线性方程组迭代法—雅克比迭代法C++

article/2025/4/22 3:09:54

此例子使用三个变量、三个方程的情况，如需讨论多个的情况，使用vector稍加修改即可。
使用的方程组如下：
在这里插入图片描述

每次迭代的值如下：

程序流程图：

程序代码：

/********雅克比迭代法*********
*1.Xi为每一步迭代的初值,X1=0，Yi存放迭代结果
*2.迭代精度e，max|Xi-Yi|,1<=i<=n
*3.最大迭代次数N
*4.输入数据：
*(1)Bi:方程组的值
*(2)Aij:方程组变量系数
*/
#include<iostream>
#include<math.h>
using namespace std;
double sum(double x[4],double a[4][4],int temp)
{double res(0);double temp_res=0;//当i=j时，令其为0for(int j=1;j<4;j++){if(temp==j){res+=temp_res;//cout<<"res:"<<res<<endl;}else{res+=a[temp][j]*x[j];//cout<<"res:"<<res<<endl;}}return res;
}//the max in an array
double max(double *array)
{double max_var = array[1];for(int i=1;i<4;i++){if(max_var<array[i]){max_var=array[i];}}//cout<<"max_var:"<<max_var<<endl;return max_var;
}void interation(unsigned N,double a[4][4],double *b,double *x,unsigned k,double e)
{int i=1;double Y[4]={0,};for(;i<4;i++){Y[i]=(b[i]-sum(x,a,i))/a[i][i];//cout<<"Y"<<i<<":"<<Y[i]<<endl;}double max_val[4];for(i=1;i<4;i++){max_val[i]=fabs(x[i]-Y[i]);}double the_max=max(max_val);//get the max in the arraycout<<"the max:"<<the_max<<endl;if(the_max<e){cout<<"---Iterating successfully-----"<<endl;cout<<"----------------------------"<<endl;cout<<"The values are as follows:"<<endl;for(i=1;i<4;i++){cout<<"Y["<<i<<"]:"<<Y[i]<<"  ";}cout<<endl;}else{if(k==N){cout<<"--!---Iteration failed---!--"<<endl;}else{k=k+1;x=Y;//x array=Y arrayinteration(N,a,b,x,k,e);}}
}
int main()
{//store coefficients//double coeff[4][4]={{0,0,0,0},{0,10,-1,-2},{0,-1,10,-2},{0,-1,-1,5}};double coeff[4][4];//store equations' value//double values[4]={0,7.2,8.3,4.2};double values[4];double Y[4];//Y[i]unsigned i,j;double e;//precisionunsigned N;//iteration timesunsigned k=1;//flag//先初始化array为0memset(Y,0,sizeof(Y));cout<<"---------------Jacobi Interaction----------------"<<endl;cout<<"----Three equations and three variables X---------"<<endl;cout<<"Please input equations' coefficients and values:"<<endl;cout<<"Please input coeffients from first equation to last:"<<endl;for(i=1;i<4;i++){for(j=1;j<4;j++)cin>>coeff[i][j];}cout<<"---------------------------------------------"<<endl;cout<<"Please input equations' values:"<<endl;for(i=1;i<4;i++){cin>>values[i];}cout<<"-----------------------------------------------"<<endl;cout<<"Please input precision e:"<<endl;cin>>e;cout<<"----------------------------------------------"<<endl;cout<<"Please input interation times N:"<<endl;cin>>N;cout<<"----Inputting end....--------------------------"<<endl;//迭代初值为0double x[4]={0,0,0,0};interation(N, coeff,values,x, k, e);cout<<"----------------"<<endl;cout<<"Hello Boker..2021/5/11/17:00"<<endl;return 0;
}