Question

我想让程序遍历从00000000到11111111的每个可能的二进制数，并且我要计算连续＆＃34;运行的数量＆＃34;一些。例）00000001和11100000都算作单次运行 00001010和11101110都算作两次运行

问题是，它是否忽略了AND掩码部分，我不知道为什么。

{
    static void Main(string[] args)
    {
        //Start
        int stuff = BitRunner8();

        //Display
        Console.Write(stuff);
        Console.ReadKey();
    }
    public static int BitRunner8()
    {
        int uniRunOrNot = 0;
        int uniRunCount = 0;
        int uniRunTotal = 0;

        //iterating from numbers 0 to 255
        for (int x = 0; x < 255; x++)
        {
            //I use 128 as my AND mask because 128 is 10000000 in binary
            for ( int uniMask = 128; uniMask != 0; uniMask >>= 1)
            {

                //This is the if statement that doesn't return true ever
                if ((x & uniMask) != 0)

                {
                    //If the and mask is true, and ther were no previous ones before it, add to the the uniRunCount
                    if (uniRunOrNot == 0)
                    {
                        //Total count of the runs
                        uniRunCount++;
                    }
                    // Making it so that if two consective ones are in a row, the 'if' statement right above would return false,
                    //so that it wouldn't add to the uniRunCount
                    uniRunOrNot++;
                }
                else
                {
                    //add the total number of runs to uniRunTotal, and then reset both uniRunOrNot, and uniRunCount
                    uniRunTotal += uniRunCount;
                    uniRunOrNot = uniRunCount = 0;
                }
            }
        }
        //Divide the final amount by 256 total numbers
        uniRunTotal /= 256;
        return uniRunCount;
    }
}

Answer 1

问题是您的代码忽略了包含最低有效位的运行。只有在发现零位时，您的代码才会更新uniRunTotal。当最低有效位为非零时，uniRunCount永远不会添加到总数中。

在循环后添加代码以添加uniRunCount来解决此问题。

您还可以通过应用 sentinel 策略解决此问题：从另一端计数位，并使用9位而不是8位，因为第9位始终为零：

for (int uniMask = 1; uniMask <= 256; uniMask <<= 1)

使用c＃计算二进制数中连续数的数量

1 个答案: