计算不是N周期的不同正常括号序列的数量

In this article, we're going to delve into an intriguing problem from the realm of combinatorics and string processing: "Counting distinct regular bracket sequences which are not N periodic". This problem involves generating distinct valid bracket sequences and then filtering out sequences that are N-periodic. We'll discuss the problem, provide a C++ code implementation of a brute-force approach, and explain a test case.

Understanding the Problem Statement

给定一个整数N，任务是计算长度为2N的不是N周期的不同的正则括号序列的数量。如果一个序列可以表示为字符串S重复M次，其中S的长度为N且M>1，则该序列是N周期的。

A regular bracket sequence is a string that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters. For example, the sequence "(())" is regular, while ")(" and "(()" are not.

方法

由于问题的复杂性，直接的数学解决方案可能不可行。相反，我们需要使用编程方法来生成括号序列，并计算符合我们条件的括号序列的数量。

We will use a recursive function to generate all possible bracket sequences of length 2N. While generating the sequences, we'll keep track of two important parameters: the balance of the brackets (the number of open brackets should never be less than the number of closed brackets) and the number of open brackets (should not exceed N).

为了筛选出N周期序列，我们将检查每个生成的序列。如果该序列是长度为N的较小序列的重复，我们将从计数中排除它。

C++ Implementation

这是一个用C++解决问题的蛮力方法。该方法生成所有可能的括号序列，并检查每个序列是否是N周期的，如果不是则增加计数。这个解决方案适用于小规模的输入，因为它具有指数时间复杂度，并且对于较大的输入不具有良好的可扩展性。

Example

#include <bits/stdc++.h>
using namespace std;

// Function to check if string is N periodic
bool isPeriodic(string s, int N) {
   string sub = s.substr(0, N);
   for (int i = N; i < s.size(); i += N) {
      if (sub != s.substr(i, N)) return false;
   }
   return true;
}

// Function to generate all bracket sequences
void generateBrackets(string s, int open, int close, int N, int &count) {
   // If sequence is complete
   if (s.size() == 2*N) {
      if (!isPeriodic(s, N)) count++;
      return;
   }
   
   // Add an open bracket if we have some left
   if (open < N) generateBrackets(s + "(", open + 1, close, N, count);
   
   // Add a close bracket if it doesn't make the sequence invalid
   if (close < open) generateBrackets(s + ")", open, close + 1, N, count);
}

int main() {
   int N = 3;
   int count = 0;
   generateBrackets("", 0, 0, N, count);
   cout << "Count of sequences: " << count;
   return 0;
}

Output

Count of sequences: 5

Test Case

Let's consider a test case with N = 3. This code will generate all 5 distinct regular bracket sequences of length 6 that are not 3-periodic: ((())), (()()), (())(), ()(()), ()()().

Conclusion

本文介绍了一种暴力解决非N周期的不同正则括号序列计数问题的方法。虽然这种方法可以处理小规模的输入，但需要注意的是，由于需要生成和检查所有可能的括号序列，该问题具有指数复杂度，因此不适用于大规模输入。

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