二进制字符串的最大拆分数，以使每个子字符串可被给定的奇数整除

给定一个二进制字符串str ，任务是计算最大可能的拆分，以使每个子字符串可以被给定的奇数K整除。

例子：

Input: str = “110111001”, K = 9
Output: 2
Explanation:
The two possible substrings are “11011” and “1001”. The equivalent decimal values are 27 and 9 respectively which are divisible by 9.

Input: str = “10111001”, K = 5
Output: 2
Explanation:
The two possible substrings are “101” and “11001”. The equivalent decimal values are 5 and 25 respectively which are divisible by 5.

方法：为了解决此问题，我们从字符串的末尾开始遍历并生成所遍历长度的总和。一旦总和整除K，我们通过1和复位和增加数量为0，穿越着，重复同样的过程。完整遍历字符串，如果将sum重置为0，则count的值将给出所需的最大可能分割数。否则，请打印“不可能”，因为所有段都不能被K整除。

下面的代码是上述方法的实现：

C++

// C++ Program to split
// a given binary string
// into maximum possible
// segments divisible by
// given odd number K
  
#include 
using namespace std;
// Function to calculate
// maximum splits possible
void max_segments(string str, int K)
{
    int n = str.length();
    int s = 0, sum = 0, count = 0;
    for (int i = n - 1; i >= 0; i--) {
        int a = str[i] - '0';
        sum += a * pow(2, s);
  
        s++;
        if (sum != 0 && sum % K == 0) {
            count++;
            sum = 0;
            s = 0;
        }
    }
    if (sum != 0)
        cout << "-1" << endl;
    else
        cout << count << endl;
}
  
// Driver code
int main()
{
    string str = "10111001";
    int K = 5;
    max_segments(str, K);
    return 0;
}

Java

// Java code to split a given
// binary string into maximum
// possible segments divisible
// by given odd number K 
import java.io.*; 
import java.util.*;
  
class GFG{
      
// Function to calculate 
// maximum splits possible 
static void rearrange(String str, int K) 
{ 
    int n = str.length(); 
    int s = 0, sum = 0, count = 0;
      
    for(int i = n - 1; i >= 0; i--)
    { 
       int a = str.charAt(i) - '0'; 
       sum += a * Math.pow(2, s); 
       s++; 
         
       if (sum != 0 && sum % K == 0) 
       { 
           count++; 
           sum = 0; 
             s = 0; 
       } 
    }
   
    if (sum != 0) 
        System.out.println("-1");    
    else
        System.out.println(count);
}
  
// Driver code 
public static void main(String[] args) 
{ 
    String str = "10111001"; 
    int K = 5;
      
    rearrange(str, K); 
} 
} 
  
// This code is contributed by coder001

Python3

# Python3 program to split
# a given binary string
# into maximum possible
# segments divisible by
# given odd number K
  
# Function to calculate
# maximum splits possible
def max_segments(st, K):
  
    n = len(st)
    s, sum, count = 0, 0, 0
      
    for i in range(n - 1, -1, -1):
        a = ord(st[i]) - 48
        sum += a * pow(2, s)
        s += 1
          
        if (sum != 0 and sum % K == 0):
            count += 1
            sum = 0
            s = 0
              
    if (sum != 0):
        print("-1")
    else:
        print(count)
  
# Driver code
if __name__ == "__main__":
      
    st = "10111001"
    K = 5
    max_segments(st, K)
  
# This code is contributed by chitranayal

C#

// C# program to split a given 
// binary string into maximum 
// possible segments divisible by 
// given odd number K 
using System;
  
class GFG{
      
// Function to calculate 
// maximum splits possible 
static void max_segments(string str, int K)
{
    int n = str.Length;
    int s = 0;
    int sum = 0;
    int count = 0;
      
    for(int i = n - 1; i >= 0; i--)
    {
       int a = str[i] - '0';
       sum += a * (int)Math.Pow(2, s);
       s++;
         
       if (sum != 0 && sum % K == 0)
       {
           count++;
           sum = 0;
             s = 0;
       }
    }
    if (sum != 0)
    {
        Console.Write("-1");
    }
    else
    {
        Console.Write(count);
    }
}
  
// Driver code 
public static void Main() 
{
    string str = "10111001";
    int K = 5;
      
    max_segments(str, K);
}
}
  
// This code is contributed by sayesha

输出：

2