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);}
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[【技术文档】]一元四次方程的简单修正算法