Function: smallest positive integer - c++

Question

#include <iostream>
using namespace std;
int enough(int goal)
{
int C { };
int i { };
if (goal > 1)
{
while (((C += i) <= goal) && ((goal - i) <= (i + 1)))
{
i += 1;
}
}
else
{
i = 1;
}
return i;
}
int main()
{
cout << enough(21);
return 0;
}
So the purpose of this function is for it to return the smallest positive integer that can be summed consecutively from 1 before the cumulative sum becomes greater than the parameter "goal".
So, for example:
cout << enough(9) << endl;
will print 4 because 1+2+3+4 > 9 but 1+2+3<9
cout << enough(21) << endl;
will print 6 'cause 1+2+ . . .+6 > 21 but 1+2+ . . . 5<21
cout << enough(-7) << endl;
will print 1 because 1 > -7 and 1 is the smallest positive integer
cout << enough(1) << endl;
will print 1 because 1 = 1 and 1 is the smallest positive integer
My logic at first was using just while ((C += i) <= goal), but that went wrong, for example, for the parameter 21, where you get C = 20, which passes the test and ends up in i being augmented by 1 (resulting in the return value i = 7, which is clearly wrong).
Therefore I decided to create a test which tested both C and i, but that went wrong because the code isn't considering goal - i and i + 1 as separate tests for the while circuit, but I believe it is actually altering the value of ints goal and i, which screws everything up.
Any ideas where I went wrong?

Answer 1

Your approach is unnecessarily verbose. There is a closed-form (i.e. O(1)) solution for this.
The sum S of the arithmetic progression 1, 2, ..., n is
S = n * (n + 1) / 2
Rearranging this (completing the square), rejecting the spurious root, and rounding appropriately to fit the requirements of the question yields the result
n = std::ceil((-1 + std::sqrt(1 + 8 * S)) / 2)
This will not work for negative n, and possibly 0 too depending on the specific (and unspecified) requirements.
Or if you must use an O(N) approach, then
int enough(int goal){
int i = 0;
for (int total = 0; (total += ++i, total) < goal; );
return i;
}
will do it, which returns 1 for goal <= 1.

Answer 2

You could perhaps try to simplify your original O(N) approach by only having one condition in your while-loop, i.e., no &&:
#include <iostream>
using namespace std;
int enough(int goal)
{
if (goal <= 0)
{
return 1;
}
int total{};
int i{};
while (total < goal)
{
total += ++i;
}
return i;
}
int main()
{
cout << "enough(9): " << enough(9) << endl;
cout << "enough(21): " << enough(21) << endl;
cout << "enough(-7): " << enough(-7) << endl;
cout << "enough(1): " << enough(1) << endl;
cout << "enough(0): " << enough(0) << endl;
return 0;
}
Output:
enough(9): 4
enough(21): 6
enough(-7): 1
enough(1): 1
enough(0): 1