C#中矩陣運算辦法實例剖析。本站提示廣大學習愛好者：（C#中矩陣運算辦法實例剖析）文章只能為提供參考，不一定能成為您想要的結果。以下是C#中矩陣運算辦法實例剖析正文
本文實例講述了C#中矩陣運算辦法。分享給年夜家供年夜家參考。詳細剖析以下：
1、測試情況：
主機:XP
開辟情況:VS2008
2、功效:
在C#中完成矩陣運算
3、源代碼：
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
//矩陣數據構造
//二維矩陣
class _Matrix
{
public int m;
public int n;
public float[] arr;
//初始化
public _Matrix()
{
m = 0;
n = 0;
}
public _Matrix(int mm,int nn)
{
m = mm;
n = nn;
}
//設置m
public void set_mn(int mm,int nn)
{
m = mm;
n = nn;
}
//設置m
public void set_m(int mm)
{
m = mm;
}
//設置n
public void set_n(int nn)
{
n = nn;
}
//初始化
public void init_matrix()
{
arr = new float[m * n];
}
//釋放
public void free_matrix()
{
//delete [] arr;
}
//讀取i,j坐標的數據
//掉敗前往-31415,勝利前往值
public float read(int i,int j)
{
if (i >= m || j >= n)
{
return -31415;
}
//return *(arr + i * n + j);
return arr[i * n + j];
}
//寫入i,j坐標的數據
//掉敗前往-1,勝利前往1
public int write(int i,int j,float val)
{
if (i >= m || j >= n)
{
return -1;
}
arr[i * n + j] = val;
return 1;
}
};
//二維運算類
class _Matrix_Calc
{
//初始化
public _Matrix_Calc()
{
}
//C = A + B
//勝利前往1,掉敗前往-1
public int add(ref _Matrix A,ref _Matrix B,ref _Matrix C)
{
int i = 0;
int j = 0;
//斷定能否可以運算
if (A.m != B.m || A.n != B.n ||
A.m != C.m || A.n != C.n)
{
return -1;
}
//運算
for (i = 0;i < C.m;i++)
{
for (j = 0;j < C.n;j++)
{
C.write(i,j,A.read(i,j) + B.read(i,j));
}
}
return 1;
}
//C = A - B
//勝利前往1,掉敗前往-1
public int subtract(ref _Matrix A,ref _Matrix B, ref _Matrix C)
{
int i = 0;
int j = 0;
//斷定能否可以運算
if (A.m != B.m || A.n != B.n ||
A.m != C.m || A.n != C.n)
{
return -1;
}
//運算
for (i = 0;i < C.m;i++)
{
for (j = 0;j < C.n;j++)
{
C.write(i,j,A.read(i,j) - B.read(i,j));
}
}
return 1;
}
//C = A * B
//勝利前往1,掉敗前往-1
public int multiply(ref _Matrix A, ref _Matrix B, ref _Matrix C)
{
int i = 0;
int j = 0;
int k = 0;
float temp = 0;
//斷定能否可以運算
if (A.m != C.m || B.n != C.n ||
A.n != B.m)
{
return -1;
}
//運算
for (i = 0;i < C.m;i++)
{
for (j = 0;j < C.n;j++)
{
temp = 0;
for (k = 0;k < A.n;k++)
{
temp += A.read(i,k) * B.read(k,j);
}
C.write(i,j,temp);
}
}
return 1;
}
//行列式的值,只能盤算2 * 2,3 * 3
//掉敗前往-31415,勝利前往值
public float det(ref _Matrix A)
{
float value = 0;
//斷定能否可以運算
if (A.m != A.n || (A.m != 2 && A.m != 3))
{
return -31415;
}
//運算
if (A.m == 2)
{
value = A.read(0,0) * A.read(1,1) - A.read(0,1) * A.read(1,0);
}
else
{
value = A.read(0,0) * A.read(1,1) * A.read(2,2) +
A.read(0,1) * A.read(1,2) * A.read(2,0) +
A.read(0,2) * A.read(1,0) * A.read(2,1) -
A.read(0,0) * A.read(1,2) * A.read(2,1) -
A.read(0,1) * A.read(1,0) * A.read(2,2) -
A.read(0,2) * A.read(1,1) * A.read(2,0);
}
return value;
}
//求轉置矩陣,B = AT
//勝利前往1,掉敗前往-1
public int transpos(ref _Matrix A,ref _Matrix B)
{
int i = 0;
int j = 0;
//斷定能否可以運算
if (A.m != B.n || A.n != B.m)
{
return -1;
}
//運算
for (i = 0;i < B.m;i++)
{
for (j = 0;j < B.n;j++)
{
B.write(i,j,A.read(j,i));
}
}
return 1;
}
//求逆矩陣,B = A^(-1)
//勝利前往1,掉敗前往-1
public int inverse(ref _Matrix A, ref _Matrix B)
{
int i = 0;
int j = 0;
int k = 0;
_Matrix m = new _Matrix(A.m,2 * A.m);
float temp = 0;
float b = 0;
//斷定能否可以運算
if (A.m != A.n || B.m != B.n || A.m != B.m)
{
return -1;
}
/*
//假如是2維或許3維求行列式斷定能否可逆
if (A.m == 2 || A.m == 3)
{
if (det(A) == 0)
{
return -1;
}
}
*/
//增廣矩陣m = A | B初始化
m.init_matrix();
for (i = 0;i < m.m;i++)
{
for (j = 0;j < m.n;j++)
{
if (j <= A.n - 1)
{
m.write(i,j,A.read(i,j));
}
else
{
if (i == j - A.n)
{
m.write(i,j,1);
}
else
{
m.write(i,j,0);
}
}
}
}
//高斯消元
//變換下三角
for (k = 0;k < m.m - 1;k++)
{
//假如坐標為k,k的數為0,則行變換
if (m.read(k,k) == 0)
{
for (i = k + 1;i < m.m;i++)
{
if (m.read(i,k) != 0)
{
break;
}
}
if (i >= m.m)
{
return -1;
}
else
{
//交流行
for (j = 0;j < m.n;j++)
{
temp = m.read(k,j);
m.write(k,j,m.read(k + 1,j));
m.write(k + 1,j,temp);
}
}
}
//消元
for (i = k + 1;i < m.m;i++)
{
//取得倍數
b = m.read(i,k) / m.read(k,k);
//行變換
for (j = 0;j < m.n;j++)
{
temp = m.read(i,j) - b * m.read(k,j);
m.write(i,j,temp);
}
}
}
//變換上三角
for (k = m.m - 1;k > 0;k--)
{
//假如坐標為k,k的數為0,則行變換
if (m.read(k,k) == 0)
{
for (i = k + 1;i < m.m;i++)
{
if (m.read(i,k) != 0)
{
break;
}
}
if (i >= m.m)
{
return -1;
}
else
{
//交流行
for (j = 0;j < m.n;j++)
{
temp = m.read(k,j);
m.write(k,j,m.read(k + 1,j));
m.write(k + 1,j,temp);
}
}
}
//消元
for (i = k - 1;i >= 0;i--)
{
//取得倍數
b = m.read(i,k) / m.read(k,k);
//行變換
for (j = 0;j < m.n;j++)
{
temp = m.read(i,j) - b * m.read(k,j);
m.write(i,j,temp);
}
}
}
//將右邊方陣化為單元矩陣
for (i = 0;i < m.m;i++)
{
if (m.read(i,i) != 1)
{
//取得倍數
b = 1 / m.read(i,i);
//行變換
for (j = 0;j < m.n;j++)
{
temp = m.read(i,j) * b;
m.write(i,j,temp);
}
}
}
//求得逆矩陣
for (i = 0;i < B.m;i++)
{
for (j = 0;j < B.m;j++)
{
B.write(i,j,m.read(i,j + m.m));
}
}
//釋放增廣矩陣
m.free_matrix();
return 1;
}
};
namespace test
{
public partial class Form1 : Form
{
double zk;
double xkg, pkg, kk, xk, pk, q, r;
public Form1()
{
InitializeComponent();
xk = 0;
pk = 0;
q = 0.00001;
r = 0.0001;
int i = 0;
int j = 0;
int k = 0;
_Matrix_Calc m_c = new _Matrix_Calc();
//_Matrix m1 = new _Matrix(3,3);
//_Matrix m2 = new _Matrix(3,3);
//_Matrix m3 = new _Matrix(3,3);
_Matrix m1 = new _Matrix(2, 2);
_Matrix m2 = new _Matrix(2, 2);
_Matrix m3 = new _Matrix(2, 2);
//初始化內存
m1.init_matrix();
m2.init_matrix();
m3.init_matrix();
//初始化數據
k = 1;
for (i = 0;i < m1.m;i++)
{
for (j = 0;j < m1.n;j++)
{
m1.write(i,j,k++);
}
}
for (i = 0;i < m2.m;i++)
{
for (j = 0;j < m2.n;j++)
{
m2.write(i,j,k++);
}
}
m_c.multiply(ref m1,ref m2, ref m3);
//output.Text = Convert.ToString(m3.read(1,1));
output.Text = Convert.ToString(m_c.det(ref m1));
}
/*
private void button1_Click(object sender, EventArgs e)
{
zk = Convert.ToDouble(input.Text);
//時光方程
xkg = xk;
pkg = pk + q;
//狀況方程
kk = pkg / (pkg + r);
xk = xkg + kk * (zk - xkg);
pk = (1 - kk) * pkg;
//輸入
output.Text = Convert.ToString(xk);
}
private void textBox1_TextChanged(object sender, EventArgs e)
{
}
* */
}
}
願望本文所述對年夜家的C#法式設計有所贊助。
