UVA 10025 The ? 1 ? 2 ? ... ? n = k problem

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

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 }