st-l5.cpp

FIND ROOT OF EQUATION WITH DIVIDUAL IN TWO METHOD

#include <iostream.h>
#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
#include <math.h>
#include <iomanip.h>
#include <ctype.h>
#include <string.h>
typedef float(*pFunc)(float);
  ////Root of equation////
float root(pFunc F,float a,float b,float eps)
 {
  float Fa,Fc,c;
  Fa = (*F)(a);
  c= (a+b)/2;
  Fc=(*F)(c);
  if (fabs(Fc)<eps) return c;
  else if(Fa*Fc>0) return root(F,c,b,eps);
	   else return root(F,a,c,eps);
 }
 ////Function f1////
float f1(float x)
 {
  return pow(x,2)-16;
 }
 ////Function f2////
float f2(float x)
 {
  return pow(x,3)+ 2*pow(x,2)- 8;
 }
 ////Function f3////
float f3(float x)
 {
  return pow(x,4) - 2*pow(x,3) - 2;
 }
 ////Check real number for string str////
int check_num(char str[20])
  {
	 int i,k=0,OK=1;
	 if ((str[0]=='+')|(str[0]=='-')|(isdigit(str[0])))
	   {
		 for(i=1;str[i];i++)
		   {
			 if(!((isdigit(str[i]))|(str[i]==',')|(str[i]=='.')))
			   {
				 OK=0;break;
			   }
			 else
			   {
				 if((str[i]==',')|(str[i]=='.')) k++;
				 if ((k > 1)|(str[strlen(str)-1]==',')|(str[strlen(str)-1]=='.'))
				  {
					OK=0;break;
				  }
			   }
		   } //end of for
		   if ((k==1)&&((str[0]=='+')|(str[0]=='-'))&&(!isdigit(str[1])))
		   OK = 0;
	   }
	 else OK=0;
	 return OK;
  }
  ////Input data////
void input(pFunc F,float &a,float &b,float &eps)
 {
   char str[20];
   int OK;
   cout <<"   \n------------------------INPUT---------------------\n";
	   //// Input a ////
   do
	{
	  cout <<" \n-Input value of a : a = ";gets(str);
	  OK=check_num(str);
	  if (!OK) cout <<"  Error input value of a. Please try again!\n";
	}
   while(!OK);
   a = atof(str);
	  //// Input b ////
   do
	{
	  cout <<" \n-Input value of b : b = ";gets(str);
	  OK=check_num(str);
	  if(OK)
		{
		  b = atof(str);
		  if ((b <= a)|((*F)(b)*(*F)(a)>0)) OK=0;
		}
	  if (!OK) cout <<"  Error input value b > a(f(b)*f(a)<0). Please try again!\n";
	}
   while(!OK);
   b = atof(str);
	  //// Input precision ////
   do
	{
	  cout <<" \n-Input value of precision(Eps > 0) : Eps = ";gets(str);
	  OK=check_num(str);
	  if(OK)
		{
		  eps = atof(str);
		  if (eps <= 0) OK=0;
		}
	  if (!OK) cout <<"  Error input value precision(Eps > 0). Please try again!\n";
	}
   while(!OK);
   eps = atof(str);
 }
 ////MAIN////
void main()
 {
  float a,b,eps,rootF;
  int choice,k;
  do
  {
   clrscr();
   cout <<"Program to find roots of equation with dividual in two method on interval (a,b).\n" ;
   cout <<" Select an equation : \n\n";
   cout <<"  1.x^2 - 16       = 0\n\n";
   cout <<"  2.x^3 + 2x^2 - 8 = 0\n\n";
   cout <<"  3.x^4 - 2x^3 - 2 = 0\n\n";
   cout <<"  4.Press any key to exit...\n\n";
   cout <<"  You select : ";choice = getch();
   switch (choice) {
	case '1' : cout <<"  1.x^2 - 16       = 0\n";
			   input(f1,a,b,eps);
			   rootF = root(f1,a,b,eps);
			   break;
	case '2' : cout <<"  2.x^3 + 2x^2 - 8 = 0\n";
			   input(f2,a,b,eps);
			   rootF = root(f2,a,b,eps);
			   break;
	case '3' : cout <<"  3.x^4 - 2x^3 - 2 = 0\n";
			   input(f3,a,b,eps);
			   rootF = root(f3,a,b,eps);
			   break;
	default  :
			  cout <<" exit!";
			  getch();
			  exit(1);
   }
   cout <<"\n---ROOT OF EQUATION WITH DIVIDUAL IN TWO METHOD---\n";
   cout <<setiosflags(ios::showpoint) <<setprecision(5);
   cout <<"\n Root of equation : x = " <<rootF;
   cout <<"\n\n Do you want to continue ?(Y/N)";
   k=getch();
  }
  while((k=='y')|(k=='Y'));
 } ////END OF MAIN