这题穷举举到大概N=210就很慢了，不过这个题没有让我们给出相应的式子，参考了几个文章，学习到了这种方法：每一次把一个加号改为减号的时候 此式子的值比原式减小了一个偶数，因此，我们只需要判断 1+2+...+k=S S-N是不是偶数就可以了 。
题目及AC代码如下：
| 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
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 int main(void) 5 { 6 int cases; 7 scanf("%d", &cases); 8 for (int i = 0; i < cases; i++) 9 { 10 int n; 11 scanf("%d", &n); 12 n = abs(n); 13 int k = 1; 14 while (1) 15 { 16 int S = (k + 1)*k / 2; 17 if ((S - n) % 2==0 && S-n>=0) break; 18 else k++; 19 } 20 printf("%d ", k); 21 if (i < cases - 1) printf(" "); 22 } 23 }