Question

C语言中用递归算法实现二分法求方程在（0，5）的近似解？

#include <stdio.h>#include <math.h>int main() { double calculate(float y1,float y2,float fy1,float fy2); float x1,x2,fx1,fx2; double a; printf("enter x1 & x2:"); scanf("%f,%f",&x1,&x2); fx1=(((x1+3)*x1-8)*x1+12)*x1-10; fx2=(((x2+3)*x2-8)*x2+12)*x2-10; a=calculate(x1,x2,fx1,fx2); printf("x=%lf\n",a); return 0;} double calculate(float y1,float y2,float fy1,float fy2) { float fy0,y0; double a; y0=(y1+y2)/2; fy0=(((y0+3)*y0-8)*y0+12)*y0-10; printf("%f ",y0); if(fabs(fy0)>=1e-10) { if ((fy0*fy1)<0) { y2=y0; fy2=fy0; } else { y1=y0; fy1=fy0; } a=calculate(y1,y2,fy1,fy2); } else { a=y0; printf("%6.2f",a); } return(a);}

Answer 1

重点改动：

scanf("%lf,%lf",&x1,&x2);

#include <stdio.h>

#include <math.h>

int main()

{ double calculate(double y1,double y2,double fy1,double fy2);

  double x1,x2,fx1,fx2;

  double a;

  printf("enter x1 & x2:");

  scanf("%lf,%lf",&x1,&x2);

  fx1=(((x1+3)*x1-8)*x1+12)*x1-10;

  fx2=(((x2+3)*x2-8)*x2+12)*x2-10;

  printf("%f %f %f %f\n",x1,x2,fx1,fx2);

  a=calculate(x1,x2,fx1,fx2);

  printf("x=%lf\n",a);

  return 0;

}

double calculate(double y1,double y2,double fy1,double fy2)

{ double fy0,y0;

  double a;

  y0=(y1+y2)/2;

  fy0=(((y0+3)*y0-8)*y0+12)*y0-10;

  printf("%f ",y0);

  if(fabs(fy0)>=1e-10)

  { if ((fy0*fy1)<0)

    { y2=y0;

      fy2=fy0;

    }

    else

    { y1=y0;

      fy1=fy0;

    }

    a=calculate(y1,y2,fy1,fy2);

  }

  else

  { a=y0;

    printf("%6.2f",a);

  }

  return(a);

}