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法律学矩阵乘法算法
本节将介绍如何将两个矩阵相乘。如果满足此条件，才能执行矩阵乘法。假设有两个矩阵 A 和 B，并且它们的维度分别为 A (m x n) 和 B (p x q)，只有当 n = p 时才能找到结果矩阵。然后，结果矩阵 C 的顺序为 (m x q)。
算法
matrixMultiply(A, B): Assume dimension of A is (m x n), dimension of B is (p x q) Begin if n is not same as p, then exit otherwise define C matrix as (m x q) for i in range 0 to m - 1, do for j in range 0 to q – 1, do for k in range 0 to p, do C[i, j] = C[i, j] + (A[i, k] * A[k, j]) done done done End
示例
#include<iostream>
using namespace std;
int main() {
int product[10][10], r1=3, c1=3, r2=3, c2=3, i, j, k;
int a[3][3] = {
{2, 4, 1},
{2, 3, 9},
{3, 1, 8}
};
int b[3][3] = {
{1, 2, 3},
{3, 6, 1},
{2, 4, 7}
};
if (c1 != r2) {
cout<<"Column of first matrix should be equal to row of second matrix";
} else {
cout<<"The first matrix is:"<<endl;
for(i=0; i<r1; ++i) {
for(j=0; j<c1; ++j)
cout<<a[i][j]<<" ";
cout<<endl;
}
cout<<endl;
cout<<"The second matrix is:"<<endl;
for(i=0; i<r2; ++i) {
for(j=0; j<c2; ++j)
cout<<b[i][j]<<" ";
cout<<endl;
}
cout<<endl;
for(i=0; i<r1; ++i)
for(j=0; j<c2; ++j) {
product[i][j] = 0;
}
for(i=0; i<r1; ++i)
for(j=0; j<c2; ++j)
for(k=0; k<c1; ++k) {
product[i][j]+=a[i][k]*b[k][j];
}
cout<<"Product of the two matrices is:"<<endl;
for(i=0; i<r1; ++i) {
for(j=0; j<c2; ++j)
cout<<product[i][j]<<" ";
cout<<endl;
}
}
return 0;
}输出
The first matrix is: 2 4 1 2 3 9 3 1 8 The second matrix is: 1 2 3 3 6 1 2 4 7 Product of the two matrices is: 16 32 17 29 58 72 22 44 66
更新于: 2019-08-27
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