[Ty Meador]

I am an amateur computer programmer with 2 years experience.The following source code returns the square root of a number as a square root and a multiplier (ie sqrt45 = 3 * sqrt5). It works perfectly aside from a few.....anomalies.... I have written the program (which is project 9-2 in "Introduction to Computer Science using C++", this is not a school project). I have attached the source code (below) as well as a picture of the issue. Thanks in advance for your help!

//Header Files
#include <iostream>; 	//Required for i/o
#include <iomanip>;  	//Required for i/o Manipulators
#include <conio.h>;    	//Required for clrscr (Clear Screen)
#include <dos.h>;      	//Required for DOS commands
#include <stdio.h>;	   	//
#include <math.h>;	   	//Required for Math Library Functions
#include <ctype.h>;		/*Required for Character
						Analization Library Functions*/
using namespace std;
/*Programmer: 	Ty Meador
Name:			Project 9-2
Purpose: 		Calculate the Square root of a number as
				a  Multiplier and a Square root

Function Prototypes*/
void calceve(int incoming_integer);

//Globals
//double PI = 3.1415926535897

main()
{
		system("cls");
		int integer = 0;
		
		while(integer < 1)
		{
			cout<<"Please Input a Positive Integer: ";
			cin>>integer;
			if(integer > (32766))
			{
				system("cls");
				cout<<"Number Too Large\n";
				integer = 0;
			}
		}
		calceve(integer); //Passes input to function calc(int);
		cout<<'\n';
		system("pause");
return 0;
}

void calceve(int incoming_integer)
{
	double root = sqrt(incoming_integer), multiplier = 0;
	int i = 0,i2;       //i is outer loop control, i2 is inner
	int report = 2;
	do                  // this loop checks to see if the number is a perfect square, if so
	{                   // theres no reason to execute the next loop
		i++;
		if(i == root)
		{
			i = root;
			i2 = 1;
			report = 1;
			break;
		}
	}while(i < root);
	
	if (report == 2)//this checks to see if the previous loop determined
	{				//the number to be a perfect square, if not, GO!;
		i = 0;		//used 'i' as loop control for last loop, had to reset;
		do//outer loop runs until control multiplier == root;
		{
		
			i++;
			i2=0;
			if (i > incoming_integer)//if the outer loop becomes larger than
			{						 //the number input, the number cannot
				i = 1;				 //be simplified, so break loop, and set
				i2 = incoming_integer;//variables for proper output
				report=0;
				break;
			}
			while(i2 < incoming_integer-1)//inner loop runs to 1 less
			{							  //than input, to keep all answers from
				i2++;					  //being "1 * sqrt(input)"
				multiplier = i * sqrt(i2);
				if((multiplier >= root))//if multiplier >= root, break inner loop
				{						//because they cannot be equal
					break;
				}
			}	
		}while(multiplier != root);
	}
	cout<<endl<<incoming_integer<<" = "<<i<<" * sqrt("<<i2<<')';
	if(!report)
	{
		cout<<" OR "<<sqrt(incoming_integer);
		cout<<"\n\n(Cannot Be Simplified to a\nSquareroot and Multiplier)\n";
	}
}

Attached File(s)

•  Anomalie.bmp (712.2K)
  Number of downloads: 87

[Dogstopper]

Welcome to </dream.in.code>!

As this is your first post here, I'll go ahead and move this topic to the C++ forum, where it belongs. Next time though, please post in the correct forum. Thanks!

[NinjaCoder]

Here's an algorithm to break down the square root without using factorization. I'll leave it to you to break this down

#include <iostream>
#include <cmath>

int main()
{
	int number, power;
	std::cout<<"Enter Number : ";
	std::cin>>number;
	
	for(power = ceil(sqrt(number)); power > 1 && number % (int)pow(power,2)!=0; power--){
	}
	
	std::cout<<"sqrt("<<number<<") = "<<power<<"* sqrt("<<(number/pow(power,2))<<")"<<std::endl;
	std::cin.ignore();
	std::cin.get();
	return 0;
}

[Ty Meador]

NinjaCoder, on 23 May 2011 - 10:23 PM, said:

Here's an algorithm to break down the square root without using factorization. I'll leave it to you to break this down

#include <iostream>
#include <cmath>

int main()
{
	int number, power;
	std::cout<<"Enter Number : ";
	std::cin>>number;
	
	for(power = ceil(sqrt(number)); power > 1 && number % (int)pow(power,2)!=0; power--){
	}
	
	std::cout<<"sqrt("<<number<<") = "<<power<<"* sqrt("<<(number/pow(power,2))<<")"<<std::endl;
	std::cin.ignore();
	std::cin.get();
	return 0;
}

This was way more efficient than my method, of at the very least, it looks alot neater in code. Now i just have to look everything up and see just how it works. THANKS ALOT!!!!