给定一组 A 包含 N 正整数，我怎么能找到在最小整数> = 0 的，可以使用可获得的集合中的所有的元素。每个元素可以是可以是的加或减以总。 几个例子来说明这一点。

Given a set A containing n positive integers, how can I find the smallest integer >= 0 that can be obtained using all the elements in the set. Each element can be can be either added or subtracted to the total. Few examples to make this clear.

A = [2,1，3]

结果= 0 （2 + 1 - 3）

Result = 0 (2 + 1 - 3)

A = [1，2，0]

结果= 1 （1 + 2 + 0）

Result = 1 (-1 + 2 + 0)

A = [1，2，1，7，6]

结果= 1 （1 + 2 - 1 - 7 + 6）

Result = 1 (1 + 2 - 1 - 7 + 6)

推荐答案

您可以通过使用布尔整数规划解决这个问题。有几种算法（如戈莫里或分支定界）和自由库（例如LP-解决）可用。

You can solve it by using Boolean Integer Programming. There are several algorithms (e.g. Gomory or branch and bound) and free libraries (e.g. LP-Solve) available.

计算列表的总和，并调用它第双冠王列表中的号码。再说了一倍数字是A，B，C。然后你有以下方程组：

Calculate the sum of the list and call it s. Double the numbers in the list. Say the doubled numbers are a,b,c. Then you have the following equation system:

Boolean x,y,z 

a*x+b*y+c*z >= s

Minimize ax+by+cz!

布尔变量指示相应数量应该增加（如果真）或相减（当假的）。

The boolean variables indicate if the corresponding number should be added (when true) or subtracted (when false).

我要指出的转化问题，可以被看作是背包问题还有：

I should mention that the transformed problem can be seen as "knapsack problem" as well:

Boolean x,y,z 

-a*x-b*y-c*z <= -s

Maximize ax+by+cz!