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1、精选优质文档-倾情为你奉上用高斯消去法解线性方程组。设有n元线性方程组如下：a00x0+a01x1+a0,n-1xn-1=b0a10x0+a11x1+a1,n-1xn-1=b1An-1,0x0+an-1,1x1+an-1,n-1xn-1=bn-1算法：对于k从0开始到n-2结束，进行以下步骤：首先，第k行第k列的数不能为0，若为0，则与下面不为0的行交换。将k行所有数除以第k行第k列的数.其次，进行消去：aij=aij-aik*akj,j,i=k+1,k+2,，n-1bi=bi-aik*bk,i=k+1,k+2,n-1最后，回代过程xn-1=bn-1/an-1n-1xi=bi-aijxj,I=
2、n-2,1,0定义一个矩阵类Matrix作为基类，然后由矩阵类派生出线性方程组类Linequ。程序清单：#include#includeusing namespace std;class Matrix /定义矩阵类public:Matrix(int dims=2) /构造函数index=dims; /保护数据赋值MatrixA=new doubleindex*index; /动态内存分配Matrix() delete MatrixA; /内存释放void setMatrix(double *rmatr) /设置矩阵值for(int i=0;iindex*index;i+)*(MatrixA+i
3、)=rmatri; /矩阵成员赋初值void printM(); /显示矩阵protected:int index;double* MatrixA;class Linequ:public Matrix /线性方程组类public:Linequ(int dims=2):Matrix(dims) sums=new doubledims;solu=new doubledims; Linequ();void setLinequ(double*a,double*b); /方程赋值void printL(); /显示方程int Solve(); /高斯消元法求解void showX(); /输出方程的解p
4、rivate:double *sums;double *solu;void Matrix:printM() /显示矩阵的元素coutThe Matrix is:endl;for(int i=0;iindex;i+)for(int j=0;jindex;j+)cout*(MatrixA+i*index+j) ;coutendl;Linequ:Linequ()deletesums;deletesolu;void Linequ:setLinequ(double *a,double *b) /方程赋值setMatrix(a);for(int i=0;iindex;i+)sumsi=bi;void Li
5、nequ:printL() /显示方程coutThe Line eqution is:endl;for(int i=0;iindex;i+)for(int j=0;jindex;j+)cout*(MatrixA+i*index+j) ;cout sumsiendl;void Linequ:showX() /输出方程的解coutThe Result is:endl;for(int i=0;iindex;i+)coutXi=soluiendl;int Linequ:Solve() /解线性方程组int l,k,i,j,is,p,q,m=0;double d,t;l=1;for(k=0;k=inde
6、x-2;k+) /消去过程d=0.0;for(i=k;i=index-1;i+)for(j=k;jd)d=t;is=i;if(d+1.0=1.0)l=0;if(l=0)coutfailendl;return(0);d=MatrixAk*index+k;if(d=0)m+;d=MatrixA(k+m)*index+k;is=k+m;if(is!=k)for(j=k;j=index-1;j+)p=k*index+j;q=is*index+j;t=MatrixAp;MatrixAp=MatrixAq;MatrixAq=t;t=sumsk;sumsk=sumsis;sumsis=t;d=MatrixA
7、k*index+k; for(j=k;j=index-1;j+) p=k*index+j;MatrixAp=MatrixAp/d; sumsk=sumsk/d; for(i=k+1;i=index-1;i+) for(j=k+1;j=index-1;j+) p=i*index+j; MatrixAp=MatrixAp-MatrixAi*index+k*MatrixAk*index+j; sumsi=sumsi-MatrixAi*index+k*sumsk; d=MatrixA(index-1)*index+index-1;if(fabs(d)+1.0=1.0) coutfail=0;i-) t=0.0; for(j=i+1;j=index-1;j+) t=t+MatrixAi*index+j*soluj; solui=sumsi-t;int main() /主函数double a=2,-0.5,-0.5,0, -0.5,2,0,-0.5, -0.5,0,2,-0.5, 0,-0.5,-0.5,2;double b4=0,3,3,0;Linequ equl(4);equl.setLinequ(a,b);equl.printL();if(equl.Solve()equl.showX();elsecoutFailendl;return 0;运行结果：专心-专注-专业