1019. General Palindromic Number (20)

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Although palindromic numbers are most often considered in the decimal system, the concept of palindromicity can be applied to the natural numbers in any numeral system. Consider a number N > 0 in base b >= 2, where it is written in standard notation with k+1 digits ai as the sum of (aibi) for i from 0 to k. Here, as usual, 0 <= ai < b for all i and ak is non-zero. Then N is palindromic if and only if ai = ak-i for all i. Zero is written 0 in any base and is also palindromic by definition.

Given any non-negative decimal integer N and a base b, you are supposed to tell if N is a palindromic number in base b.

Input Specification:

Each input file contains one test case. Each case consists of two non-negative numbers N and b, where 0 <= N <= 109 is the decimal number and 2 <= b <= 109 is the base. The numbers are separated by a space.

Output Specification:

For each test case, first print in one line "Yes" if N is a palindromic number in base b, or "No" if not. Then in the next line, print N as the number in base b in the form "ak ak-1 ... a0". Notice that there must be no extra space at the end of output.

Sample Input 1:
27 2
Sample Output 1:
Yes
1 1 0 1 1
Sample Input 2:
121 5
Sample Output 2:
No
4 4 1



#include <iostream>
#include <vector>
using namespace std;

vector<int> trans(int n, int b)
{
    vector<int> a;
    while(n)
    {
        a.push_back(n%b);
        n /= b;
    }
    return a;
}

bool ispalnum(vector<int> a)
{
    for(int i=0; i<a.size()/2; i++)
    {
        if(a[i] != a[a.size()-1-i])
            return false;
    }
    return true;
}
int main()
{
    int n, b;
    cin >> n >> b;
    vector<int> a;
    a = trans(n, b);
    if(ispalnum(a))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    for(int i=a.size()-1; i>=0; i--)
    {
        if(i != 0)
            cout << a[i] << " ";
        else
            cout << a[i] << endl;
    }
    if(n == 0)
        cout << "0" << endl;
}


回文树(Palindrome Tree),也称为字典树或自动机树,是一种用于高效查找字符串中所有子串是否为回文的方法的数据结构。在C++中,我们可以使用前缀树(Trie)为基础,对每个节点添加一个标记,表示其后缀是否为回文。 以下是一个简单的回文树实现步骤: 1. **定义节点**: 创建一个`Node`类,包含一个字符数组`children`用于存储子节点、一个布尔值`is_palindrome`标记当前节点及其所有后缀是否为回文,以及一个指向父节点的指针。 ```cpp class Node { public: char c; bool is_palindrome; Node* children[26]; // 26 for ASCII alphabet, adjust if needed Node(Node* parent); }; ``` 2. **构造函数**: 初始化字符和指向父节点的指针。 ```cpp Node::Node(Node* parent) : parent(parent), c('\0'), is_palindrome(false) { for (int i = 0; i < 26; ++i) children[i] = nullptr; } ``` 3. **插入操作**: 插入单词时,从根节点开始遍历,如果遇到某个字符不存在于子节点中,则新建一个子节点。同时,检查新插入的子串是否为回文。 ```cpp void insert(Node*& root, const string& word) { Node* node = &root; for (char c : word) { int idx = c - 'a'; // 根据ASCII调整 if (!node->children[idx]) { node->children[idx] = new Node(node); } node = node->children[idx]; if (word.size() > 1 && word[word.size() - 1] == c) { // 遍历过程中检查末尾是否成对出现 node->is_palindrome = true; } } } ``` 4. **查询操作**: 检查给定的子串是否在树中并返回所有回文后缀。 ```cpp vector<string> findPalindromes(Node* root, const string& prefix) { vector<string> result; dfs(root, prefix, result); return result; } void dfs(Node* node, const string& prefix, vector<string>& result) { if (!node) return; if (node->is_palindrome && !prefix.empty()) { result.push_back(prefix); } for (int i = 0; i < 26; ++i) { if (node->children[i]) dfs(node->children[i], prefix + node->c, result); } } ``` 5. **构建回文树**: 对输入的一系列单词进行插入,构建完整个树。 ```cpp Node* buildPalindromeTree(const vector<string>& words) { Node* root = new Node(nullptr); for (const auto& word : words) { insert(root, word); } return root; } ```
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