Function solve

Spoj_problems/NSTEPS/solution.cpp:9–23

Here, in this method, we will solve this problem using a odd and even pattern. And, we will check if the given x,y points lies on line x=y or x-y=2 */

	And, we will check if the given x,y points lies on line x=y or x-y=2
*/
int solve(int x, int y){
	if(x-y==0 && y%2==0){			//line is x=y and even pattern
		return y*2;
	}
	else if(x-y ==0 && y%2 !=0 ){		//line is x=y and odd pattern
		return (y-1)*2 + 1;
	}
	else if(x-y == 2 && y%2 != 0){		//line is x-y=2 and odd pattern
		return y*2 + 1;
	}
	else if(x-y == 2 && y%2 == 0){		//line is x-y=2 and even pattern
		return y*2 + 2;
	}
	else return -1;				//no point is there
}
int main()
{
	jaadu;