一元三次方程-盛金公式求解

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原理参考-百度百科（http://baike.baidu.com/link?url=eA-bEvbcOBM2XmA4rzIG-lgci4MQdQcr7lCzCHBW-qG-qcPaDNovXp_jYxS2FUjlrOh1obH_D3Yv6ME2JYOxPyCgKhHIaXCHe-k9e69_c2tUxMTY90Z5-Oh9pZ-8LRf_）

先说一下中间出现的问题，开三次方的时候不能直接pow(x,1.0/3.0); 要分x大于零小于零，pow只接受大于零的参数。

另外，由于工程中只考虑实数，这里复数没有实现。

//cubicSolute.h

#pragma onceclass cubicSolute{public:cubicSolute(void);~cubicSolute(void);void setCoefficient(double a,double b,double c,double d)//设方程系数{this->a =a;this->b =b;this->c =c;this->d =d;}void judge();double getX1(){return X1;}double getX2(){return X2;}double getX3(){return X3;}private:double a,b,c,d;double A,B,C;double delt;//判别式double X1,X2,X3;};

//cubicSolute.cpp

#include "cubicSolute.h"#include <math.h>cubicSolute::cubicSolute(void){}cubicSolute::~cubicSolute(void){}//Δ根的判别式void cubicSolute::judge(){A = pow(b,2)-3*a*c;B = b*c-9*a*d;C = pow(c,2)-3*b*d;delt = pow(B,2)-4*A*C;//当A=B=0时，公式1.+定理6if ((A==0&&B==0)||(delt==0&&A==0)){X1 = -b/(3*a);X2 = X1;X3 = X1;}//Δ=B2－4AC>0时，公式2if (delt>0){double Y1 = A*b+3*a*((-B+pow(delt,0.5))/2);double Y2 = A*b+3*a*((-B-pow(delt,0.5))/2);//解决负数开三方出问题double Y1_three,Y2_three;if (Y1<0)Y1_three = - pow(-Y1,1.0/3.0);elseY1_three = pow(Y1,1.0/3.0);if (Y2<0)Y2_three = - pow(-Y2,1.0/3.0);elseY2_three = pow(Y2,1.0/3.0);X1 = (-b-(Y1_three+Y2_three))/(3*a);//X1,X2为复数，这里不需要，未实现。}//当Δ=B2－4AC=0时，公式3if (delt==0&&(A!=0)){double K = B/A;X1 = -b/a + K;X2 = -K/2.0;X3 = X2;}//当Δ=B2－4AC<0时，公式4if (delt<0){double T = (2*A*b-3*a*B)/(2*A*pow(A,0.5));//（A>0，－1<T<1）double theita = acos(T);X1 = (-b-2*pow(A,0.5)*cos(theita/3.0))/(3*a);X2 = (-b+pow(A,0.5)*(cos(theita/3.0)+pow(3,0.5)*sin(theita/3.0)))/(3*a);X3 = (-b+pow(A,0.5)*(cos(theita/3.0)-pow(3,0.5)*sin(theita/3.0)))/(3*a);}}

测试ok

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