Keep the rhythm with the count

closed account (ybo2y60M)
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#include "stdafx.h"
#include <iostream>
#include <cmath>

using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
	int X = 0;

	double a, b, c, xMin, xMax;

	double y = 0;

	cout << "#1(A): ";
	cin >> a;

	cout << "\n#2(B): ";
	cin >> b;

	cout << "\#3(C): ";
	cin >> c;


	cout << "Enter Xmin" << endl;
	cin >> xMin;

	cout << "Enter Xmax" << endl;
	cin >> xMax;


	y = a + b + c + X;

	for (int count = xMin; count <= xMax; count++)
	{
		cout << count << "\t" << y << "\n";
	}

	return 0;
}



The program is meant to perform the quadratic formula, everything is fine till I tried to get this output from the bottom section of the code

a+b+c = 3

xMin = 2, xMax = 6

Output

2----5

3----6

4----7

5----8

6----9

Instead

it's just

2---0

3----0

4 ---0

5----0

6----0

For something to that affect

Adding a link to xMin and xMAx doesn't seem to help either.

I'm not sure on how to proceed to achieving this result.
Last edited on
Did you mean
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for (X= xMin; X<= xMax; X++)
	{
		cout << X << "\t" << y+X<< "\n";
	}
?

Sorry, but I don't understand your program at all -..-
Maybe cause of my math s**ks...
Last edited on
line 32; y = a + b + c + X; should be moved inside the loop, so that it is executed each time.

Also the quadratic formula looks more like:
ax2 + bx + c
rather than
a + b + c + x
I see what's it.

And refer to what Chervil said,
Maybe you should use,
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for (X= xMin; X<= xMax; X++)
	{
		cout << X << "\t" << (a*X*X+b*X+c) << "\n";
	}
Last edited on
closed account (ybo2y60M)
Thank you lsk & Chervil.

That did the trick, gotta love the little nuances in coding.

Cheers
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