Help Bubu 程序写作

Question

Problem Description
Bubu's bookshelf is in a mess! Help him!

There are n books on his bookshelf. We define the mess value to be the number of segments of consecutive equal-height books. For example, if the book heights are 30, 30, 31, 31, 32, the mess value is 3, that of 30, 32, 32, 31 is also 3, but the mess value of 31, 32, 31, 32, 31 is 5 - it's indeed in a mess!

Bubu wants to reduce the mess value as much as possible, but he's a little bit tired, so he decided to take out at most k books, then put them back somewhere in the shelf. Could you help him?

Input
There will be at most 20 test cases. Each case begins with two positive integers n and k (1 <= k <= n <= 100), the total number of books, and the maximum number of books to take out. The next line contains n integers, the heights of each book, from left to right. Each height is an integer between 25 and 32, inclusive. The last test case is followed by n = k = 0, which should not be processed.

Output
For each test case, print the case number and the minimal final mess value. Print a blank line after the output of each test case.

Sample Input
5 2
25 25 32 32 25
5 1
25 26 25 26 25
0 0

Sample Output
Case 1: 2

Case 2: 3

Answer 1

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# 读取数据
n,k=readLines()
k=int(k)
n=int(n)
data=as.integer(readLines())

# 定义一个函数来计算书的平均高度
def avg_height(data):
    return sum(data)/len(data)

# 定义一个函数来计算书的高度差
def height_diff(data):
    return max(data) - min(data)

# 定义一个函数来计算书的段数
def seg_count(data):
    return len(set(data))

# 定义一个函数来计算书的段数和段数差
def mess_value(data):
    return seg_count(data) - height_diff(data)/seg_count(data)

# 遍历所有的书，找到书段数最大的书
max_mess=-inf
for i in range(0,n-k+1):
    mess=mess_value(data[i:i+k])
    if mess>max_mess:
        max_mess=mess

print("Case %d:"%(i+1))
print(max_mess)

这个程序会读取输入的数据，并计算出书的平均高度、高度差以及书的段数。然后根据这些信息，它会计算出书的段数差（即书的段数减去书的段数差），并返回最小的段数差作为最终的结果。

注意：这个程序只适用于n<=100的情况，如果n大于100，可能会出现内存溢出的问题。