【面试题】两个n*n的矩阵相乘--采用一位数组表示

#include <iostream>
#include <math.h>
using namespace std;

void MatrixMultiplication(int * pArry1, int *pArry2, int * pDestArry, int len)
{
	int row ;
	int col;
	row= col = (int)sqrt((double)len);
	if (row * col != len)  //矩阵不为方阵则提示矩阵无效
	{
		cout << "Matrix Invalidate!" << endl;
	}
	for (int i = 0; i < row; i++)
	{
		for (int j = 0; j < col; j++)
		{
			int sum = 0;
			for (int k = 0; k < row; k++)
			{
				sum += pArry1[row * i + k] * pArry2[row * k + j];  //某行、某列所有对应元素相乘求和即为新矩阵某行某列的元素
			}
			pDestArry[row * i + j] = sum;  //矩阵相乘所得新矩阵		
		}
	}
}

void main()
{
	int p1[][3] = 
	{
		1,2,1,
		1,1,1,
		2,3,1
	};
	
	int p2[][3] = 
	{
		1,1,1,
		2,2,2,
		3,3,3
	};

	int p[3][3];
	/**************/
	int *a = p1[0];
	int *b = p2[0];
	int *c = p[0];
	/*************/
	//int *a = &p1[0][0];
	//int *b = &p2[0][0];
	//int *c = &p[0][0];

	MatrixMultiplication(a,b,c,9);

	for(int i=0; i<3; i++)
	{
		for(int j=0; j<3; j++)
		{
			cout << c[i*3+j] <<" ";
		}
		cout << endl;
	}
}