方程式

 

方程式答案：

http://firstyear.chem.usyd.edu.au/calculators/cubic.shtml 

 

算法描述：

http://www.vibrationdata.com/math/cubic.pdf 

 

http://terpconnect.umd.edu/~nsw/ench250/cubiceq.htm

http://wenku.baidu.com/view/596bac0f6c85ec3a87c2c559.html

 

//////////////////函数--用卡丹公式解一元三次方程/////////////////void fun(double a,double b,double c,double d, double *real_y1,double *real_y2,double *real_y3, double *imag_y1,double *imag_y2,double *imag_y3){ double p,q,r,u,v,g,h,fai; p=(3.0*a*c-b*b)/(3*a*a); q=(2.0*pow(b,3.0)-9*a*b*c+27.0*a*a*d)/(27.0*pow(a,3.0)); r=b/(3.0*a); h=pow(q/2.0,2.0)+pow(p/3.0,3.0); g=sqrt(h); if(h>=0) { if(-q/2.0+g<0) { u=-pow(fabs(-q/2.0+g),1.0/3.0); } else { u=pow((-q/2.0+g),1.0/3.0); } if(-q/2.0-g<0) { v=-pow(fabs(-q/2.0-g),1.0/3.0); } else { v=-pow((-q/2.0-g),1.0/3.0); } if(h==0) { *real_y1=u+v-r; *imag_y1=0; *real_y2=-(u+v)/2-r; *imag_y2=0; *real_y3=-(u+v)/2-r; *imag_y3=0; } else { *real_y1=u+v-r; *imag_y1=0; *real_y2=-(u+v)/2; *imag_y2=sqrt(3.0)*(u-v)/2; *real_y3=-(u+v)/2; *imag_y3=-sqrt(3.0)*(u-v)/2; } } else { fai=acos((-q/2)/(sqrt(pow(fabs(p),3)/27))); *real_y1=2*sqrt(fabs(p)/3.0)*cos(fai/3.0)-r; *real_y2=-2*sqrt(fabs(p)/3.0)*cos((fai+3.1415926)/3.0)-r; *real_y3=-2*sqrt(fabs(p)/3.0)*cos((fai-3.1415926)/3.0)-r; *imag_y1=0; *imag_y2=0; *imag_y3=0; }}//////////////////////////////////主函数////////////////////////////////void main(){double a,b,c,d;double real_x1,real_x2,real_x3;double *preal_x1=&real_x1;double *preal_x2=&real_x2;double *preal_x3=&real_x3;double imag_x1,imag_x2,imag_x3;double *pimag_x1=&imag_x1;double *pimag_x2=&imag_x2;double *pimag_x3=&imag_x3;cout<<"请输入方程的系数a,b,c,d:"<<"/n"<<endl;cout<<"系数a=";cin>>a;cout<<endl;cout<<"系数b=";cin>>b;cout<<endl;cout<<"系数c=";cin>>c;cout<<endl;cout<<"系数d=";cin>>d;cout<<endl;fun(a,b,c,d,preal_x1,preal_x2,preal_x3,pimag_x1,pimag_x2,pimag_x3);cout<<" "<<" "<<"根的实部"<<" "<<"根的虚部"<<"/n"<<endl;printf("x1 %.5f %.5f/n/n",real_x1,imag_x1);printf("x2 %.5f %.5f/n/n",real_x2,imag_x2);printf("x3 %.5f %.5f/n/n",real_x3,imag_x3);}///////////////////////////结束/////////////////////////////

void fun2(double a1, double b1, double c1, double a2, double b2, double c2){ double x, y, d, e, f; d = a1*b2 - b1*a2; e = c1*b2-b1*c2; f = a1*c2-c1*a2; x = e/d; y = f/d; if (d == 0) { printf("infinite number of solutions"); } else { printf("x=%f, y=%f", x, y); }}