The
 ? 1 ? 2 ? ... ? n = k problem 
The problem
Given the following formula, one can set operators ‘+‘ or ‘-‘ instead of each ‘?‘, in order to obtain a given k
? 1 ? 2 ? ... ? n = k
For example: to obtain k = 12 , the expression to be used will be:
- 1 + 2 + 3 + 4 + 5 + 6 - 7 = 12
with n = 7
The Input
The first line is the number of test cases, followed by a blank line.
Each test case of the input contains integer k (0<=|k|<=1000000000).
Each test case will be separated by a single line.
The Output
For each test case, your program should print the minimal possible n (1<=n) to obtain k with the above formula.
Print a blank line between the outputs for two consecutive test cases.
Sample Input
2

12

-3646397
Sample Output
7

2701
Alex Gevak
September 15, 2000 (Revised 4-10-00, Antonio Sanchez)


   题意：通过在1到x之间添加+或者-运算，使得最后的结果等于n，输出最小的x的

值。最初想着使用BFS，提交时TLE，因为n的值太大，看到别人用到的一种数学做

法，很简单。

  思路：通过1到i相加得到的和sum大于等于n，自i开始求（sum-n）%2 == 0 ，等式

不成立时sum = sum + i （i++），等式成立时输出i的值，这是的i就是要求的值。

因为sum存的是1到i的和，如果在i的时候出现“-”号 就相当于在n的基础上加了2个

i，所以差值被2整除的值所对应的就是要求的值。 


#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main()
{
    int T;
    scanf("%d",&T);
    long long int n;
    while(T--)
    {
        long long int sum = 0;
        scanf("%lld",&n);
        if(n<0)
        {
            n = -n;
        }
        int ii  = 0;
        for(int i=1;;i++)
        {
            sum += i;
            if(sum >= n)
            {
                ii = i;
                break;
            }
        }
        if((sum - n)%2 == 0)
        {
            if(T>0)
            {
                printf("%d\n\n",ii);
            }
            else
            {
                printf("%d\n",ii);
            }
            continue;
        }
        for(int j=ii+1;;j++)
        {
            sum += j;
            if((sum-n)%2 == 0)
            {
                if(T>0)
                {
                    printf("%d\n\n",j);
                }
                else
                {
                    printf("%d\n",j);
                }
                break;
            }
        }
    }
    return 0;
}