多维数组，如何一次显示else语句？

Question

我有这个由3个Matrix[3][3]组成的矩阵。我的任务是显示包含至少两个质数的矩阵的行数，我已经做了这个质数工作，但问题是，我必须显示此消息一旦cout << "No row containing at least two prime numbers" << endl;。由于迭代，如果任何一行不包含两个质数或只有一行包含两个质数，它会向我显示三次/两次。如果没有消息或有消息，我希望消息只显示一次。

质数代码..。

cout << "ROW NUMBER CONTAINING AT LEAST TWO PRIME NUMBERS" << endl;
    for (int i = 0; i < 3; i++)
    {
        Count = 0;
        for (int j = 0; j < 3; j++)
        {
            Prime = 0;
            for (int k = 2; k < Matrix[i][j]; k++)
            {
                if (Matrix[i][j] % k == 0)
                {
                    Prime = 1;
                    break;
                }
            }
            if (Prime == 0)
            {
                Count += 1;
            }
        }
        if (Count == 2)
        {
            cout << "ROW NUMBER " << i + 1 << endl;
        }
        else
        {
            cout << "NO ROW CONTAINING AT LEAST TWO PRIME NUMBERS" << endl;
        }
    }

Answer 1

使用像标志一样的整数。

    int flag = 0;
    cout << "ROW NUMBER CONTAINING AT LEAST TWO PRIME NUMBERS" << endl;
    for (int i = 0; i < 3; i++)
    {
        Count = 0;
        for (int j = 0; j < 3; j++)
        {
            Prime = 0;
            for (int k = 2; k < Matrix[i][j]; k++)
            {
                if (Matrix[i][j] % k == 0)
                {
                    Prime = 1;
                    break;
                }
            }
            if (Prime == 0)
            {
                Count += 1;
            }
        }
        if (Count == 2)
        {
            cout << "ROW NUMBER " << i + 1 << endl;
        }
        else
        {
            flag++;
        }
    }
    if(flag == 3)
    {
        cout << "NO ROW CONTAINING AT LEAST TWO PRIME NUMBERS" << endl;
    }

Answer 2

只需添加另一个计数变量

int row_count = 0;
for (int i = 0; i < 3; i++)
{

    ...

    if (Count == 2)
    {
        cout << "ROW NUMBER " << i + 1 << endl;
        row_count += 1;
    }
}
if (row_count == 0)
    cout << "NO ROW CONTAINING AT LEAST TWO PRIME NUMBERS" << endl;

这不是世界上最漂亮的代码，但它可以完成这项工作。

顺便说一句，至少有两个质数，所以

if (Count == 2)

应该是

if (Count >= 2)