Well at least with (i*i) you wont get function overheads, and its less work to type.
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Well at least with (i*i) you wont get function overheads, and its less work to type.
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...
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...