uva 10025 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

题目大意：给出一个数， 判断它需要从1~m（可用+、-）填充的算式求得。求m

解题思路：首先， 1~m的和一定要 > n(如果全+都无法表示的话， 更别说有-）其次，找到一个m,使得1~m的和>n之后，因为sum >= n, 所以要减掉一个数， 比如在数字k前面加个-号， 相当于sum - 2 * k，也就是说每次减掉只能是偶数， 那么就要求sum % 2 == n % 2.(负数同理）

#include 
int main() {
	int cnt;
	long long sum, i, num;
	scanf("%d", &cnt);
	while (cnt--) {
		scanf("%lld", &num);
		sum = 0;
		if (num < 0) num = -num;
		for (i = 1; ; i++){	
			sum = sum + i;	
			if (sum >= num && (sum-num) % 2 == 0) {
				printf("%lld\n", i);
				break;
			}
		}
		if(cnt)	printf("\n");
	}
	return 0;
}