Maths Programming Help

This is a discussion on Maths Programming Help within the C++ Programming forums, part of the General Programming Boards category; Hey, I new to C++ and I was trying to make a program that solved a maths problem for me ...

  1. #1
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    Question Maths Programming Help

    Hey, I new to C++ and I was trying to make a program that solved a maths problem for me (I wanted to find out how many times the anwer would be a positive integer, given that n is a positive integer):
    n+17
    ------ // That's a dividing sign
    n-7

    I tried a few methods, but couldn't get what I want.
    Here's my latest attempt:
    Code:
    #include <stdafx.h>
    #include <iostream>
    
    using namespace std;
    
    
    float main()
    {
    	float x, y, z;
    	for ( float n = 0; n < 1000; n = n + 14){
    		x = n - 7;
    		y = n + 17;
    		z = n / x;
    		if (z > 0){
    			cout<<y <<endl;
    			n = n - 17;
    			x = x - 7;
    		}
    		else {
    			cout<<y <<" is not a positive integer" <<endl;
    		}
    	}
    	cin.get();
    	return 0;
    }
    I'm using the Visual C++ Express 2005 program if tht helps. I thought I might make the variables floats to make sure the decimals aren't truncated in an int thinking the program will see them as positive integers anyway. Thanks!

  2. #2
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    If you want to know whether (n+17)/(n-7) is an integer, just check whether (n+17)&#37;(n-7) == 0 using integer arithmetic. The % is the modulus (remainder) operator.

    Edit: BTW, it's "int main()", not "float main()".
    Last edited by robatino; 04-20-2007 at 09:19 PM.

  3. #3
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    Thanks, I tried it once and it didn't work:
    Code:
    int main()
    {
    	for (int n = 0; n < 1000; n = n + 14){	
    		if ((n + 17)%(n - 7) == 0){
    			cout<<n <<endl;
    		}
    	}
    	cin.get();
    	return 0;
    }
    so then I assigned x the value of (n + 17)%(n - 7) here:

    Code:
    int main()
    {
    	for (int n = 0; n < 1000; n = n + 14){
    		int x = (n + 17)%(n - 7);
    		if ( x == 0){
    			cout<<n <<endl;
    		}
    	}
    	cin.get();
    	return 0;
    }
    It still didn't work.
    Note: There are very few answers to (n+17)/(n-7) which is why the computer might not be printing out anything.

  4. #4
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    On the one hand, you know that if n > 7, then (n+17)/(n-7) is greater than 1. On the other hand, as n increases, it decreases, and as n goes to infinity, it approaches 1, so for large enough n it will be less than 2 and so will never be an integer again. Keeping this in mind, you can probably work out by hand what all the possible values of n are.

  5. #5
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    > It still didn't work.
    Have you tried

    int n = someValueWhichIsKnownToWork;
    int x = (n + 17)&#37;(n - 7);

    To check the code does actually produce the right answer given correct input?
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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  6. #6
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    Robatino, yeh I wanted to try and make a program that would find those for me, as I haven't learnt how to do it yet (at school). And in one of the different attempts the computer printed -10, and as the condition was set to < 10,000,000 in the for loop (I think) (I set it that high because I wanted to see if the numbers would ever come down to -1, before it was just 1000), it kept printing -10 then started printing -9 and so forth until -2, each time taking more loops to change the number, then started printing -1.9 and so forth until -1.1 taking even more time per number, and finally reached -1 (whole) (probably because the computer couldn't calculate any lower decimals).

    Salem, as I said above, I wanted to find the answer with the help of a computer as I didn't know how to do it by hand.

    Anyway thanks for your help, I'll just leave it unless someone knows maybe a differen way of doing this. Thanks again!
    Last edited by UPNPAD; 04-21-2007 at 12:14 AM. Reason: Just a spelling mistake

  7. #7
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    Code:
    #include <iostream>
    using namespace std;
    
    int main() {
    	int n = 8;
    	while ((n+17)/(n-7) >= 2) {
    		if ((n+17)&#37;(n-7) == 0) cout << n << "\n";
    		n++;
    	}
    	return 0;
    }
    Just a pretty simple loop if you do some basic math to determine where it will converge and where to start from. Obviously you don't want to start from 7 because you will get divide by 0, anything less will result in a negative number which isn't wanted. Keep looping until it's less than 2 because the fraction goes to 1 for inf, and anything between 1 and 2 can't be a positive integer. (Basically what everyone said above )

    Btw, why did you increment n by 14 each time. The first "n" is actually 8 (25/1 = 25). Also when I ran it none were multiples of 14 - they were 8,9,10,11,13,15,19,31.
    Last edited by 0rion; 04-23-2007 at 03:19 AM.
    The cost of software maintenance increases with the square of the programmer's creativity.

  8. #8
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    It's easier to do some of this by hand - for example, if (x+17)/(x-7) == 2, it's trivial to solve for x to get x == 31. So one can just check by hand (or in one's head) values of n between 8 and 30 (of course n == 31 is also a solution).
    Last edited by robatino; 04-23-2007 at 07:14 AM.

  9. #9
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    Stay away from floats and doubles if you need to work with real mathematical fractions.

    You had the correct solution before, except you were doing n = n + 14
    Code:
    #include <iostream>
    
    int main()
    {
    	for (int n = 0; n < 1000; ++n){	
    		// Careful to not divide by 0
    		if (n != 7 && (n + 17)&#37;(n - 7) == 0){
             std::cout<< n <<std::endl;
    		}
    	}
       std::cin.get();
    	return 0;
    }
    Callou collei we'll code the way
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