11B - Jumping Jack【思维】

B. Jumping Jack

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get to the point x. In order to train, he has decided that he'll first jump by only one unit, and each subsequent jump will be exactly one longer than the previous one. He can go either left or right with each jump. He wonders how many jumps he needs to reach x.

Input

The input data consists of only one integer x ( - 109 ≤ x ≤ 109).

Output

Output the minimal number of jumps that Jack requires to reach x.

Examples

input

2

output

3

input

6

output

3

input

0

output

0

全部当成正的来算，效果一样

先试着将x从0一直增加，当增加到x>=n时，进行讨论

1，x=n，那么所得即所求

2，如果多出的部分为偶数，那么只需要把这个偶数的一半的那个奇数前的符号变换符号，转化为情况1

3，如果多于的部分为奇数，那就继续向后加，一直加到差值为偶数，转化为情况2

所以代码如下：

#include
#include
using namespace std;
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		n=abs(n);
		int ans=ceil((-1+sqrt(1+8.0*n))/2.0); 
		int d=ans*(ans+1)/2-n;
		while(d&1)
		{
			++ans;d+=ans;
		}
		printf("%d\n",ans);
	}
	return 0;
}