一元四次方程的简单修正算法

//本程序基于.NET平台，编译环境是Microsoft Visual C++ .NET
//对于VC++6.0平台的修改，只须把stdafx.h换成iostream.h，然后去掉主函数里的“using namespace std”
#include "stdafx.h"
#include "math.h"
using namespace std;
double du,dv;
void solu_w(double b,double c)
{
double tem,tem1;
tem1=b*b-4*c;
double m,n;
m=(-1)*b/2;
if(tem1>=0)
{
tem=sqrt(tem1);
n=tem/2;
cout<<"/n方程根为："<<'/n'<<"/t X1 = "<<m+n<<'/n'<<"/t X2 = "<<m-n<<endl;
}
else
{
tem=sqrt((-1)*tem1);
n=tem/2;
cout<<"/n方程根为："<<'/n'<<"/t X1 = "<<m<<" + "<<n<<" i"<<'/n'<<"/t X2 = "<<m<<" - "<<n<<" i"<<'/n'<<endl;
}
}
void get_du_dv(double a,double b,double c,double d,double e,double f)
{
dv=(b*d-a*e)/(c*e-b*f);
du=(c*d-a*f)/(b*f-c*e);
}
void main()
{
cout<<"输入四次方程的各个系数"<<endl;
double a[5];
double temp;
for(int t=4;t>=0;t--)
{
cout<<"/t a["<<t<<"]=";
cin>>temp;
a[t]=temp;
}
cout<<"/n方程为：/n/t "<<a[4]<<" (X*X*X*X) + "<<a[3]<<" (X*X*X) + ";
cout<<a[2]<<" (X*X) + "<<a[1]<<" (X) + "<<a[0]<<" = 0"<<'/n'<<endl;
double u,v;
cout<<"输入因子的常数项和一次项系数："<<endl;
cout<<"/t u = ";
cin>>u;
cout<<"/t v = ";
cin>>v;
cout<<"/n因子式为：/n/t w(X) = "<<"(X*X) + "<<u<<" (X) + "<<v<<endl;
double p0,p1,p2,r0,r1;
cout<<"/n输入修正次数： ";
int count1;
cin>>count1;
for(count1;count1>=0;count1--)
{
p2=a[4];
p1=a[3]-u*p2;
p0=a[2]-v*p2-u*p1;
r0=a[1]-v*p1-u*p0;
r1=a[0]-v*p0;
//下面一行加在程序中，将导致的问题是：如果修正次数太大，程序运行时间大大延长，主要时间消耗在屏幕显示上。
//cout<<"/t p(X) = "<<p2<<" (X*X) + "<<p1<<" (X) + "<<p0<<" = 0"<<endl;
double pp3,pp2,pp1,pp0,r0v,r1v,r0u,r1u;
pp3=(-1)*p2;
pp2=(-1)*p1;
pp1=(-1)*p0;
pp0=0;
r0v=u*p2-p1;
r1v=v*p2-p0;
r0u=r1v-u*r0v;
r1u=(-1)*v*r0v;
get_du_dv(r0,r0u,r0v,r1,r1u,r1v);
u=u+du;
v=v+dv;
}
solu_w(u,v);
}