解线性方程组（一）——克拉默法则求解（C++）-编程知识

解线性方程组（一）——克拉默法则求解（C++）

克拉默法则

解线性方程组最基础的方法就是使用克拉默法则，需要注意的是，该方程组必须是线性方程组。
假设有方程组如下：
$\begin{cases} a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n=b_1\\ a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n=b_2\\ \cdots \qquad \qquad\cdots \qquad \qquad \cdots \\ a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n=b_n\\ \end{cases}$
将其转换为矩阵形式
$A\vec{x}=\vec{b}$
$\begin{bmatrix} {a_{11}}&{a_{12}}&{\cdots}&{a_{1n}}\\ {a_{21}}&{a_{22}}&{\cdots}&{a_{2n}}\\ {\vdots}&{\vdots}&{\ddots}&{\vdots}\\ {a_{m1}}&{a_{m2}}&{\cdots}&{a_{mn}}\\ \end{bmatrix} \begin{bmatrix} {x_{1}}\\ {x_{2}}\\ {\vdots}\\ {x_{n}}\\ \end{bmatrix}= \begin{bmatrix} {b_{1}}\\ {b_{2}}\\ {\vdots}\\ {b_n} \end{bmatrix}$
根据克拉默法则有：
$x_i=\frac{|A_i|}{|A|}$
$A_i= \begin{bmatrix} {a_{11}}&{\cdots}&{a_{1i-1}}&{b1}&{a_{1i+1}}&{\cdots}&{a_{1n}}\\ {a_{21}}&{\cdots}&{a_{2i-1}}&{b2}&{a_{2i+1}}&{\cdots}&{a_{2n}}\\ {\vdots}&{\vdots}&{\ddots}&{\vdots}\\ {a_{n1}}&{\cdots}&{a_{ni-1}}&{bn}&{a_{ni+1}}&{\cdots}&{a_{nn}}\\ \end{bmatrix}$

代码

实现方法可以参考行列式求值（C++）、插值（一）——多项式插值（C++）

//克拉默法则求解线性方程组
/*
5x1+2x2-2x3=1
x1+4x2+x3=2
x1-2x2+4x3=-1
*/
#include<iostream>
#include<cmath>
//使用代数余子式进行求解
double determinant_value(double **D,int n)
{if(n==1){return D[0][0];}else if(n==0){throw "error: determinant is empty";}char flag = 1;//符号位double D_value = 0.0;double **D_temp = new double*[n];for (int i = 0; i < n;i++){D_temp [i]= new double[n];for(int j=0;j<n;j++){D_temp[i][j]=D[i][j];}}// 转为上三角for (int i = 0; i < n; i++){if (D_temp[i][i] == 0){int j = i + 1;// 将主对角线上的值转为非0值for (; j < n; j++){if (D_temp[j][i] != 0){double temp;for (int k = 0; k < n; k++){temp = D_temp[i][k];D_temp[i][k] = D_temp[j][k];D_temp[j][k] = temp; // 交换两行}flag = -flag;break;}}if (j == n){return D_value;}}// 将主对角线下面的值转为0for (int j = i + 1; j < n; j++){double temp = D_temp[j][i] / D_temp[i][i];for (int k = i; k < n; k++){D_temp[j][k] -= temp * D_temp[i][k];}}}// 计算行列式的值D_value = 1.0;for (int i = 0; i < n; i++){D_value *= D_temp[i][i];}D_value*=flag;for (int i = 0; i < n;i++){delete[] D_temp[i];}delete [] D_temp;return D_value;
}
//克拉默法则求解线性方程组
void Kramer(double **A,double *b,double *x,int n)
{double **A_i=new double*[n];double A_value=determinant_value(A,n);if(A_value==0){std::cout<<"该方程组不是线性方程组"<<std::endl;exit(0);}for (int i = 0; i < n; i++){A_i[i] = new double[n];for (int j = 1; j < n; j++){A_i[i][j] = A[i][j];}}//下面是为了求D_i，每次只需要修改两列数据for (int i = 0; i < n;i++){if(i==0){for(int j = 0;j < n;j++){A_i[j][0]=b[j];}}else{for(int i2 = 0;i2 < n;i2++){A_i[i2][i-1]=A[i2][i-1];A_i[i2][i]=b[i2];}}// for(int i2 = 0;i2 < n;i2++)// {//     for (int j2 = 0;j2 < n;j2++)//     {//         std::cout<<A_i[i2][j2]<<" ";//     }//     std::cout<<std::endl;// }//求多项式系数x[i] = determinant_value(A_i, n) / A_value;}for (int i = 0; i < n; i++){delete[] A_i[i];}delete[] A_i;
}
int main()
{int n;//矩阵维度std::cout<<"请输入矩阵维度：";std::cin>>n;double *x=new double[n];double **A = new double *[n];std::cout<<"请输入矩阵A的元素："<<std::endl;for(int i = 0;i < n;i++){A[i] = new double [n];for(int j = 0;j < n;j++){std::cin>>A[i][j];}}double *b = new double [n];std::cout<<"请输入矩阵b的元素："<<std::endl;for(int i = 0;i < n;i++){std::cin>>b[i];}Kramer(A, b, x, n);for (int i = 0; i < n; i++){std::cout<<"x"<<i+1<<"="<<x[i]<<" ";}for (int i = 0; i < n; i++){delete[] A[i];}delete[] A;delete [] b;delete[] x;return 0;
}