如何在cplex IBM ilog中找到数组的长度和数组的和，然后求出两个diff数组的和的乘积？

int[] array = { 1, 2, 3, 4, 5 };

int length = array.Length;

int sum = 0;
for (int i = 0; i < length; i++)
{
    sum += array[i];
}

// 输出数组的长度和数组的和
Console.WriteLine("数组的长度：" + length);
Console.WriteLine("数组的和：" + sum);

int[] diffArray1 = { 1, 2, 3 };
int[] diffArray2 = { 4, 5, 6 };

int length1 = diffArray1.Length;
int length2 = diffArray2.Length;

int sum1 = 0;
int sum2 = 0;

for (int i = 0; i < length1; i++)
{
    sum1 += diffArray1[i];
}

for (int i = 0; i < length2; i++)
{
    sum2 += diffArray2[i];
}

int sumOfDiffs = sum1 + sum2;
int product = sumOfDiffs * sumOfDiffs;

// 输出两个diff数组的和的乘积
Console.WriteLine("两个diff数组的和的乘积：" + product);