c++解线性方程组的迭代法.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 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 #includeiostream #includemath.h #include iomanip using namespace std; main() { double s=0; double max(double array[10]); double a[10][10]={ {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 b[10]={7,5,-13,2,6,-12,14,-4,5,-5}; double c[10]={0,0,0}; double x[10]={0,0,0,0,0,0,0,0,0,0}; double x0[10]={0,0,0,0,0,0,0,0,0,0}; int i,k,j; double r,sum=0; cout输入精度：endl; cins; for(k=1;;k++) { for(i=0;i10;i++) { for(j=0;j10;j++) { sum=a[i][j]*x0[j]+sum; } x[i]=x0[i]+((b[i]-sum)/a[i][i]); c[i]=fabs(x[i]-x0[i]); sum=0; } r=max(c); if(rs) {for(i=0;i10;i++) coutxi = x[i]setprecision(10)endl; cout迭代次数：kendl; return 0; } else for(i=0;i10;i++) x0[i]=x[i]; } } double max(double array[10]) { double a=array[0]; int i; for(i=1;i10;i++) { if(aarray[i]) a=array[i]; } return a; } GS迭代法 #includeiostream #includemath.h #include iomanip using namespace std; main() { double s=0; double max(double array[3]); double a[10][10]={ {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 b[10]={7,5,-13,2,6,-12,14,-4,5,-5}; double c[10]={0,0,0}; double x[10]={0,0,0,0,0,0,0,0,0,0}; double x0[10]={0,0,0,0,0,0,0,0,0,0}; int