求助取N个,等于固定值算法.

[czyoooo]

ID TOTAL
1 100
2 200
3 200
4 300
5 400
6 500


取其中几个累加等于1000,的所有ID
例如:
100+200+200+500=1000, 得到1,2,3,6
200+300+500=1000,得到2,4,6或3,4,6
100+400+500=1000,得到1,5,6
......



求这样的函数.

[vrhero]

子集合加总问题...贪婪算法，动态规划，自己google去...

[czyoooo]

都是C++的,C#的有吗?

[threenewbee]

懒得专门为你这个写算法了。。。

给你一个排列组合再筛选的的代码。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            List<int> list = new List<int>() { 100, 200, 300, 400, 500, 600 };
            list.Combo(1, list.Count)
                .Where(x => x.Sum() == 1000).ToList().ForEach(x => Console.WriteLine(String.Join("", x)) );
        }
    }

    static class ListCombo
    {
        public static IEnumerable<List<T>> Combo<T>(this List<T> sourcelist, int min, int max)
        {
            for (int i = min; i <= max; i++)
                foreach (var item in _Combo(sourcelist, i))
                    yield return item;
        }

        static IEnumerable<List<T>> _Combo<T>(this List<T> sourcelist, int len)
        {
            if (len <= 0)
            {
                yield return new List<T>();
            }
            else
            {
                int[] pos = new int[len];
                for (int i = 0; i < len; i++) pos[i] = i;
                while (pos[0] < sourcelist.Count - len)
                {
                    List<T> newlist = new List<T>();
                    for (int i = 0; i < len; i++) newlist.Add(sourcelist[pos[i]]);
                    for (int i = len - 1; i >= 0; i--)
                    {
                        if (pos[i] < sourcelist.Count - len + i)
                        {
                            pos[i]++;
                            for (int j = i + 1; j <= len - 1; j++)
                                pos[j] = pos[i] + j - i;
                            break;
                        }
                        else
                        {
                            continue;
                        }
                    }
                    yield return newlist;
                }
                List<T> last = new List<T>();
                for (int i = sourcelist.Count - len; i < sourcelist.Count; i++)
                    last.Add(sourcelist[i]);
                yield return last;
            }
        }
    }
}

[czyoooo]

这样的写法2的N次方.内存能够?
如果记录数量为２００．恐怖呀．

[chenchenyangll]

[Quote=引用 4 楼 czyoooo 的回复:]

这样的写法2的N次方.内存能够?
如果记录数量为２００．恐怖呀．
[/Quote]

用背包

[jshi123]

static void Main(string[] args)
{
	int[] values = { 100, 200, 200, 300, 400, 500 };
	var testdata = values.Select((n, i) => new { id = i + 1, total = n }).ToArray();
	var total = 1000;

	Func<IEnumerable<int[]>, int[], IEnumerable<int[]>> comb = (x, y) =>
		from a in x
		from b in y
		where a.Count() == 0 || a.Last() < b
		select a.Concat(new[] { b }).ToArray();

	var indices = Enumerable.Range(0, testdata.Length).ToArray();
	var list = new int[][] { new int[] { } }.AsEnumerable();
	var result = Enumerable.Range(1, testdata.Length)
		.SelectMany(i => list = comb(list, indices))
		.Where(c => c.Sum(i => testdata[i].total) == total)
		.Select(c => c.Select(i => testdata[i]));

	foreach (var c in result)
		Console.WriteLine("({0})\t {1} = {2}",
			string.Join(", ", c.Select(d => d.id)),
			string.Join(" + ", c.Select(d => d.total)),
			total);
}