编写程序，可以实现m*n矩阵和n*p矩阵相乘。m,n,p均小于10，矩阵元素为整数。

分析：

首先我们可以根据题意写出函数头。可以定为void MatrixMutiply(int m,int n,int p,long lMatrix1[MAX][MAX],long lMatrix2[MAX][MAX],long lMatrixResult[MAX][MAX])，其中lMatrix1和lMatrix2分别是输入的m*n矩阵和n*p矩阵，lMatrixResult是输出的m*p矩阵。

因为m，n和p都是未知量，要进行处理的矩阵大小是变量。但我们可以定义比较大的二维数组，只使用其中的部分数组元素。

矩阵相乘的算法比较简单，输入一个m*n矩阵和一个n*p矩阵，结果必然是m*p矩阵，有m*p个元素，每个元素都需要计算，可以使用m*p嵌套循环进行计算。

根据矩阵乘法公式：



可以用循环直接套用上面的公式计算每个元素。嵌套循环内部进行累加前，一定要注意对累加变量进行清零。-Write a program, you can achieve the m* n matrix and n* p matrix multiplied. m, n, p are less than 10, the matrix elements are integers. Analysis: First of all, we can write the function header according to the meaning of the questions. As void MatrixMutiply (int m, int n, int p, long lMatrix1 [MAX] [MAX], long lMatrix2 [MAX] [MAX], long lMatrixResult [MAX] [MAX]) of which lMatrix1 and lMatrix2, are input m* N matrix and n* p matrix, lMatrixResult is the output of m* p matrix. M, n and p are unknown quantity, the size of the matrix to be processed is variable. However, we can define a large two-dimensional arrays, which use only part of the array elements. M* p nested relatively simple matrix multiplication algorithm, the input of an m* n matrix, and an n* p matrix, the inevitable result is a matrix of m* p, there are m* p elements, each element needs to be calculated and can be used loop calculation. Matrix multiplication formula: cycle directly apply the above formula to calculate ea