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

1.思考

• 一开始想用数学的方法进行简化处理的……但是没有想到很好的方法……
• 然后就直接用循环的方式来发糖果；
• 首先，建立一个长度为num_people初始值为0的vector，相当于每个人都维护自己的糖果数量；
• 其次，用c表示当前剩下的糖果数，用n表示下一次要发的糖果数；
• 然后，每次发放糖果之后，对应人(n-1)%num_people的糖果数增加n，剩余糖果数c=c-n，下次发放的糖果数n++；
• 最后，若剩余的糖果数c < 要发放的糖果数n，则将剩余的糖果数c都给当前这个人，并跳出循环。

2.实现

Runtime: 0ms(100%)

Memory: 8.5MB(100%)

class Solution {public:    vector
    
      distributeCandies(int candies, int num_people) {        vector
     
       res(num_people, 0);        int c = candies, n = 1;        while(c>0){            if(c-n>0){                res[(n-1)%num_people] += n;            }            else{                res[(n-1)%num_people] += c;                break;            }            c -= n;            n++;        }        return res;    }};