日期:2014-05-20  浏览次数:21138 次

用java递归求一个平均数多少种可能的算法
在做一个考试系统,用户输入了考试要出多少道题目,和平均难度系数,我想得出根据这个平均系数能产生多少种可能,难度系数是1-5;比如用户选择3道选择题,平均系数为3,应该显示1-3-5、1-4-4、2-2-5、2-3-4,3-3-3,谢了

------解决方案--------------------
Java code

    public static void main(String[] args) {
        List<List<Integer>> re = op(1, 5, 3, 3);
        if (re == null) {
            System.out.println("error");
            return;
        }
        for (List<Integer> tl : re) {
            for (Integer n : tl) {
                System.out.print(n + ",");
            }
            System.out.println("");
        }
    }

    static List<List<Integer>> op(final int min, final int max, final int age, final int count) {
        List<List<Integer>> tn = new ArrayList<List<Integer>>();
        {//初始化
            int sum = age * count;
            //确定最大边界
            int bmax = sum - min * (count - 1);
            bmax = Math.min(bmax, max);
            //最小边界
            int bmin = sum - max * (count - 1);
            bmin = Math.max(bmin, min);
            if (bmax < bmin) {
                return null;
            }
            for (int i = bmin; i <= bmax; i++) {
                List<Integer> nl = new ArrayList<Integer>();
                nl.add(i);
                tn.add(nl);
            }
        }
        for (int i = 1; i < count; i++) {
            List<List<Integer>> btn = new ArrayList<List<Integer>>();
            for (int j = 0, jm = tn.size(); j < jm; j++) {
                List<Integer> nl = tn.get(j);
                int sum = count * age;
                for (Integer item : nl) {
                    if (item == null) {
                        break;
                    }
                    sum -= item;
                }
                //确定最大边界
                int bmax = sum - min * (count - i - 1);
                bmax = Math.min(bmax, max);
                //最小边界
                int bmin = sum - max * (count - i - 1);
                bmin = Math.max(bmin, min);
                for (int z = bmin; z <= bmax; z++) {
                    List<Integer> bnl = new ArrayList<Integer>();
                    bnl.addAll(nl);
                    bnl.add(z);
                    btn.add(bnl);
                }
            }
            tn = btn;
        }
        //排序
        for (List<Integer> l : tn) {
            Collections.sort(l);
        }
        Collections.sort((List) tn, new Comparator<List<Integer>>() {

            public int compare(List<Integer> o1, List<Integer> o2) {
                for (int i = 0; i < o1.size(); i++) {
                    if (!o1.get(i).equals(o2.get(i))) {
                        return o1.get(i) - o2.get(i);
                    }
                }
                return 0;
            }
        });
        //去重复
        for (int i = 1; i < tn.size(); i++) {
            List<Integer> a = tn.get(i - 1);
            List<Integer> l = tn.get(i);
            if (a.get(0).equals(l.get(0))) {
                if (a.containsAll(l) && l.containsAll(a)) {
                    tn.remove(i);
                    i--;
                }
            }
        }
        return tn;
    }

------解决方案--------------------
Java code
    public static void test4(int[] array, int begin, int num, int cVal, int val, List<Integer> result)
    {
        if (result.size()==num)
        {
            if (cVal==val)
            {
                System.out.println(result);    
            }
            return;    
        }
        for (int i = begin; i<array.length; i++)
        {
            
            if (cVal<val&&result.size()<num)
            {
                result.add(array[i]);
                cVal += array[i];
                test4(array,i+1, num, cVal, val, result);
                cVal -= result.remove(result.size()-1);
            }
        }
    }

    public static void main(String[] args)
    {
        int[] array = { 1, 2, 3, 4 ,5,6,7,8,9};
        test4(array, 0, 3, 0, 11, new ArrayList());
    }