A block of 64 contiguous LAPs is reserved for inquiry operations; one LAP common to all devices is reserved for general inquiry, the remaining 63 LAPs are reserved for dedicated inquiry of specific classes of devices (see Assigned Numbers). The same LAP values are used regardless of the contents of UAP and NAP. Consequently, none of these LAPs can be part of a user BD_ADDR. The reserved LAP addresses are 0x9E8B00 to 0x9E8B3F. The general inquiry LAP is 0x9E8B33. All addresses have the LSB at the rightmost position, hexadecimal notation. The default check initialization (DCI) is used as the UAP whenever one of the reserved LAP addresses is used. The DCI is defined to be 0x00 (hexadecimal).
时间: 2024-03-07 12:52:27 浏览: 17
有一组64个连续的LAP保留用于查询操作;其中一个LAP是所有设备共用的,用于通用查询,剩下的63个LAP保留用于针对特定设备类别的专用查询(请参阅分配的编号)。无论UAP和NAP的内容如何,都使用相同的LAP值。因此,这些LAP都不能成为用户BD_ADDR的一部分。保留的LAP地址为0x9E8B00至0x9E8B3F。通用查询LAP为0x9E8B33。所有地址的最低有效位位于最右边的位置,使用十六进制表示。当使用保留的LAP地址之一时,使用默认检查初始化(DCI)作为UAP。DCI被定义为0x00(十六进制)。
相关问题
A+subsequence+of+a+number+is+a+series+of+(not+necessarily+contiguous)+digits+of+the+number.+For+exam
你的问题是:A+subsequence+of+a+number+is+a+series+of+(not+necessarily+contiguous)+digits+of+the+number.+For+exam。这不是一个完整的问题,它似乎被截断了。请问你想问什么?
如果你想问的是什么是数字的子序列,那么数字的子序列是指从原数字中选出一些数字,这些数字不一定要相邻,但它们在原数字中的相对顺序必须与原数字中的相对顺序相同。
如果你有其他问题,请继续提出。
outputstandard output You are given a string s consisting of the characters 0, 1 and/or ?. Let's call it a pattern. Let's say that the binary string (a string where each character is either 0 or 1) matches the pattern if you can replace each character ? with 0 or 1 (for each character, the choice is independent) so that the strings become equal. For example, 0010 matches ?01?, but 010 doesn't match 1??, ??, or ????. Let's define the cost of the binary string as the minimum number of operations of the form "reverse an arbitrary contiguous substring of the string" required to sort the string in non-descending order. You have to find a binary string with the minimum possible cost among those that match the given pattern. If there are multiple answers, print any of them.
This is an interesting problem! To solve it, you can use dynamic programming. Let dp[i][j] be the minimum cost to sort the first i characters of the pattern such that the i-th character is j (0 or 1). The base case is dp[0][0]=dp[0][1]=0, since an empty string is already sorted.
Then, to compute dp[i][j], you can consider the last character of the pattern, which can be either 0, 1, or ?. If it's 0 or 1, you can simply match it with the i-th character of the binary string, and update dp[i][j] accordingly. If it's ?, you need to try both possibilities (0 or 1) and take the minimum cost.
The final answer is the minimum of dp[n][0] and dp[n][1], where n is the length of the pattern.
Here's the Python code:
```
s = input().strip()
n = len(s)
dp = [[float('inf')]*2 for _ in range(n+1)]
dp[0][0] = dp[0][1] = 0
for i in range(1, n+1):
if s[i-1] == '0':
dp[i][0] = min(dp[i-1][0], dp[i-1][1]+1)
elif s[i-1] == '1':
dp[i][1] = min(dp[i-1][1], dp[i-1][0]+1)
else:
dp[i][0] = min(dp[i-1][0], dp[i-1][1]+1)
dp[i][1] = min(dp[i-1][1], dp[i-1][0]+1)
ans = [''] * n
if dp[n][0] <= dp[n][1]:
i, j = n, 0
while i > 0:
if s[i-1] == '0':
ans[i-1] = '0'
i, j = i-1, 0
elif s[i-1] == '1':
ans[i-1] = '1'
i, j = i-1, 1
else:
if dp[i][j] == dp[i-1][0]:
ans[i-1] = '0'
i, j = i-1, 0
else:
ans[i-1] = '1'
i, j = i-1, 1
else:
i, j = n, 1
while i > 0:
if s[i-1] == '0':
ans[i-1] = '0'
i, j = i-1, 0
elif s[i-1] == '1':
ans[i-1] = '1'
i, j = i-1, 1
else:
if dp[i][j] == dp[i-1][1]:
ans[i-1] = '1'
i, j = i-1, 1
else:
ans[i-1] = '0'
i, j = i-1, 0
print(''.join(ans))
```
Hope this helps!