. 输入格式: Each input file only contains one test case. The first line contains two integer n, m (1 ≤ n ≤ 106, 1 ≤ m ≤ 109), indicating the length of the track sequence and the number of the current magnetic head. The second line contains n integers a1​,a2​,...,an​ (1 ≤ ai​ ≤ 109), indicating the track sequence. 输出格式： For each test case output an integer indicating the sum of the magnetic head's moving distance. 输入样例： 9 100 55 58 39 18 90 160 150 38 184 输出样例： 248

以下是一个实现上述功能的Python程序：

n, m = map(int, input().split())
a = list(map(int, input().split()))

dist = 0
prev = m
while len(a) > 0:
    # 找到离当前位置最近的磁道
    min_idx = 0
    for i in range(1, len(a)):
        if abs(a[i] - prev) < abs(a[min_idx] - prev):
            min_idx = i
    # 更新移动距离和当前位置
    dist += abs(a[min_idx] - prev)
    prev = a[min_idx]
    # 从序列中移除当前磁道
    a.pop(min_idx)

print(dist)

首先，我们使用`map()`函数将输入的字符串转换为整数类型，并将它们存储在变量`n`和`m`中。接下来，我们使用`list()`和`map()`函数将输入的磁道序列转换为整数列表`a`。然后，我们使用一个循环来迭代所有磁道，直到序列为空为止。在每次循环中，我们找到当前位置最近的磁道，计算移动距离并更新当前位置。最后，我们输出移动距离的总和。

需要注意的是，在实际应用中，可能需要使用更高效的算法来解决这个问题。该实现的时间复杂度为$O(n^2)$，对于大型数据集可能会导致超时。

相关问题

The task of this problem is simple: insert a sequence of distinct positive integers into a hash table, and output the positions of the input numbers. The hash function is defined to be H(key)=key%TSize where TSize is the maximum size of the hash table. Quadratic probing (with positive increments only) is used to solve the collisions. Note that the table size is better to be prime. If the maximum size given by the user is not prime, you must re-define the table size to be the smallest prime number which is larger than the size given by the user. Input Specification: Each input file contains one test case. For each case, the first line contains two positive numbers: MSize (≤10 4 ) and N (≤MSize) which are the user-defined table size and the number of input numbers, respectively. Then N distinct positive integers are given in the next line. All the numbers in a line are separated by a space. Output Specification: For each test case, print the corresponding positions (index starts from 0) of the input numbers in one line. All the numbers in a line are separated by a space, and there must be no extra space at the end of the line. In case it is impossible to insert the number, print "-" instead.

这道题的任务很简单：将一串不同的正整数插入哈希表中，并输出输入数字的位置。哈希函数的定义为 H(key)=key%TSize，其中 TSize 是哈希表的最大大小。使用二次探测（只增加正数）来解决碰撞。注意，表格大小最好为质数。如果用户给出的最大大小不是质数，则必须重新定义表格大小为大于用户给出大小的最小质数。

输入格式：每个输入文件包含一个测试用例。对于每种情况，第一行包含两个正数：MSize（≤10 4）和N（≤MSize），它们是用户定义的表大小和输入数字的数量。然后，在下一行中给出N个不同的正整数。一行中的所有数字由空格分隔。

输出格式：对于每个测试用例，在一行中打印输入数字的相应位置（索引从0开始）。一行中的所有数字由空格分隔，行末不能有多余的空格。如果无法插入数字，则打印“-”。

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again! Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number. Input Specification: Each input contains one test case. Each case contains one positive integer with no more than 20 digits. Output Specification: For each test case, first print in a line "Yes" if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or "No" if not. Then in the next line, print the doubled number. Sample Input: 1234567899 Sample Output: Yes 2469135798

To solve this problem, we can first check if the number of digits in the given number is even or odd. If it is odd, then we can immediately output "No" and the doubled number. Otherwise, we can count the number of occurrences of each digit in the original number and the doubled number. If they match, then the doubled number consists of only a permutation of the digits in the original number; otherwise, it does not.

Here's the Python code:

num = input().strip()
double_num = str(int(num) * 2)

if len(num) % 2 == 1:
    print("No")
    print(double_num)
else:
    num_count = [0] * 10
    double_count = [0] * 10
    for digit in num:
        num_count[int(digit)] += 1
    for digit in double_num:
        double_count[int(digit)] += 1
    if num_count == double_count:
        print("Yes")
    else:
        print("No")
    print(double_num)

Note that we convert the doubled number to a string so that we can count the occurrences of each digit using a list of counts. We also use the `strip()` method to remove any leading or trailing whitespace in the input.