Question

因此我需要使用随机数制作2维数组，并将数字排除在乘法表中。它应该如下所示：

我现在的代码如下：

$(function () {

// Ensure if our request context doesn't have an ID we at least get something back.
function getContextRequestID(settings, defaultIfNotSet)
{
    if (typeof settings.context.requestID !== "undefined")
    {
        return settings.context.requestID;
    }
    return defaultIfNotSet;
}

function writeToJsResult(method, settings)
{
    var DefaultRequestID = -1;

    var requestID = getContextRequestID(settings, DefaultRequestID);

    $(".js-result").append('<p>' + method + ': Ajax request for page -->  ' + settings.url + ' with ID of: ' + settings.context.requestID + '</p>');
}

$(document).ajaxSend(function (event, jqxhr, settings) {
    writeToJsResult("Starting", settings);
});

$(document).ajaxError(function (event, req, settings) {
    writeToJsResult("Error", settings);
});

$(document).ajaxComplete(function (event, req, settings) {
    writeToJsResult("Complete", settings);
});

/* triggering all Ajax requests */
$('.js-btn-json').on('click', function () {
    loadBBC();
    loadBBC();
});

function loadBBC()
{
    loadSite("http://www.bbc.com");
}

function loadCNN()
{
    loadSite("http://www.cnn.com");
}

function loadSite(url)
{
    // Not sure how you want to correlate them. For now using random. 
    // Perhaps a global running counter
    // Perhaps a GUID/UUID
    // http://stackoverflow.com/questions/105034/create-guid-uuid-in-javascript
    var requestID = getRandomInt(0, 100);

    $.ajax({
        url: url,
        context: {
            requestID: requestID
        }
    });
}

function getRandomInt(min, max) {
    return Math.floor(Math.random() * (max - min + 1)) + min;
}

编辑:::这对我有用！

Random random = new Random();
int[,] array = new int[10, 10];
for (int i = 0; i < 10; i++)
{
    for (int j = 0; j < 10; j++)
    {
        array[i, j] = random.Next();
        Console.Write("\t{0}\t{1}" ,i,j);
    }

Answer 1

正如ThatBrianDude所建议的

static void Main(string[] args)
    {
      Random random = new Random();

      int[,] array = new int[10, 10];

      for (int i = 0;
           i < 10;
           i++)
      {

        for (int j = 0;
             j < 10;
             j++)
        {
          array[i,
                j] = random.Next();
          Console.Write("\t{0}",
                        array[i,
                              j]);
        }
        Console.Write("\n");
      }
      Console.ReadKey();
    }

Answer 2

可以像这样生成一个10x10的随机数数组：

var rnd = new Random();

var randomArray = Enumerable.Range(0,10).Select(x => {
    return Enumerable.Range(0,10).Select(y => rnd.Next()).ToArray();
}).ToArray();

实例：http://rextester.com/GWTAK54347

Answer 3

我认为这应该适合你：

audioManager.setStreamVolume(AudioManager.STREAM_VOICE_CALL, i, 0); // i = seekbar progress

c＃2维数组随机数

3 个答案: