LeetCode 1103. Distribute Candies to People 二分求和推公式-优快云博客

本文介绍了一种糖果分配算法，该算法将一定数量的糖果按特定规则分配给一排人数不等的人。通过迭代过程，每个人获得的糖果数逐渐增加，直到所有糖果分配完毕。文章提供了两个示例，详细展示了算法的执行流程和最终的糖果分配结果。

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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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求和反应有些慢，二分反应还有些慢，其实二分按照套路（https://blog.youkuaiyun.com/taoqick/article/details/100026169）来写，就两种情况，注意left和right的范围，最后返回的时候left-1和right其实是一样的，因为是<=：

第一个>=target（数组元素<target的时候left=mid+1，返回l）和最后一个<target（数组元素<target的时候left=mid+1，返回l-1或者r）
第一个>target（数组元素<=target的时候left=mid+1，返回l）和最后一个<target（数组元素<=target的时候left=mid+1，返回l-1或者r）

上代码：

class Solution:
    def f(self,x,n):
        return ((n*(n+1)*(x+1))+(n*n*(x+1)*x))>>1
    def get_K(self, candies, n):
        l,r = 0,100000
        while(l<=r):
            mid = l+(((r-l))>>1) #bug1: mid = l+((r-l))>>1
            cur = self.f(mid, n)
            if (cur <= candies):
                l = mid+1
            else:
                r = mid-1
        return r
    def distributeCandies(self, candies: int, num_people: int):
        res = [0 for i in range(num_people)]
        K,n = self.get_K(candies, num_people),num_people
        left = candies-self.f(K,n)
        print("K={0} left={1}".format(K,left))

        for j in range(n):
            smaller = min(left,j+1+(K+1)*n)
            res[j] = (K+1)*(j+1)+((n*(K+1)*K)>>1)+smaller
            left-=smaller
        return res

s = Solution()
print(s.distributeCandies(7,4))