java – 所有自然数,总和为N,反转总和为1

a + b + c + ... = N
1/a + 1/b + 1/c + ... = 1

a,b,c不一定是唯一的.

我已经提出了Java中的以下代码.它适用于简单的情况,但对于N>已经非常缓慢. 1000.

如何重写该方法,即使数百万的工作速度也很快？也许,我应该放弃递归,或者用我想念的数学技巧来切断一些分支？

SSCEE：

private final static double ONE = 1.00000001;

public List<Integer> search (int number) {
    int bound = (int)Math.sqrt(number) + 1;
    List<Integer> list = new ArrayList<Integer>(bound);

    if (number == 1) {
        list.add(1);
        return list;
    }

    for (int i = 2; i <= bound; i++) {
        list.clear();
        if (simulate(number,i,list,0.0)) break;
    }

    return list;
}


//TODO: how to reuse already calculated results?
private boolean search (int number,int n,List<Integer> list,double sum) {
    if (sum > ONE) {
        return false;
    }

    //would be larger anyway
    double minSum = sum + 1.0 / number;
    if (minSum > ONE) {
        return false;
    }

    if (n == 1) {
        if (minSum < 0.99999999) {
            return false;
        }

        list.add(number);
        return true;
    }

    boolean success = false;
    for (int i = 2; i < number; i++) {
        if (number - i > 0) {
            double tmpSum = sum + 1.0 / i;
            if (tmpSum > ONE) continue;

            list.add(i);
            success = search(number - i,n - 1,tmpSum);
            if (!success) {
                list.remove(list.size() - 1);
            }

            if (success) break;
        }
    }

    return success;
}