用c#来编程：创建一个3*3的矩阵（二维数组），求出最大值

Question

显示在lable中，结果显示在另一个lable中

Answer 1

static void Main(string[] args)
{
int[,] aa = new int[3, 5];
Random a = new Random();
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 5; j++)
{

aa[i, j] = a.Next(0, 100);
}
}

for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 5; j++)
{
Console.Write(aa[i, j].ToString() + " ");
}
Console.WriteLine();
}
int k = aa[0, 0];
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 5; j++)
{
if (k < aa[i, j])
{
k = aa[i, j];
}
}
}
Console.Write("最大值为:" + k.ToString());

}本回答被提问者采纳

Answer 2

int a[3,3] = new int[3,3]
{
21,4,5,56,5,7,67,7,8,8
};

int iMax = int.MinValue;
for(int i = 0; i< 3 ; i++)
{
for(int i = 0; i< 3 ; i++)
{
if(a[i,j] > iMax)
iMax = a[i,j];
}
}

console.Writeline("最大的值是 {0}。", iMax);