alternating sum

This is a discussion on alternating sum within the C++ Programming forums, part of the General Programming Boards category; Well at least with (i*i) you wont get function overheads, and its less work to type....

  1. #16
    Dr Dipshi++ mike_g's Avatar
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    Well at least with (i*i) you wont get function overheads, and its less work to type.

  2. #17
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    Hi all,
    Thank you for all the helpful pointers. I couldn't get the sign=-sign idea to work, I guess, I have no idea how. Can somebody be more specific on how to do that?
    Here's the code and the output though.. I still got warnings but it compiled anyway..

    Code:
    #include <iostream>
    #include <cmath>
    using namespace std;
    
    int main()
    {
    	double i = 0;
    	int sum = 0, n = 0;
    	int elements = 0;
    	cout << "Enter a value for n where n >= 1: ";
    	cin >> n;
    
    	for (i = 1; i < n; ++i)
    	{
    		elements = (pow(-1.0,i-1))*(i * i);
    		sum += elements;
    		cout << elements << " + ";
    	}
    	elements = (pow(-1.0,i-1))*(i * i);
    	sum += elements;
    	cout << elements << " = " << sum << endl;
    }

    Code:
    Enter a value for n where n >= 1: 4
    1 + -4 + 9 + -16 = -10
    Press any key to continue...

  3. #18
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by alyeska View Post
    Hi all,
    Thank you for all the helpful pointers. I couldn't get the sign=-sign idea to work, I guess, I have no idea how. Can somebody be more specific on how to do that?
    Well, that's pretty much it. The (-1)^i-1 goes back and forth between 1 and -1. So you start it off as 1; then after you use it, you set sign=-sign; and now it's -1. Then after you use it, you set sign=-sign; and now it's 1 again.

  4. #19
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by Elysia View Post
    I would say that depends.
    In Microsoft's implementation, at least, when you do power to an integer, it does a simple multiplication loop. Though if you do power against a double, it will enter complicated assembly code with lots of instructions to get the result.
    Oh, I hadn't noticed that until you said today.
    I guess it's only approx 10-20 times slower on recent versions of MSVC then. It's also O(log n) instead of O(1), as it takes longer the bigger the power is. Still no reason to use it when there are simpler ways.
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  5. #20
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by alyeska View Post
    Hi all,
    Thank you for all the helpful pointers. I couldn't get the sign=-sign idea to work, I guess, I have no idea how. Can somebody be more specific on how to do that?
    Well in your case you change sign to elements, but you basically write the line of code exactly as given.

    The warnings you are getting are about conversion from double to int. The above would eliminate that.
    My homepage
    Advice: Take only as directed - If symptoms persist, please see your debugger

    Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"

  6. #21
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    Here is what I came up with. There is definitely a better way to write this but I will go for it this time. Thank you everyone, especially tabstop and iMalc.

    Code:
    #include <iostream>
    #include <cmath>
    using namespace std;
    
    int main()
    {
    	double i = 0;
    	int sum = 0, n = 0;
    	int elements = 0;
    	cout << "Enter a value for n where n >= 1: ";
    	cin >> n;
    
    	for (i = 1; i < n; ++i)
    	{
    		elements = (pow(-1.0,i-1))*(i * i);
    		sum += elements;
    		if (elements % 2 != 0)
    			cout << " + " << elements << " ";
    		else
    			cout << elements;
    	}
    	elements = (pow(-1.0,i-1))*(i * i);
    	sum += elements;
    	if (elements % 2 != 0)
    		cout << " + " << elements << " = " << sum << endl;
    	else
    		cout << elements << " = " << sum << endl;
    }
    Code:
    Enter a value for n where n >= 1: 4
     + 1 -4 + 9 -16 = -10
    Press any key to continue...

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