与其他二进制字符串的XOR为0的循环置换计数

给定两个二进制字符串$A$和$B$ 。让$S$设置字符串的所有循环排列$B$ 。任务是查找集合中有多少个字符串$S$与XOR时$A$给$0$结果。

例子：

Input: A = “101”, B = “101”
Output: 1
S = {“101”, “011”, “110”}
Only “101” XOR “101” = 0

Input: A = “111”, B = “111”
Output: 3

方法：串联$B$和$B$因此它包含所有循环排列的子字符串。现在减少问题以查找模式的出现次数$A$在字符串$B$可以使用Z算法(用于线性时间模式搜索)解决。

下面是上述方法的实现：

C++

// C++ program to find bitwise XOR between binary
// string A and all the cyclic permutations
// of binary string B
#include 
using namespace std;
  
// Implementation of Z-algorithm
// for linear time pattern searching
void compute_z(string s, int z[])
{
    int l = 0, r = 0;
    int n = s.length();
    for (int i = 1; i <= n - 1; i++) {
        if (i > r) {
            l = i, r = i;
            while (r < n && s[r - l] == s[r])
                r++;
            z[i] = r - l;
            r--;
        }
        else {
            int k = i - l;
            if (z[k] < r - i + 1) {
                z[i] = z[k];
            }
            else {
                l = i;
                while (r < n && s[r - l] == s[r])
                    r++;
                z[i] = r - l;
                r--;
            }
        }
    }
}
  
// Function to get the count of the cyclic
// permutations of b that given 0 when XORed with a
int countPermutation(string a, string b)
{
    // concatenate b with b
    b = b + b;
  
    // new b now contains all the cyclic
    // permutations of old b as it's sub-strings
    b = b.substr(0, b.size() - 1);
  
    // concatenate pattern with text
    int ans = 0;
    string s = a + "$" + b;
    int n = s.length();
  
    // Fill z array used in Z algorithm   
    int z[n];
    compute_z(s, z);
  
    for (int i = 1; i <= n - 1; i++) {
  
        // pattern occurs at index i since
        // z value of i equals pattern length
        if (z[i] == a.length())
            ans++;
    }
    return ans;
}
  
// Driver code
int main()
{
    string a = "101";
    string b = "101";
  
    cout << countPermutation(a, b) << endl;
  
    return 0;
}

Java

// Java program to find bitwise XOR between binary
// string A and all the cyclic permutations
// of binary string B
  
public class GFG{
      
    // Implementation of Z-algorithm
    // for linear time pattern searching
    static void compute_z(String s, int z[])
    {
        int l = 0, r = 0;
        int n = s.length();
        for (int i = 1; i <= n - 1; i++) {
            if (i > r) {
                l = i;
                r = i;
                while (r < n && s.charAt(r - l) == s.charAt(r))
                    r++;
                z[i] = r - l;
                r--;
            }
            else {
                int k = i - l;
                if (z[k] < r - i + 1) {
                    z[i] = z[k];
                }
                else {
                    l = i;
                    while (r < n && s.charAt(r - l) == s.charAt(r))
                        r++;
                    z[i] = r - l;
                    r--;
                }
            }
        }
    }
      
    // Function to get the count of the cyclic
    // permutations of b that given 0 when XORed with a
    static int countPermutation(String a, String b)
    {
        // concatenate b with b
        b = b + b;
        // new b now contains all the cyclic
        // permutations of old b as it's sub-strings
        b = b.substring(0, b.length() - 1);
      
        // concatenate pattern with text
        int ans = 0;
        String s = a + "$" + b;
        int n = s.length();
      
        // Fill z array used in Z algorithm   
        int z[] = new int[n];
        compute_z(s, z);
      
        for (int i = 1; i <= n - 1; i++) {
      
            // pattern occurs at index i since
            // z value of i equals pattern length
            if (z[i] == a.length())
                ans++;
        }
        return ans;
    }
      
      
    // Driver Code
     public static void main(String []args){
           
               String a = "101";
               String b = "101";
               System.out.println(countPermutation(a, b)) ;
     }
     // This code is contributed by ANKITRAI1
}

Python3

# Python 3 program to find bitwise XOR 
# between binary string A and all the 
# cyclic permutations of binary string B
  
# Implementation of Z-algorithm
# for linear time pattern searching
def compute_z(s, z):
    l = 0
    r = 0
    n = len(s)
    for i in range(1, n, 1):
        if (i > r):
            l = i
            r = i
            while (r < n and s[r - l] == s[r]):
                r += 1
            z[i] = r - l
            r -= 1
      
        else:
            k = i - l
            if (z[k] < r - i + 1):
                z[i] = z[k]
              
            else:
                l = i
                while (r < n and s[r - l] == s[r]):
                    r += 1
                z[i] = r - l
                r -= 1
      
# Function to get the count of the cyclic
# permutations of b that given 0 when XORed with a
def countPermutation(a, b):
      
    # concatenate b with b
    b = b + b
  
    # new b now contains all the cyclic
    # permutations of old b as it's sub-strings
    b = b[0:len(b) - 1]
  
    # concatenate pattern with text
    ans = 0
    s = a + "$" + b
    n = len(s)
  
    # Fill z array used in Z algorithm 
    z = [0 for i in range(n)]
    compute_z(s, z)
  
    for i in range(1, n, 1):
          
        # pattern occurs at index i since
        # z value of i equals pattern length
        if (z[i] == len(a)):
            ans += 1
      
    return ans
  
# Driver code
if __name__ == '__main__':
    a = "101"
    b = "101"
  
    print(countPermutation(a, b))
      
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find bitwise XOR between 
// binary string A and all the cyclic 
// permutations of binary string B 
using System;
  
class GFG
{
  
// Implementation of Z-algorithm 
// for linear time pattern searching 
public static void compute_z(string s, 
                             int[] z)
{
    int l = 0, r = 0;
    int n = s.Length;
    for (int i = 1; i <= n - 1; i++)
    {
        if (i > r)
        {
            l = i;
            r = i;
            while (r < n && s[r - l] == s[r])
            {
                r++;
            }
            z[i] = r - l;
            r--;
        }
        else
        {
            int k = i - l;
            if (z[k] < r - i + 1)
            {
                z[i] = z[k];
            }
            else
            {
                l = i;
                while (r < n && s[r - l] == s[r])
                {
                    r++;
                }
                z[i] = r - l;
                r--;
            }
        }
    }
}
  
// Function to get the count of the 
// cyclic permutations of b that
// given 0 when XORed with a 
public static int countPermutation(string a, 
                                   string b)
{
    // concatenate b with b 
    b = b + b;
      
    // new b now contains all the cyclic 
    // permutations of old b as it's sub-strings 
    b = b.Substring(0, b.Length - 1);
  
    // concatenate pattern with text 
    int ans = 0;
    string s = a + "$" + b;
    int n = s.Length;
  
    // Fill z array used in Z algorithm 
    int[] z = new int[n];
    compute_z(s, z);
  
    for (int i = 1; i <= n - 1; i++)
    {
  
        // pattern occurs at index i since 
        // z value of i equals pattern length 
        if (z[i] == a.Length)
        {
            ans++;
        }
    }
    return ans;
}
  
// Driver Code 
public static void Main(string[] args)
{
    string a = "101";
    string b = "101";
    Console.WriteLine(countPermutation(a, b));
}
}
  
// This code is contributed by Shrikant13

输出：

1