Codeforces:d1. choosing carrots (easy version) [max min problem + control one side to make the other side as close as possible + thinking]

analysis

Must let a Array divided by p After array
The value across is the smallest
We can traverse the minimum value , Then let each value close to the minimum value to complete greed
Obviously, the minimum value can be taken from [1,a[0]]
And then the others pi We can set it to pi = min(k, ai // vi) vi Is the current minimum
This means to let pi Take the maximum , Make it as close as possible vi Not more than k

ac code

import sys
input = sys.stdin.readline


for _ in range(int(input())):
    n, k = list(map(int, input().split()))
    a = list(map(int, input().split()))
    # greedy
    if k > a[-1]:
        print(0)
        continue

    # try the minn
    mincost = 0xffffffff
    minn = a[0]
    for i in range(1, minn + 1):
        p = []
        # make pi max to greedly get the min cost
        for aa in a:
            if i != 0:
                p.append(min(k, aa // i))
            else:
                p.append(k)
        q = [a[i] // p[i] for i in range(n)]
        mincost = min(mincost, max(q) - min(q))
    print(mincost)

summary

Worst value problem
thinking
Consider fixing one , Let the other as close as possible
This is the simplification of thinking