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### 题目描述

We distribute some number of candies, to a row of n = num_people people in the following way:

We then give 1 candy to the first person, 2 candies to the second person, and so on until we give n candies to the last person.

Then, we go back to the start of the row, giving n + 1 candies to the first person, n + 2 candies to the second person, and so on until we give 2 * n candies to the last person.

This process repeats (with us giving one more candy each time, and moving to the start of the row after we reach the end) until we run out of candies. The last person will receive all of our remaining candies (not necessarily one more than the previous gift).

Return an array (of length num_people and sum candies) that represents the final distribution of candies.

### 例子

#### 例子 1

Input: candies = 7, num_people = 4
Output: [1,2,3,1]
Explanation:
On the first turn, ans[0] += 1, and the array is [1,0,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0,0].
On the third turn, ans[2] += 3, and the array is [1,2,3,0].
On the fourth turn, ans[3] += 1 (because there is only one candy left), and the final array is [1,2,3,1].

#### 例子 2

Input: candies = 10, num_people = 3
Output: [5,2,3]
Explanation:
On the first turn, ans[0] += 1, and the array is [1,0,0].
On the second turn, ans[1] += 2, and the array is [1,2,0].
On the third turn, ans[2] += 3, and the array is [1,2,3].
On the fourth turn, ans[0] += 4, and the final array is [5,2,3].

#### Note

• 1 <= candies <= 10^9
• 1 <= num_people <= 1000

### 解题思路

• 糖果可以完整地发几次 m_step，根据等差数列求和公式可以计算出 m_stepsqrt(2 * candies) sqrt(2 * candies) - 1 中的一个
• 针对每个人再计算糖果能给他完整地发几次 turnsturns = m_step / num_people，如果最后一轮还能发到（即 i < m_step % num_people）则还要加上一次
• 当我们知道每个人可以发几次的时候，可以通过等差数列求和计算出一个给他发多少颗
• 最后将剩下的糖果给下一个人即可

• 时间复杂度：O(n)
• 空间复杂度：O(n) 输出所需空间