数值方法高斯列主元消元法试验报告.doc

1）用高斯列主元消元法求解下面的方程组#include #include #include using namespace std;int main()/double a45=1,-1,1,-4,2, 5,-4,3,12,4, 2, 1,1,11,3, 2,-1,7,-1,0;/int i,j,k;int Line,Row;double temp45;/中间量Row=4;Line=5;/Inintial/Result/double X5;cout最初的矩阵为:endl;for(i=0;i4;i+)for(j=0;j5;j+)coutaij ;coutendl;coutendl;/for(k=0;kRow-1;k+)/for(i=k+1;iRow;i+)/tempik=aik/akk;for(j=k;jLine;j+)aij-=tempik*akj;/cout经过消元后的增广矩阵为:endl;for(i=0;i4;i+)for(j=0;j5;j+)coutaij ;coutendl;cout=0;i-)/double temp_new;temp_new=0;for(j=i+1;j=Row-1;j+)aiRow-=aij*ajRow;aiRow/=aii;/Print;cout最后的解为:endl;for(i=0;iRow;i+)/coutXi+1=aiRowendl;/return 0;1.2 列主元方法#include #include #include using namespace std;void Print(double a5)int i,j;for(i=0;i4;i+)for(j=0;j5;j+)coutaij ;coutendl;coutendl;int main()/double a45=1,-1,1,-4,2, 5,-4,3,12,4, 2, 1,1,11,3, 2,-1,7,-1,0;/double b45=0.3e-15,int i,j,k,n;int Line,Row;double temp45;/中间量Row=4;Line=5;/Inintial/cout最初的矩阵为:endl;Print(a);/the main process is underint kk;/flagsdouble max; /bool flag=false;double t;/the temp of change;for(k=0;kRow-1;k+)/search the max_num/flag=false;max=akk;kk=k;for(i=k;imax)max=aik;kk=i;/flag=true;/change the linefor(j=0;jLine;j+)t=akj;akj=akkj;akkj=t;cout第k+1次换行结果:endl;Print(a);coutendl;cout第k+1次消元结果:endl;/消元的过程for(i=k+1;iRow;i+)/tempik=aik/akk;for(j=k;j=0;i-) / double temp_new; temp_new=0; for(j=i+1;j=n;j+) ain+1-=aij*ajn+1; ain+1/=aii;/cout经过消元后的增广矩阵为:endl;Print(a);/Print;cout最后的解为:endl;for(i=0;iRow;i+)/coutXi+1=aiRowendl;/return 0;2）分别用列主元消元法与不选主元消元法求解，分析对结果的影响#include #include #include using namespace std;void Print(double b5)int i,j;for(i=0;i4;i+)for(j=0;j5;j+)coutbij ;coutendl;coutendl;int main()double b45=0.3e-15,59.14,3,1,59.17, 5.291,-6.130,-1,2,46.78,11.2,9,5,2,1,1,2,1,1,2; int i,j,k,n;int Line,Row;double temp45;/中间量Row=4;Line=5;/Inintial/cout最初的矩阵为:endl;Print(b);/the main process is underint kk;/flagsdouble max; /bool flag=false;double t;/the temp of change;for(k=0;kRow-1;k+)/search the max_num/flag=false;max=bkk;kk=k;for(i=k;imax)max=bik;kk=i;/flag=true;/change the linefor(j=0;jLine;j+)t=bkj;bkj=bkkj;bkkj=t;cout第k+1次换行结果:endl;Print(b);coutendl;cout第k+1次消元结果:endl;/消元的过程for(i=k+1;iRow;i+)/tempik=bik/bkk;for(j=k;j=0;i-) / double temp_new; temp_new=0; for(j=i+1;j=n;j+) bin+1-=bij*bjn+1; bin+1/=bii;/cout经过消元后的增广矩阵为:endl;Print(b);/Print;cout最后的解为:endl;for(i=0;iRow;i+)/coutXi+1=biRowendl;/return 0;2、 用迭代法求解 （迭代法收敛速度实验）注意修改不同的A 、B的数组3、 /雅可比迭代法/* auther luozhengxiao*/#include #include #include using namespace std;/void Print(double x)for(int i=0;i3;i+)coutsetprecision(8)fixedxiendl;int main()/double B13=-3,2,4; double B23=100,-200,345; double A33=6,2,-1, 1,4,-2,-3,1,4; double x3,x_old3,temp;int i,j,k;for(i=0;i3;i+)cout请输入第i+1个数:;coutt xixi;/x_oldi=xi;int n;coutn;/for(k=0;kn;k+