我如何将指数除以大量

Question

我有一个排序算法

def partition(competitors, left, right):
    pivot = competitors[left]
    i = left + 1
    j = right - 1
    while True:
        if (i <= j and competitors[j] > pivot):
            j -= 1
        elif (i <= j and competitors[i] < pivot):
            i += 1
        elif (competitors[j] > pivot) or (competitors[i] < pivot):
            continue
        if i <= j:
            competitors[i], competitors[j] = competitors[j], competitors[i]
        else:
            competitors[left], competitors[j] = competitors[j], competitors[left]
            return j


def quick_sort(competitors, left, right):
    if ((right - left) > 1):
        p = partition(competitors, left, right)
        quick_sort(competitors, left, p)
        quick_sort(competitors, p + 1, right)


def transformation(competitors):
    competitors[1] = - int(competitors[1])
    competitors[2] = int(competitors[2])

    return [competitors[1], competitors[2], competitors[0]]


if __name__ == '__main__':
    number = int(input())
    competitors = [transformation(input().split()) for _ in range(number)]
    quick_sort(competitors, left=0, right=len(competitors))
    print(*(list(zip(*competitors))[2]), sep="\n")

这里

pivot = competitors[left]

审阅者说，我可以得到一个支点除以数组中的第一个索引和最后一个索引。我尝试了很多选择，其中之一就是

pivot = competitors[left // right]

但是输出是错误的

在这里描述任务

**让我们得到左和右两个指针，分别出现在段的左端和右端。然后，我们将左指针向右移动，直到它触发小于枢轴的元素的警报。类似地，我们将右指针移向左侧，而它位于超过引用one.As的元素上，结果表明，所有元素都准确地属于第一组，而右侧则属于第二组。带有指针的元素是无序的。让我们交换它们(大多数编程语言​​使用swap()函数)并将指针提前到下一个元素。我们将重复这一行动，直到左、右碰撞。**

Input:
    5
    alla 4 100
    gena 6 1000
    gosha 2 90
    rita 2 90
    timofey 4 80

Output:
    gena
    timofey
    alla
    gosha
    rita

And my output is: gena
                  timofey
                  alla
                 <rita>
                 <gosha>

请帮我找出解决这个算法的方法

Answer 1

这是一种新的算法，我用透视=竞争对手(左+右) // 2字符串更新了它

def partition(competitors, left, right):
    if right <= left:
        return
    pivot = (left + right) // 2
    part = competitors[pivot]
    begin = left
    end = right
    while begin <= end:
        while part > competitors[begin]:
            begin += 1
        while part < competitors[end]:
            end -= 1
        if begin <= end:
            competitors[begin], competitors[end] = competitors[
                end], competitors[begin]
            begin += 1
            end -= 1
    partition(competitors, left, end)
    partition(competitors, begin, right)


def quick_sort(competitors):
    partition(competitors, 0, len(competitors) - 1)
    return [line[2] for line in competitors]


if __name__ == '__main__':
    n = int(input())
    competitors = [None] * n
    for x in range(n):
        login, Pi, Fi = input().split()
        competitors[x] = (-int(Pi), int(Fi), login)
    result = quick_sort(competitors)
    print(*result, sep="\n")