拆分二进制数以使每个部分都可被2整除的方式数量 _拆分二进制数

本文概述

C ++
Java
Python3
C#

【拆分二进制数以使每个部分都可被2整除的方式数量】给定一个二进制字符串小号, 任务是找到将其拆分为多个部分的方法, 以使每个部分都能被2.
例子：

输入：S ="100"
输出：2
有两种分割字符串的方法：{"10", "0"}和{"100"}
输入：S ="110"
输出：1

方法:一个观察结果是，字符串只能在0之后拆分。因此，计算字符串中0的数量。我们称这个count为c_0。假设字符串是偶数的情况，除了最右边的0，有两种选择，要么在这个0之后切断字符串，要么不切断字符串。因此，对于偶数字符串，最终的答案是2^(c_zero - 1)。
字符串不能被分割的情况是当它以1结束时。因此，对于奇数字符串，答案将始终为零，因为最后分割部分将始终为奇数。
下面是上述方法的实现：
C ++

//C++ implementation of the approach #include < bits/stdc++.h> using namespace std; #define maxN 20 #define maxM 64//Function to return the required count int cntSplits(string s) { //If the splitting is not possible if (s[s.size() - 1] == '1' ) return 0; //To store the count of zeroes int c_zero = 0; //Counting the number of zeroes for ( int i = 0; i < s.size(); i++) c_zero += (s[i] == '0' ); //Return the final answer return ( int ) pow (2, c_zero - 1); }//Driver code int main() { string s = "10010" ; cout < < cntSplits(s); return 0; }

Java

//Java implementation of the approach class GFG {static int maxN = 20 ; static int maxM = 64 ; //Function to return the required count static int cntSplits(String s) { //If the splitting is not possible if (s.charAt(s.length() - 1 ) == '1' ) return 0 ; //To store the count of zeroes int c_zero = 0 ; //Counting the number of zeroes for ( int i = 0 ; i < s.length(); i++) c_zero += (s.charAt(i) == '0' ) ? 1 : 0 ; //Return the final answer return ( int )Math.pow( 2 , c_zero - 1 ); }//Driver code public static void main(String []args) { String s = "10010" ; System.out.println(cntSplits(s)); } }//This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach # Function to return the required count def cntSplits(s) :# If the splitting is not possible if (s[ len (s) - 1 ] = = '1' ) : return 0 ; # To store the count of zeroes c_zero = 0 ; # Counting the number of zeroes for i in range ( len (s)) : c_zero + = (s[i] = = '0' ); # Return the final answer return int ( pow ( 2 , c_zero - 1 )); # Driver code if __name__ = = "__main__" : s = "10010" ; print (cntSplits(s)); # This code is contributed by AnkitRai01

//C# implementation of the approach using System; class GFG {static int maxN = 20; static int maxM = 64; //Function to return the required count static int cntSplits(String s) { //If the splitting is not possible if (s[s.Length - 1] == '1' ) return 0; //To store the count of zeroes int c_zero = 0; //Counting the number of zeroes for ( int i = 0; i < s.Length; i++) c_zero += (s[i] == '0' ) ? 1 : 0; //Return the final answer return ( int )Math.Pow(2, c_zero - 1); }//Driver code public static void Main(String []args) { String s = "10010" ; Console.WriteLine(cntSplits(s)); } }//This code is contributed by 29AjayKumar

输出如下：