Codeforces 11B Jumping Jack(数学)

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.

Sample test(s)

input

2

output

3

input

6

output

3

input

0

output

0

题意：

Jack在数轴上跳。他第i次跳能够往左或往右跳i个单位。如今问你他最少花多少步跳到点x。

思路：

能够无论x正负把x变成正数。这样是等效的。然后他最快的跳法当然是往一个方向一直跳。假设这样恰好能到x。这样肯定是步数最小的解。假设不能恰好跳到。我们设按这样的方法跳第一个比x大的位置为y。

假设y-x是偶数的话。我们还是能够达到最优步数。

就是先将第(y-x)/2步往左跳。其他不变。那么恰好也能到x。假设为奇数。变成下一个偶数即可了。为什么能够这样呢。由于无论如何。x=(±1)+(±2)+(±3)+.......(±n)。所以要想跳到必须奇偶性一致。比y小一定不行、所以仅仅有在y右边找了。

具体见代码：

#include<bits/stdc++.h>
int main()
{
    int tp,x,ans;
    while(~scanf("%d",&x))
    {
        x=x<0?-x:x,ans=ceil((-1+sqrt(1+8.0*x))/2),tp=ans*(ans+1)/2-x;
        while(tp&1) ans++,tp+=ans;
        printf("%d
",ans);
    }
    return 0;
}