uva 10025 The ? 1 ? 2 ? ... ? n = k problem（算式规律）

 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

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<stdio.h>int main(){int t, n, sum, i;scanf("%d", &t);while (t--){scanf("%d", &n);if (n < 0)n = -n;for (sum = 0, i = 1; ; i++){sum += i;if (sum >= n && (sum - n) % 2 == 0){printf("%d\n", i);break;}}if (t)printf("\n");}return 0;}