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📜  在 N 个男孩中分配 C 个糖果，使得收到的最大和最小糖果之间的差异为 K(1)

📅  最后修改于: 2023-12-03 15:23:13.769000             🧑  作者: Mango

在 N 个男孩中分配 C 个糖果，使得收到的最大和最小糖果之间的差异为 K

问题背景

在某一天，班主任手里有 $C$ 颗糖果，要分给 $N$ 个男孩，并且要求这些男孩收到的糖果数量，最大值与最小值的差不超过 $K$。假设每个男孩都要分到至少一颗糖果，那么如何分配呢？

问题分析

如果我们知道了最大值和最小值，那么问题就简单了。我们首先分配最小值，然后剩下的 $C-N$ 颗糖果，分配给 $N$ 个男孩中的 $N-1$ 个，每个男孩分配的糖果数量都增加了 $1$。最后，再将多余的糖果分配给最小值。这样就能保证差不超过 $K$。

那么，如何求出最大值和最小值呢？我们可以采用二分法。假设最小值为 $low$，最大值为 $high$。则我们可以从 $mid = (low + high) / 2$ 开始，依次检查是否存在一种分配方案，使得最小值为 $low$，最大值为 $mid$，且差不超过 $K$。如果有，那么就将 $mid$ 设为新的下界；如果没有，则将 $mid$ 设为新的上界。

注意，为了方便起见，我们可以先将 $N$ 个男孩按照升序排列。

算法实现

下面是一个参考实现，使用了二分法来快速搜索最小值和最大值。

def can_allocate(candies, n, k, low, high):
    # Check if we can allocate `candies` with `n` boys,
    # such that the difference between the maximum and minimum
    # number of candies is at most `k`, given that the minimum
    # number of candies is `low` and the maximum number of 
    # candies is `high`.
    diff = 0
    max_candies = low
    min_candies = low
    i = 0
    j = n - 1
    while i <= j and candies[i] <= low:
        min_candies += 1
        i += 1
    while i <= j and candies[j] >= high:
        max_candies += 1
        j -= 1
    if max_candies - min_candies > k:
        return False
    while i <= j:
        if candies[j] - high > k - diff:
            diff += candies[j] - high - (k - diff)
            high = candies[j] - k + diff
        elif candies[i] - low > k - diff:
            diff += low - candies[i] + k - diff
            low = candies[i] + k - diff
        else:
            diff += candies[j] - high
            high = candies[j]
            max_candies += 1
            j -= 1
            diff += low - candies[i]
            low = candies[i]
            min_candies += 1
            i += 1
        if max_candies - min_candies > k:
            return False
    return True


def allocate(candies, n, k, c):
    # Searches for the minimum value `x` such that we can 
    # allocate `c` candies to `n` boys, such that the difference 
    # between the maximum and minimum number of candies is at most 
    # `k`, given that the minimum number of candies is `x`.
    low = 1
    high = max(candies)
    while low < high:
        mid = (low + high) // 2
        if can_allocate(candies, n, k, mid, mid + c // n):
            high = mid
        else:
            low = mid + 1
    return low
参考文献