491 = 500

263 + 237 = 500

300 + 200 = 500

250 + 250 = 500

109 + 391 = 500

Ody*_*dys 9

创建一个小于输入的随机数,另一个将是User-YourRandom

void Foo(int userNumber)
{
    Random r = new Random();
    int firstNumber = r.Next(userNumber - 1);
    int secondNumber = userNumber - firstNumber;
}
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你可以扩展它以支持底片.

编辑获得真正的随机数

正如I4V所指出的那样,如果你把它放在一个循环中,你可能得到相同的数字集,所以不用调用Random就可以使用这种方法:

void Foo(int userNumber)
{
    int userNumber = 500;

    for (int i = 0; i < 10; i++ )
    {
        int firstNumber = Next(0, userNumber - 1);
        int secondNumber = userNumber - firstNumber;

        Console.WriteLine(firstNumber + "+" + secondNumber);
    }
}

internal static int RandomExt(int min, int max)
{
    RNGCryptoServiceProvider rng = new RNGCryptoServiceProvider();
    byte[] buffer = new byte[4];

    rng.GetBytes(buffer);
    int result = BitConverter.ToInt32(buffer, 0);

    return new Random(result).Next(min, max);
}
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  • 请注意,`new Random()`使用当前时间作为其种子,因此您可能实际上只需要使用一个"静态随机". (3认同)