Codeforces Beta Round #11-B. Jumping Jack

原题链接

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

首先对输入的x取绝对值，再用(1+a)*a/2<= x，求出最大的a, 若(1+a)*a/2 == x直接输出a,否则 e = x - (1+a)*a/2;

1.如果a + 1-e是偶数那么可以在1到a中取一个数(-e+a+1)/2往左走，输出a+1

2.如果-e+a+1是奇数,那么-e+2*a+3是偶数的话在1到a+1中取一个数(-e+2*a+3)/2往左走,输出a+2

3.若-e+2*a+3是奇数，那么-e+3*a+6一定偶数，那么输出a+3

#include 
#define maxn 505
#define MOD 1000000007
using namespace std;
typedef long long ll;

int main(){
	
	ll a, b;
	
	scanf("%I64d", &a);
	a = abs(a);
	b = (-1 + sqrt(1+8 * a)) / 2;
	ll h = (1 + b) * b / 2;
	if(h == a){
		printf("%I64d\n", b);
	}
	else{
		ll e = a - h;
		e = b + 1 - e;
		b++;
		if(e % 2 == 0){
			printf("%I64d\n", b);
		}
		else if((e + b + 1) % 2 == 0){
			printf("%I64d\n", b+1);
		}
		else{
			printf("%I64d\n", b+2);
		}
	}
	return 0;
}