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

2

12

-3646397

Sample Output

7

2701

#include<cstdio>
#include<cstdlib>
#include<cmath>
int main(void)
{
    int cases;
    scanf("%d", &cases);
    for (int i = 0; i < cases; i++)
    {
        int n;
        scanf("%d", &n);
        n = abs(n);
        int k = 1;
        while (1)
        {
            int S = (k + 1)*k / 2;
            if ((S - n) % 2==0 && S-n>=0) break;
            else k++;
        }
        printf("%d
", k);
        if (i < cases - 1) printf("
");
    }
}