計算不是N週期的不同正常括號序列的數量

In this article, we're going to delve into an intriguing problem from the realm of combinatorics and string processing: "Counting distinct regio bracket sequences which are not N perdicm. 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 expexp a# testase.

Understanding the Problem Statement

給定一個整數N，任務是計算長度為2N的不是N週期的不同的正則括號序列的數量。如果一個序列可以表示為字串S重複M次，其中S的長度為N且M>1，則該序列是N週期的。

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 ""())" (" and "(()" are not.

方法

由於問題的複雜性，直接的數學解決方案可能不可行。相反，我們需要使用程式方法來產生括號序列，並計算符合我們條件的括號序列的數量。

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, openets brackets the number of closed brackets) and the number of open brackets (should not exceed N).

為了篩選出N週期序列，我們將檢查每個產生的序列。如果該序列是長度為N的較小序列的重複，我們將從計數中排除它。

C Implementation

這是一個用C 解決問題的蠻力方法。此方法產生所有可能的括號序列，並檢查每個序列是否是N週期的，如果不是則增加計數。這個解決方案適用於小規模的輸入，因為它具有指數時間複雜度，並且對於較大的輸入不具有良好的可擴展性。

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

本文介紹了一種暴力解決非N週期的不同正規括號序列計數問題的方法。雖然這種方法可以處理小規模的輸入，但需要注意的是，由於需要產生和檢查所有可能的括號序列，因此該問題具有指數複雜度，因此不適用於大規模輸入。

以上是計算不是N週期的不同正常括號序列的數量的詳細內容。更多資訊請關注PHP中文網其他相關文章！