1103. Distribute Candies to People

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

解题要点：

新建一个空list，长度为人数数量。由于发放糖果的数量为递增+1，且知道list固定长度，可以通过mod来找出超出长度后的index。每发放一次糖果，便要在总数里减去相应糖果，直到发放完毕。

class Solution(object):
    def distributeCandies(self, candies, num_people):
        """
        :type candies: int
        :type num_people: int
        :rtype: List[int]
        """
        n = num_people
        res = [0] * n
         
        c = 0
        while candies > 0:
            res[c % n] += min(candies, c+1)
            c += 1
            candies -= c
        
        return res