设M是在范围[1的整数; 10亿。

Let M be an integer in range [1; 1,000,000,000].

一个的分解的的M是一组独特的整数，其总和等于M.

A decomposition of M is a set of unique integers whose sum is equal to M.

一个分解的奇的，如果它仅包含奇数。

A decomposition is odd if it contains only odd integers.

的M甲分解的最大的，如果有是M没有其他分解更大集合中的大小。

A decomposition of M is maximal if there is no other decomposition of M greater in size of the set.

写一个函数：

int[] maxOddDecomposition(int M)

这是返回一个数组M的的最大单分解的数组中的数字应该是升序排列。如果M没有任何奇数分解，阵列应为空。如果有一个以上的正确答案，该函数可以返回它们中的任何

that returns an array with a maximal odd decomposition of M. The numbers in array should be in ascending order. If M does not have any odd decomposition, an array should be empty. If there is more than one correct answer, the function may return any of them.

例如，M = 6有四个分解：

For example, M = 6 has four decompositions:

6 = 1 + 2 + 3
  = 1 + 5
  = 2 + 4
  = 6

只有 1 + 5 是一个奇怪的分解，因而最大奇数分解。我们应该在阵列它回报，使得数组[0] = 1 和数组[1] = 5 。

Only 1 + 5 is an odd decomposition, thus is the maximal odd decomposition. We should return it in array such that array[0] = 1 and array[1] = 5.

预计最坏情况下的时间和空间复杂度为O（开方（M））。

Expected worst-case time and space complexity is O(sqrt(M)).

我已经试过什么：

由于时间复杂度，必须SQRT（M），它提醒中号算法，在这里我们从1迭代到的sqrt（M）的幼稚因式分解的箱。没有进一步的想法出现了，虽然。只有它必须是非常快，只有开方（M）的步骤。

Since the time complexity has to be sqrt(M) it reminded me of naive factorization of M algorithm, where we iterate from 1 to sqrt(M). No further thoughts appeared though. Only that it must be really fast, only sqrt(M) steps.

所以，我做了一些例子。如何找到20例如一个答案？什么是奇数小于20？ 1 + 3 + 5 + 7 + ......我们已经有16所以，我们可以添加4，但4是偶数。

So, I did some examples. How to find an answer for 20 for example? What are the odd numbers less than 20? 1 + 3 + 5 + 7 + ... we already have 16. So, we could add 4, but 4 is even.

那么，让我们来替换7（7 + 4）= 11，我们正在做的：1 + 3 + 5 + 11，我注意到的是，初始序列一向楼（开方（M））的元素，完美的。让我们code它在伪code：

So, let's replace 7 with (7 + 4) = 11 and we are done: 1 + 3 + 5 + 11. What I noticed is that the initial sequence had always floor(sqrt(M)) elements, perfect. Let's code it up in pseudo-code:

int len = floor(sqrt(M));
int result[] = new int[len];
int sum = 0;
for (i = 0; i < len - 1; i++) {
    result[i] = 1 + 2*i;
    sum += result[i];
}
result[len - 1] = M - sum;
return result;

我做了一个特殊的情况下，M = 2，返回一个空数组。我认为就是这样，finito。

I did a special case for M = 2, returning an empty array. I thought that's it, finito.

我没有注意到，它打破​​3，因为它给了 1 + 2 ，而不是 3 的。而对于5，给 1 + 3 + 1 ，而不是 5 。而更多的。

I didn't notice that it breaks for 3, because it gives 1 + 2 instead of 3. And for 5, gives 1 + 3 + 1, instead of 5. And for many more.

你将如何解决呢？

推荐答案

下面是一个确定性的解决问题的办法。设M = {1，3，5，...，2 * K-3，2 * K-1，R}，其中为r = 2 * K + 1中。明显的最大分解是不将有超过数第（k + 1）。

Here is a deterministic solution to the problem. Suppose M = {1, 3, 5, ..., 2*k-3, 2*k-1, r} where r <= 2*k + 1. It is 'obvious' that the maximal decomposition is not going to have more numbers than (k+1).

我们有以下的情况下，对于k> 3（推理和处理以前的案件是后来psented $ P $）：

We have the following cases for k > 3 (the reasoning and handling of earlier cases is presented later):

案例1，如果r是奇数并等于2 * K + 1：添加r转换到列表中从而得到的第（k + 1）元素的分解

Case 1. If r is odd and equal to 2*k+1: add r into the list thereby giving a decomposition of (k+1) elements.

案例2，如果r是偶数：代替{（2 * k-1个）中，r} {2 * K-1 + R}给出k个元素的分解

Case 2. If r is even: replace {(2*k-1), r} by {2*k-1+r} giving a decomposition of k elements.

案例3.如果r是奇数和不等于2 * K + 1：取代所述第一和最后两个元素中的序列{1，2 * K-1，R}由{2 * K + R}给人（K-1）元素的分解

Case 3. If r is odd and not equal to 2*k+1: replace the first and the last two elements in the series {1, 2*k-1, r} by {2*k+r} giving a decomposition of (k-1) elements.

请注意，当输入的形式的n个将发生的（K-1）个元素的最坏情况^ 2 +（奇数2 * K + 1）。

Note that the worst case of (k-1) elements will occur when the input is of the form n^2 + (odd number < 2*k+1).

另外请注意，（情况3）将在壳体打破的元素数小于3，例如，为5和7的分解，我们将必须特殊情况下，这些数字。同样地（案例2）将突破3和将不得不特例。没有为M = 2无解。因此，限制K> 3以上。其他一切都应该很好地工作。

Also note that (Case 3) will break in case the number of elements is less than 3. For example, the decomposition of 5 and 7. We will have to special-case these numbers. Likewise (Case 2) will break for 3 and will have to be special-cased. There is no solution for M=2. Hence the restriction k > 3 above. Everything else should work fine.

这需要O（开方（M））的步骤。

This takes O(sqrt(M)) steps.

一些C / C ++ code：

Some C/C++ code:

#include <stdio.h>

int main(int argc, char *argv[])
{
    printf("Enter M:");
    int m = 0;
    scanf("%d", &m);

    int arr[100] = {0};
    printf("The array is:\n");
    switch(m) {
        case 2:
            printf("No solution\n");
            return 0;
        case 1:
        case 3:
        case 5:
        case 7:
            printf("%d\n", m);
            return 0;
    }

    int sum = 0;
    int count = 0;
    for (int i = 1; (sum + i) < m; i+= 2) {
        arr[count++] = i;
        sum += i;
    }
    int start = 0;
    int r = m - sum;
    if (r % 2 == 0) {
        arr[count - 1] += r;
    } else if (r > arr[count - 1]) {
        arr[count++] = r;
    } else {
        start = 1;
        arr[count - 1] += r + 1;
    }

    for (int i = start; i < count; i++) {
        printf("%d\n", arr[i]);
    }

    return 0;
}

例如：

Enter M:24
The array is:
1
3
5
15

Enter M:23
The array is:
3
5
15