下面这个程序是用龙格库塔法求解微分方程的，哪位学长能帮忙把它改成用数组形式的，我不太会，想学习一下，我编的这个太繁杂可，谢谢！
/*龙格库塔法求解微分方程*/
#include   "stdio.h"
#include   "math.h"
#define   A   0                                       /*定义初值和步长*/
#define   B   0.002
#define   C   1
#define   D   1
#define   E   1

double   f(double,double,double,double);                 /*   定义函数*/
double   g(double,double,double,double);
double   e(double,double,double,double);

main()
{
    double   x0=A,   h=B,
                  y01=C,   u01=D,   v01=E,
                  j01,j02,j03,j04,k01,k02,k03,k04,l01,l02,l03,l04;
    int   i,n=10000;
    FILE   *cfPtr;
    if((cfPtr=fopen("luolunzi.dat","w"))==NULL)                     /*数据要写入的文件*/
    printf("File   could   not   be   opened\n");
    else
    {   
    for   (i=1;i <=n;i++)
      {                                                                                               /*套用公式开始计算*/   
        x0=x0+h;
        j01=f(x0,y01,u01,v01);
        k01=g(x0,y01,u01,v01);
        l01=e(x0,y01,u01,v01);
      
        j02=f(x0+h/2.0,y01+h/2.0*j01,u01+h/2.0*k01,v01+h/2.0*l01);
        k02=g(x0+h/2.0,y01+h/2.0*j01,u01+h/2.0*k01,v01+h/2.0*l01);
        l02=e(x0+h/2.0,y01+h/2.0*j01,u01+h/2.0*k01,v01+h/2.0*l01);

        j03=f(x0+h/2.0,y01+h/2.0*j02,u01+h/2.0*k02,v01+h/2.0*l02);
        k03=g(x0+h/2.0,y01+h/2.0*j02,u01+h/2.0*k02,v01+h/2.0*l02);
        l03=e(x0+h/2.0,y01+h/2.0*j02,u01+h/2.0*k02,v01+h/2.0*l02);
         
        j04=f(x0+h,y01+h*j02,u01+h*k02,v01+h*l02);
        k04=g(x0+h,y01+h*j02,u01+h*k02,v01+h*l02);
        l04=e(x0+h,y01+h*j02,u01+h*k02,v01+h*l02);
      
        y01=y01+h/6.0*(j01+2.0*j02+2.0*j03+j04);
        u01=u01+h/6.0*(k01+2.0*k02+2.0*k03+k04);
        v01=v01+h/6.0*(l01+2.0*l02+2.0*l03+l04);
        
        printf("%15.7f\t%15.7f\t%15.7f\t\n",x0,y01,u01,v01);
        fprintf(cfPtr,"%15.7f\t%15.7f\t%15.7f\t\n",x0,y01,u01,v01);
        }
      fclose(cfPtr);
}
}   

double   f(double   s1,double   q1,double   u1,double   r1)                         /*定义的微分方程*/     
{
      double   z1;
      z1=-10*q1+(10*u1);
      return   z1;
}
double   g(double   s2,double   q2,double   u2,double   r2)
{
      double   z2;
      z2=28*q2-u2-q2*r2;
      return   z2;
}
double   e(double   s3,double   q3,double   u3,double   r3)
{
double   z3;
z3=(-8/3)*r3+q3*u3;
return   z3;
}