# LeetCode 1103. 分糖果 II Distribute Candies to People

### 一、中文版

1 <= candies <= 10^9

1 <= num_people <= 1000

### 二、英文版

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.
Example 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].
Example 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].
Constraints:
1 <= candies <= 10^9
1 <= num_people <= 1000

class Solution:
def distributeCandies(self, candies: int, num_people: int) -> List[int]:
ans = [0] * num_people
candy = 0
base = 1 # 递增加1
while candy < candies:
for i in range(num_people):
if candy + base <= candies:
ans[i] += base
candy += base
base += 1
else:
ans[i] += (candies - candy)
candy += (candies - candy)
# i = 0 因为不只遍历一遍数组，所以临时糖果总数candy小于总数 candies 时会重头开始for 循环
return ans

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