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

212-3646397

Sample Output

72701

#include<stdio.h>int main(){    int i , n , j , sum , k ;    scanf("%d", &n);    for(i = 1 ; i <= n ; i++)    {        scanf("%d", &k);        if(k<0)        {k=-1*k;}         if(k > 0)        {            sum = 0;            for(j = 1 ; ; j++)            {                sum = sum + j;                if(sum ==k || (sum > k && (sum-k)%2==0 ) )                {                    printf("%d\n", j);                    break;                }            }        }        else            printf("3\n");        if(i!=n)            printf("\n");    }    return 0;}
这道题刚开始一看一点没思路。后来仔细思考了一下

他减去的都是偶数所以只要判断和比要求的数大于或等于

然后判断差值是否是偶数。

我开始的时候漏了k=0；的情况导致错误，很可惜。