1103.Distribute Candies to People

Tags: Easy Math

Links: https://leetcode.com/problems/distribute-candies-to-people/

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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

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class Solution {
public:
    vector<int> distributeCandies(int candies, int num_people) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int n = num_people;
        vector<int> res(n, 0);
        int pos = 0;
        int cnt = 1;
        while (candies > 0) {
            if (cnt >= candies) {
                res[pos] += candies; break;
            }
            candies -= cnt;
            res[pos] += cnt++;
            pos = (pos + 1) % n;
        }

        return res;
    }
};

这种方法属于暴力模拟，可能需要多次遍历数组，考虑到标签是数学，所以存在数学解法。

#include <cmath>
class Solution {
public:
    vector<int> distributeCandies(int candies, int num_people) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int n = num_people;
        vector<int> res(n, 0);

        int k = (-1 + sqrt(8 * (long long)candies + 1)) / 2;
        int t = k / n, left = k % n;

        for (int i = 0; i < n; ++i) {
            res[i] = (i + 1) * t + n * t * (t - 1) / 2;
        }

        for (int i = 0; i < left; ++i) {
            res[i] += (i + 1) + t * n;    
        }
        res[left] += candies - (k + 1) * k / 2;

        return res;
    }
};

分发过程如下：

1   2   3   ...     n
n+1 n+2 n+3  ...    n * 2

相当于有$k$堆糖果，第$k$堆有$k$个，然后$n$个人轮流去取。那么只需要求出$\sum_{i=1}^k i \leq candies$的最大整数$k$，根据求根公式可得$k$的数值，可能剩余下$candies - \frac{k(k+1)}{2}$块糖果，用left标记。共$k$堆糖，一轮一轮的取，那么取$t = k /n$轮，那么这$n$轮每个人取多少是可以计算的。然后分发left的部分即可。