Sum Problems

[Vneo]

I have to find the sum of all integer numbers lying between 1 and N inclusive.

Input
The input consists of a single integer N that is not greater than 10000 by it's absolute value.
Output
Write a single integer number that is the sum of all integer numbers lying between 1 and N inclusive.

Sample
input -3
Output -5

And this is my code

#include <iostream>
using namespace std;
int main()
{
	int a,b;
	cin >>a;
	if (a>0 && a<=10000){
		for (int i=1;i<=a;i++){
			b=b+i;
		}
	}
	else if (a<1 && a>= -10000){
		for (int i=1;i>=a;i--){
			b=b+i;
		}
	}
	cout<<b;
}

my problem is , when i submit it , it says wrong answer. but when i tested it, it seems work. Anyone can Help me?

[Peter87]

You need to initialize b.

Note that it is possible to calculate the sum without using loops.
https://en.wikipedia.org/wiki/Arithmetic_progression#Sum

[oleksii]

I think the sum of negative numbers will be wrong.
The second if statement should be: else if (a<0 && a>= -10000){...}

[Handy Andy]

Hello Vneo,

To be fair in addition to what Peater87 said you should initialize all variables before they are used, its just good practice.

When the variable is created, memory is set aside for the variable, but it contains whatever was last stored at that memory location. So, when you first use "b" it starts with whatever garbage that was at that memory location which throws the final sum off.

When you initialize "b" to zero when it is defined then the for loop will start adding to zero and not some garbage value like -8+ million as I have seen on my computer.

Hope that helps,

Andy

[jonnin]

these will help you in the long run, they are easier to type and you will see them a lot in others' code.

for adding 1,
b++;

and more generally, adding a value to a variable's existing value:
b += x;

-- exists for subtracting 1 and all the operators exist including logical for the self assignment:
--b;
b ^= 3; //xor example
b *= 3; //multiply example
etc.

[Vneo]

Thanks Handy Andy,
it seems i already know where i get wrong.