c++解线方程组的迭代法.doc

实验报告 课题3 解线性方程组的迭代法 课题3 解线性方程组的迭代法一问题提出 对线性方程组=51232346133819-21X1X2X3X4X5X6X7X8X9X10* 4, -1, 0 ,0 ,0 ,0 ,0 ,0 ,0 ,0-1, 4, -1 ,0 ,0 ,0 ,0 ,0 ,0 ,00, -1 ,4 ,-1 ,0 ,0 ,0 ,0 ,0 ,00, 0, -1, 4, -1 ,0 ,0 ,0 ,0, 00 ,0 ,0 ,-1 ,4 ,-1 ,0 ,0 ,0 ,00, 0, 0 ,0 ,-1 ,4 ,-1 ,0 ,0 ,00, 0, 0, 0 ,0 ,-1 ,4, -1 ,0 ,00 ,0, 0, 0, 0, 0, -1 ,4, -1 ,00, 0, 0, 0, 0, 0, 0, -1 ,4 ,-10, 0 ,0 ,0 ,0 ,0 ,0 ,0 ,-1 ,4分别采用jacobi迭代法，GS迭代法和SOR迭代法方法求解。二要求分别对不同的精度要求，如e=10-3,10-4,10-5,利用所需的迭代次数体会该迭代法的收敛快慢。使用SOR方法时，选取松弛因子w=0.8, 0.9 1, 1.2 等，试找出你所选用松弛因子的最佳值。编制出各种迭代法的程序并给出计算结果。三C+源代码：jacobi迭代法#include#include#include using namespace std;main()double s=0;double max(double array10); double a1010=4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4;double b10=7,5,-13,2,6,-12,14,-4,5,-5;double c10=0,0,0;double x10=0,0,0,0,0,0,0,0,0,0;double x010=0,0,0,0,0,0,0,0,0,0;int i,k,j;double r,sum=0;cout输入精度：s;for(k=1;k+)for(i=0;i10;i+)for(j=0;j10;j+) sum=aij*x0j+sum; xi=x0i+(bi-sum)/aii); ci=fabs(xi-x0i); sum=0; r=max(c); if(rs) for(i=0;i10;i+)coutxi = xisetprecision(10)endl; cout迭代次数：kendl; return 0; else for(i=0;i10;i+) x0i=xi;double max(double array10)double a=array0;int i;for(i=1;i10;i+) if(aarrayi)a=arrayi;return a;GS迭代法#include#include#include using namespace std;main()double s=0;double max(double array3); double a1010=4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4;double b10=7,5,-13,2,6,-12,14,-4,5,-5; double c10=0,0,0;double x10=0,0,0,0,0,0,0,0,0,0;double x010=0,0,0,0,0,0,0,0,0,0; int i,k,j; double r,sum=0; cout输入精度：s; for(k=1;k+)for(i=0;i10;i+)for(j=0;j10;j+)sum=aij*x0j+sum; xi=x0i+(bi-sum)/aii); ci=xi-x0i;if(ci0)ci=-ci; x0i=xi; sum=0; r=max(c); if(rs) for(i=0;i10;i+)coutxi = xisetprecision(10)endl; cout迭代次数：kendl; return 0; double max(double array10)double a=array0;int i;for(i=1;i10;i+) if(aarrayi)a=arrayi;return a;SOR方法#include#include#include using namespace std;main()double s=0,w=0;double max(double array3); double a1010=4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4,-1,0,0,0,0,0,0,0,0,-1,4;double b10=7,5,-13,2,6,-12,14,-4,5,-5; double c10=0,0,0;double x10=0,0,0,0,0,0,0,0,0,0;double x010=0,0,0,0,0,0,0,0,0,0; int i,k,j; double r,sum=0; cout输入精度：s;cout松弛因子：w; for(k=1;k+)for(i=0;i10;i+)for(j=0;j10;j+)sum=aij*x0j+sum; xi=x0i+(w*(bi-sum)/aii); ci=xi-x0i;if(ci0)ci=-ci; x0i=xi; sum=0; r=max(c); if(rs) for(i=0;i10;i+) coutxi = xisetprecision(10)endl; cout迭代次数：kendl; return 0; double max(double array10)double a=array0;int i;for(i=1;i10;i+) if(aarrayi)a=arrayi;return a;四运行结果如下：jacobi迭代法GS迭代法SOR迭代法五程序运行的结果分析由上可知