小编as2*_*2d3的帖子

我很难确定何时关系是Boyce-Codd Normal Form,以及如果不将其分解为BCNF.鉴于这个例子:

具有功能依赖性的R(A,C,B,D,E):A - > B,C - > D.

我怎么去分解它？

我采取的步骤是:

A+ = AB  
C+ = CD  
R1 = A+ = **AB**  
R2 = ACDE (since elements of C+ still exist, continue decomposing)  
R3 = C+ = **CD**  

R4 = ACE(此关系中没有FD闭包)

所以现在我知道ACE将构成整个关系,但分解的答案是:AB,CD,ACE.

我想我正在努力解决如何将关系正确地分解为BCNF形式以及如何判断你何时完成.非常感谢任何能够在解决这些问题时让我完成思考过程的人.谢谢!

我想将字符串的单个字符转换为整数,向其中添加2,然后将其转换回字符串.因此,A变为C,K变为M等.

这是我在 python3 中的代码：

import pandas as pd

tables = pd.read_html("http://http.com/")

print(tables)

我收到了这个错误：

Traceback (most recent call last):
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/urllib/request.py", line 1318, in do_open
    encode_chunked=req.has_header('Transfer-encoding'))
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/http/client.py", line 1239, in request
    self._send_request(method, url, body, headers, encode_chunked)
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/http/client.py", line 1285, in _send_request
    self.endheaders(body, encode_chunked=encode_chunked)
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/http/client.py", line 1234, in endheaders
    self._send_output(message_body, encode_chunked=encode_chunked)
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/http/client.py", line 1026, in _send_output
    self.send(msg)
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/http/client.py", line 964, in send
    self.connect()
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/http/client.py", line 1400, in connect
    server_hostname=server_hostname)
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/ssl.py", line 401, in wrap_socket
    _context=self, _session=session) …

我打算用C做这个:

#include<stdio.h>
int main() {
  int arr[5];
  arr[0] = 5;
  arr[1] = 0;
  arr[2] = 1;
  arr[3] = 3;
  arr[4] = 4;
  int max = 0;
  for(int i = 0;i < 5;i++)
    if(max < arr[i])
      max = arr[i];
  printf("%d\n", max);
  return 0;
}

这是我的代码链接:array_max.s.这是AT&T格式的汇编代码:

.data

.text
.globl _start
_start:
  movl $5, -20(%ebp)
  movl $0, -16(%ebp)
  movl $1, -12(%ebp)
  movl $3, -8(%ebp)
  movl $4, -4(%ebp)
  movl $0, %ecx
  movl $5, %eax
  loop:
    cmp $0, %eax
    je terminate
    cmp %ecx, …

我正在寻找向量优先级队列的简单 STL 实现。每个向量正好有 4 个元素。我想根据每个向量的第三个元素对我的优先级队列进行排序。具有最低第三个元素的向量应位于顶部（向量的最小优先级队列）。

如何在 C++ 中实现它？

另外，如果有人有默认优先级队列的实际 STL 实现，请提供链接。就像STL内部的官方实现一样。