用C++编写程序，实现以下问题2、题目ID Codes（POJ1146） Time Limit: 1000MS Memory Limit: 10000K 描述： It is 2084 and the year of Big Brother has finally arrived, albeit a century late. In order to exercise greater control over its citizens and thereby to counter a chronic breakdown in law and order, the Government decides on a radical measure--all citizens are to have a tiny microcomputer surgically implanted in their left wrists. This computer will contains all sorts of personal information as well as a transmitter which will allow people's movements to be logged and monitored by a central computer. (A desirable side effect of this process is that it will shorten the dole queue for plastic surgeons.) An essential component of each computer will be a unique identification code, consisting of up to 50 characters drawn from the 26 lower case letters. The set of characters for any given code is chosen somewhat haphazardly. The complicated way in which the code is imprinted into the chip makes it much easier for the manufacturer to produce codes which are rearrangements of other codes than to produce new codes with a different selection of letters. Thus, once a set of letters has been chosen all possible codes derivable from it are used before changing the set. For example, suppose it is decided that a code will contain exactly 3 occurrences of a', 2 of b' and 1 of c', then three of the allowable 60 codes under these conditions are: abaabc abaacb ababac These three codes are listed from top to bottom in alphabetic order. Among all codes generated with this set of characters, these codes appear consecutively in this order. Write a program to assist in the issuing of these identification codes. Your program will accept a sequence of no more than 50 lower case letters (which may contain repeated characters) and print the successor code if one exists or the message No Successor' if the given code is the last in the sequence for that set of characters. 输入： Input will consist of a series of lines each containing a string representing a code. The entire file will be terminated by a line consisting of a single #. 输出： Output will consist of one line for each code read containing the successor code or the words 'No Successor'. 样例输入 abaacb cbbaa # 样例输出 ababac No Successor

以下是使用C++编写的程序，可以解决该问题：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

int main()
{
    char str[55];
    while(cin >> str && str[0] != '#')
    {
        if(next_permutation(str, str + strlen(str)))
            cout << str << endl;
        else
            cout << "No Successor" << endl;
    }
    return 0;
}

首先，我们定义了一个字符数组 `str` 来存储输入的字符串。

然后，在一个循环中不断读入字符串，直到读入单独的一个 '#' 字符为止。

对于每个字符串，我们可以使用 `next_permutation` 函数来找到它的下一个排列。如果找到了下一个排列，则输出它；否则输出 "No Successor"。

需要注意的是，我们必须将 `str` 的长度作为第二个参数传递给 `next_permutation` 函数，否则它将无法正确地处理含有重复字符的字符串。

相关问题

python编写程序，要求：实现根据地址码对照表和身份证号码查询居居民常住户口所在地的功能

在Python中，实现这个功能通常涉及到数据处理、字符串操作和字典查找。你可以创建一个包含地址码对照的字典，然后通过身份证号码的一部分（如前六位代表地区编码）来查找对应的地址。以下是一个简单的示例：

# 创建地址码对照表字典（这里仅做演示，实际应用可能需要从数据库或API获取）
address_codes = {
    "110000": "北京市",
    "310100": "上海市",
    # 其他省份市区代码及对应地址...
}

def query_residence_address(id_number):
    # 身份证号码的前6位作为地区编码
    area_code = id_number[:6]
    
    if area_code in address_codes:
        return address_codes[area_code]
    else:
        return "地址信息未找到"

# 示例使用
id_number = "110101199001010030"
residence = query_residence_address(id_number)
print(f"根据身份证号码 {id_number}，常住户口所在地为：{residence}")

编写程序，实现： 1、给定一组权值，构建哈夫曼树，并求其哈夫曼编码。 2、应用1中求出的编码进行编解码操作。

编写程序实现哈夫曼树及其编码的过程通常涉及以下几个步骤：

1. **构建哈夫曼树**:
- 首先，将输入的权值作为字符与其频率的键值对存储在一个优先队列（如Python的`heapq`模块）中，最小频率的字符排在前面。
- 然后，从队列中取出两个最小频次的节点合并成一个新的节点，新节点的频率等于两原节点之和，左孩子是第一个节点，右孩子是第二个节点。
- 将新节点放回队列，重复此过程直到只剩下一个节点，这个就是哈夫曼树的根。

2. **求哈夫曼编码**:
- 根据哈夫曼树的结构，自底向上地标记路径。从叶子节点开始，每个节点记录下到达它的最短路径，对于父节点，如果它有左孩子，路径加上0；如果有右孩子，路径加上1。
- 最终，每个字符对应的编码就是从该字符到哈夫曼树根节点的路径。

3. **编解码操作**:
- 编码：对于给定的数据，遍历每个字符，按照上面求得的路径添加前缀0或1形成编码。
- 解码：反之，接受编码时，从左往右读取二进制序列，根据路径规则找到相应的字符。

以下是简化的伪代码示例：

def build_huffman_tree(weights):
    # 初始化优先队列
    priority_queue = [(weight, char) for char, weight in weights.items()]
    heapq.heapify(priority_queue)

    while len(priority_queue) > 1:
        left_weight, left_char = heapq.heappop(priority_queue)
        right_weight, right_char = heapq.heappop(priority_queue)
        combined_weight = left_weight + right_weight
        combined_node = (combined_weight, [left_char, right_char])
        heapq.heappush(priority_queue, combined_node)

    return priority_queue[0][1]

# 示例数据
weights = {'A': 5, 'B': 7, 'C': 4}

huffman_tree = build_huffman_tree(weights)
codes = {char: get_code(node) for char, node in huffman_tree.items()}  # 自定义get_code函数

def encode(message):
    return ''.join(codes[char] for char in message)

def decode(encoded_message):
    decoded_message = ''
    current_path = []
    for bit in encoded_message:
        if bit == '0':
            current_path.append('L')
        else:
            current_path.append('R')

        if current_path[-1:] == ['R', 'L']:
            decoded_message += current_path.pop()  # 拼接字符
        elif not current_path:
            break  # 找到叶节点，结束解码

    return decoded_message

# 使用例子
message = 'ABC'
encoded_message = encode(message)
print(f"Encoded: {encoded_message}")
decoded_message = decode(encoded_message)
print(f"Decoded: {decoded_message}")