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KpocaB4er16, вот нашел на Си реализацию:
Код: /* Cubic equation solution. Real coefficients case.
int Cubic(double *x,double a,double b,double c);
Parameters:
x - solution array (size 3). On output:
3 real roots -> then x is filled with them;
1 real + 2 complex -> x[0] is real, x[1] is real part of
complex roots, x[2] - non-negative
imaginary part.
a, b, c - coefficients, as described
Returns: 3 - 3 real roots;
1 - 1 real root + 2 complex;
2 - 1 real root + complex roots imaginary part is zero
(i.e. 2 real roots).
*/
#include <math.h> /* for sqrt(), fabs(), pow(), cos(), acos(). */
#define M_PI (3.141592653589793)
#define M_2PI (2.*M_PI)
int Cubic(double *x,double a,double b,double c) {
double q,r,r2,q3;
q=(a*a-3.*b)/9.; r=(a*(2.*a*a-9.*b)+27.*c)/54.;
r2=r*r; q3=q*q*q;
if(r2<q3) {
double t=acos(r/sqrt(q3));
a/=3.; q=-2.*sqrt(q);
x[0]=q*cos(t/3.)-a;
x[1]=q*cos((t+M_2PI)/3.)-a;
x[2]=q*cos((t-M_2PI)/3.)-a;
return(3);
}
else {
double aa,bb;
if(r<=0.) r=-r;
aa=-pow(r+sqrt(r2-q3),1./3.);
if(aa!=0.) bb=q/aa;
else bb=0.;
a/=3.; q=aa+bb; r=aa-bb;
x[0]=q-a;
x[1]=(-0.5)*q-a;
x[2]=(sqrt(3.)*0.5)*fabs(r);
if(x[2]==0.) return(2);
return(1);
}
} |
Если с Си туговато, говори, переделаем
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