xdfpmin.c

适合大型数值计算代码 现在网络上已经找不到了 购买需要20$

/* Driver for routine dfpmin */

#include <stdio.h>
#define NRANSI
#include "nr.h"
#include "nrutil.h"

static int nfunc,ndfunc;

float func(float x[])
{
	float x1p2sqr=SQR(2.0+x[1]);

	nfunc++;
	return 10.0*
		SQR(SQR(x[2])*(3.0-x[1])-SQR(x[1])*(3.0+x[1]))+
		x1p2sqr/(1.0+x1p2sqr);
}

void dfunc(float x[],float df[])
{
	float x1sqr=SQR(x[1]),x2sqr=SQR(x[2]),x1p2=x[1]+2.0;
	float x1p2sqr=SQR(x1p2);

	ndfunc++;
	df[1]=20.0*(x2sqr*(3.0-x[1])-x1sqr*(3.0+x[1]))*(-x2sqr-6.0*x[1]-3.0*x1sqr)+
		2.0*x1p2/(1.0+x1p2sqr)-2.0*x1p2*x1p2sqr/SQR((1.0+x1p2sqr));
	df[2]=40.0*(x2sqr*(3.0-x[1])-x1sqr*(3.0+x[1]))*x[2]*(3.0-x[1]);
}

#define NDIM 2
#define GTOL 1.0e-4

int main(void)
{
	int iter;
	float *p,fret;

	p=vector(1,NDIM);
	printf("True minimum is at (-2.0,+-0.89442719)\n");
	nfunc=ndfunc=0;
	p[1]=0.1;
	p[2]=4.2;
	printf("Starting vector: (%7.4f,%7.4f)\n",p[1],p[2]);
	dfpmin(p,NDIM,GTOL,&iter,&fret,func,dfunc);
	printf("Iterations: %3d\n",iter);
	printf("Func. evals: %3d\n",nfunc);
	printf("Deriv. evals: %3d\n",ndfunc);
	printf("Solution vector: (%9.6f,%9.6f)\n",p[1],p[2]);
	printf("Func. value at solution %14.6g\n",fret);
	free_vector(p,1,NDIM);
	return 0;
}
#undef NRANSI