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

---

從一個正換到一個負, 之間的差距是偶數, 因此我們先得到 1+ ... + n = n*(n+1)/2 >= k
只要差是偶數, 便一定可以湊得到那個差距。

#include <stdio.h>
#include <stdlib.h>
int main() {
	int t, n;
	scanf("%d", &t);
	while(t--) {
		scanf("%d", &n);
		n = abs(n);
		int k;
		for(k = 0; ; k++) {
			if(k*(k+1)/2 >= n && (k*(k+1)/2-n)%2 == 0)
				break;
		}
		printf("%d\n", n == 0 ? 3 : k);
		if(t)	puts("");
	}
    return 0;
}