Diagonal表示所有x和y坐标相同的地方如果你的数组包含：

13 8 5

339 7
4 457
5 1 74
然后对角线是粗体。

假设数组是正方形：

int sum = 0;

for(int i = 0; i < numOfArrayRows; i++)
{
    //Use the mod operator to find if the value is even.
    if(array[i][i] % 2 == 0)
        sum += array[i][i];

    //Change 0's to ones
    for(int j = 0; j < numOfArrayCols; j++)
        if(array[i][j] == 0)
            array[i][j] = 1;
}

另外，下次如果你有家庭作业的问题，请加上“家庭作业”标签：P

使用二维数组非常简单，因为你不需要任何索引魔法：

int a[N][N] = ...;
int sum = 0;
for(int i=0; i<N; ++i)
    if(a[i][i] % 2 == 0) //or a[i] & 1, if you like, just check if it's dividable by 2
        sum += a[i][i];

这段C++代码在C或C#中应该没有太大的不同，但你应该明白这一点。同样，第二个问题也很简单：

int a[M][N] = ...;
for(i=0; i<M; ++i)
    for(j=0; j<N; ++j)
        if(a[i][j] == 0)
            a[i][j] = 1;

我怀疑主对角线是从坐标0，0开始的。
要将0元素替换为1，您可以执行以下操作：
if（array[i，j] == 0）array[i，j] == 1;

这听起来像家庭作业-但我会帮助：）
因此，如果你有一个2D数组，为了找到对角线值的总和，你将知道两个值的索引将匹配，以便为你提供每个对角线值。

要迭代这些值，您可以使用一个简单的循环来对每个对角线值求和，如图所示：

//Your Sum
int sum = 0;

//This will iterate and grab all of the diagonals
//You don't need to iterate through every element as you only need
//the diagonals.
for(int i = 0; i < sizeOfArray; i++)
{
   //This will add the value of the first, second, ... diagonal value to your sum
   sum += array[i,i];
}

要将每个值设置为0到1，您可以遍历数组的每个元素并检查值是否为0，然后将该值设置为1，例如：

for(int i = 0; i < sizeOfArray; i++)
{
    for(int j = 0; j < sizeOfArray; j++)
    {
         //Check if this value is 0;

         //If it is 0, set it to 1, otherwise continue
    }
}

int[,] array = new int[,] {{1,2,3},
                           {4,5,6},
                           {7,8,9}};
    //Suppose you want to find 2,5,8
    for(int row = 0; row < 3; row++)
    {
       for(int column = 0; column < 3; column++)
       {
          if((row == 0 && column == 1) || (row == 1 && column == 1) || (row == 2 && column == 1))
          {
             Console.WriteLine("Row: {0} Column: {1} Value: {2}",row + 1, column + 1, array[row, column]);
          }
       }
    }

下面是你需要的代码，没有太多解释：

//Read the size of the array, you can get it from .Count() if you wish
        int n = Convert.ToInt32(Console.ReadLine());
        int[][] a = new int[n][];
        //Reading all the values and preparing the array (a)
        for (int a_i = 0; a_i < n; a_i++)
        {
            string[] a_temp = Console.ReadLine().Split(' ');
            a[a_i] = Array.ConvertAll(a_temp, Int32.Parse);
        }
        //performing the operation (google what diagonal matrix means)
        int PrimarySum = 0, SecondarySum = 0;
        for (int i = 0; i < n; i++)
        {
            //The If condition is to skip the odd numbers
            if (a[i][i] % 2 == 0) { PrimarySum += a[i][i]; }
            //For the reverse order
            int lastelement = a[i].Count() - 1 - i;
            if (a[i][lastelement] % 2 == 0) { SecondarySum += a[i][lastelement]; }
        }
        //Get the absolute value
        Console.WriteLine(Math.Abs(PrimarySum - SecondarySum).ToString());
        Console.ReadKey();

C#我会这样做
int[，] matrix = new int[4，4];

for (int i = 0; i < matrix.GetLength(0); i++)
        {
            for (int j = 0; j < matrix.GetLength(1); j++)
                if (matrix[i, i] == matrix[j, j]) { 
                matrix[i, j] = 1;
                }
                else
                {
                    matrix[i, j] = 0;
                }
        }