如何通过使用+或获取的数号的列表 - 操作 [英] How to get a number from a list of numbers by using + or - operations

问题描述

在我的问题，我们得到了一些数字和名单，我们应该通过行动来得到的数字给定列表中的第一个号码+或 -

In my question, we are given a number and list of numbers and we are supposed to get the first number from the given list of numbers by using the operations + or -

例如： -1是目标数 1 2 3 5号给得-1 解决办法应该是1 + 2 + 3-5 = -1 或-1-2-3 + 5 = -1

For Example: -1 is the target number 1 2 3 5 are numbers given to get -1 solution should be -1+2+3-5 = -1 or -1-2-3+5 = -1

限值目标数是从-180至180和限制编号列表是从2至20个

Limits for target number is from -180 to +180 and the limit for list of numbers is from 2 to 20

要找到什么样的算法应该使用的解决方案？如果我想使用生成所有的可能性会是有效的？而就是这个问题的任何二进制的解决方案？

To find the solution what kind of algorithm should be used? If I want to use generate all the possibilities will it be efficient? And is the any binary solution of this problem?

感谢您的帮助

推荐答案

的possibpe变种数量为 2 ^ 20 ;让我们产生从 0 所有数字以 2 ^ N 。二进制重新$ P $这些数字psentation将是00000（20零），000 ... 01,0000.10，......，1111（20的）。试想一下，每个零为负，一个是一个加号。

Number of possibpe variants is 2^20; Let's generate all numbers from 0 to 2^N. Binary representation of these numbers will be 00000(20 zeroes), 000...01,0000.10,...,1111(20 ones). Imagine, that each zero is minus and one is a plus.

    int target = -1;
    int[] numbers = new int[20];
    Arrays.fill(numbers, 0);

    numbers[0] = 1;
    numbers[1] = 2;
    numbers[2] = 3;
    numbers[3] = 5;
    for(int i=0;i<(1<<20);i++) //masks from 00...00 to 11...11 (from --...--- to ++...+++)
    {
        int sum=0;
        for(int bit=0;bit<20;bit++)
        {
            if(((1<<bit)&i)>0)
            {
                sum+=numbers[bit];
            }
            else
            {
                sum-=numbers[bit];
            }
        }
        if(sum==target)
        {
            System.out.print(target+" = ");
            for(int bit=0;bit<20;bit++)
            {
            if(((1<<bit)&i)>0)
            {
                System.out.print("+"+numbers[bit]);
            }
            else
            {
                System.out.print("-"+numbers[bit]);
            }
            }
            break;
        }
    }

输出： -1 = 1 + 2 + 3-5-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0

Output: -1 = -1+2+3-5-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0-0

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