C++实现 牛顿迭代 解非线性方程组

C++实现牛顿迭代解非线性方程组（二元二次为例）
求解{0=x*x-2*x-y+0.5;
0=x*x+4*y*y-4;
}的方程
#include<iostream>
#include<cmath>
#define N 2 // 非线性方程组中方程个数、未知量个数#define Epsilon 0.0001 // 差向量1范数的上限
#define Max 100 // 最大迭代次数
using namespace std;
const int N2=2*N;
int main()
{
void ff(float xx[N],float yy[N]); //计算向量函数的因变量向量yy[N]
void ffjacobian(float xx[N],float yy[N][N]); //计算雅克比矩阵yy[N][N]
void inv_jacobian(float yy[N][N],float inv[N][N]); //计算雅克比矩阵的逆矩阵inv
void newdundiedai(float x0[N], float inv[N][N],float y0[N],float x1[N]);
//由近似解向量x0 计算近似解向量x1
float x0[N]={2.0,0.25},y0[N],jacobian[N][N],invjacobian[N][N],x1[N],errornorm;
int i,j,iter=0;
//如果取消对x0的初始化，撤销下面两行的注释符，就可以由键盘x读入初始近似解向量for( i=0;i<N;i++)
cin>>x0[i];
cout<<"初始近似解向量："<<endl;
for (i=0;i<N;i++)
cout<<x0[i]<<" ";
cout<<endl;cout<<endl;
do
{
iter=iter+1;
cout<<"第"<<iter<<" 次迭代开始"<<endl; //计算向量函数的因变量向量y0
ff(x0,y0); //计算雅克比矩阵jacobian
ffjacobian(x0,jacobian); //计算雅克比矩阵的逆矩阵invjacobian inv_jacobian(jacobian,invjacobian); //由近似解向量x0 计算近似解向量x1 newdundiedai(x0, invjacobian,y0,x1); //计算差向量的1范数errornorm
errornorm=0;
for (i=0;i<N;i++)
errornorm=errornorm+fabs(x1[i]-x0[i]);
if (errornorm<Epsilon) break;
for (i=0;i<N;i++)
x0[i]=x1[i];
} while (iter<Max);
return 0;
}
void ff(float xx[N],float yy[N]) //调用函数
{float x,y;
int i;
x=xx[0];
y=xx[1];
yy[0]=x*x-2*x-y+0.5;
yy[1]=x*x+4*y*y-4; //计算初值位置的值
cout<<"向量函数的因变量向量是："<<endl;
for( i=0;i<N;i++)
cout<<yy[i]<<" ";
cout<<endl;
cout<<endl;
}
void ffjacobian(float xx[N],float yy[N][N])
{
float x,y;
int i,j;
x=xx[0];
y=xx[1];
//jacobian have n*n element //计算函数雅克比的值yy[0][0]=2*x-2;
yy[0][1]=-1;
yy[1][0]=2*x;
yy[1][1]=8*y;
cout<<"雅克比矩阵是："<<endl;
for( i=0;i<N;i++)
{for(j=0;j<N;j++)
cout<<yy[i][j]<<" ";
cout<<endl;
}
cout<<endl;
}
void inv_jacobian(float yy[N][N],float inv[N][N]) {float aug[N][N2],L;
int i,j,k;
cout<<"开始计算雅克比矩阵的逆矩阵："<<endl; for (i=0;i<N;i++)
{ for(j=0;j<N;j++)
aug[i][j]=yy[i][j];
for(j=N;j<N2;j++)
if(j==i+N) aug[i][j]=1;
else aug[i][j]=0;
}
for (i=0;i<N;i++)
{ for(j=0;j<N2;j++)
cout<<aug[i][j]<<" ";
cout<<endl;
}
cout<<endl;
for (i=0;i<N;i++)
{
for (k=i+1;k<N;k++)
{L=-aug[k][i]/aug[i][i];
for(j=i;j<N2;j++)
aug[k][j]=aug[k][j]+L*aug[i][j];
}
}
for (i=0;i<N;i++)
{ for(j=0;j<N2;j++)
cout<<aug[i][j]<<" ";
cout<<endl;
cout<<endl;
for (i=N-1;i>0;i--)
{
for (k=i-1;k>=0;k--)
{L=-aug[k][i]/aug[i][i];
for(j=N2-1;j>=0;j--)
aug[k][j]=aug[k][j]+L*aug[i][j];
}
}
for (i=0;i<N;i++)
{ for(j=0;j<N2;j++)
cout<<aug[i][j]<<" ";
cout<<endl;
}
cout<<endl;
for (i=N-1;i>=0;i--)
for(j=N2-1;j>=0;j--)
aug[i][j]=aug[i][j]/aug[i][i];
for (i=0;i<N;i++)
{ for(j=0;j<N2;j++)
cout<<aug[i][j]<<" ";
cout<<endl;
for(j=N;j<N2;j++)
inv[i][j-N]=aug[i][j];
}
cout<<endl;
cout<<"雅克比矩阵的逆矩阵："<<endl; for (i=0;i<N;i++)
{ for(j=0;j<N;j++)
cout<<inv[i][j]<<" ";
cout<<endl;
cout<<endl;
}
void newdundiedai(float x0[N], float inv[N][N],float y0[N],float x1[N]) {
int i,j;
float sum=0;
for(i=0;i<N;i++)
{ sum=0;
for(j=0;j<N;j++)
sum=sum+inv[i][j]*y0[j];
x1[i]=x0[i]-sum;
}
cout<<"近似解向量："<<endl;
for (i=0;i<N;i++)
cout<<x1[i]<<" ";
cout<<endl;cout<<endl;
}。