检查字符串是否可以分成两个子序列，使得和的乘积为奇数

#include <bits/stdc++.h>
using namespace std;
bool isSplitPossible(string num_str, int num_len) {
   int sub1 = 0;
   
   // Find total number of odd digits
   for (int p = 0; p < num_len; p++) {
      if ((num_str[p] - '0') % 2 != 0) {
         sub1++;
      }
   }
   
   // Return false, if sub1 is odd or 0
   if (sub1 % 2 != 0 || sub1 == 0)
   return false;
   return true;
}
int main() {
   string num_str = "3232";

   int num_len = num_str.length();
   if (isSplitPossible(num_str, num_len)) {
      cout << "Yes, it is possible to split the string into two substrings.";
   } else {
      cout << "No, it is not possible to split the string into two substrings.";
   }
   return 0;
}

public class SplitString {
   public static boolean isSplitPossible(String numStr) {
      int sub1 = 0;

      // Find total number of odd digits
      for (int p = 0; p < numStr.length(); p++) {
         if ((numStr.charAt(p) - '0') % 2 != 0) {
            sub1++;
         }
      }

      // Return false if sub1 is odd or 0
      if (sub1 % 2 != 0 || sub1 == 0) {
         return false;
      }
      return true;
   }

   public static void main(String[] args) {
      String numStr = "3232";

      if (isSplitPossible(numStr)) {
         System.out.println("Yes, it is possible to split the string into two substrings.");
      } else {
         System.out.println("No, it is not possible to split the string into two substrings.");
      }
   }
}

def is_split_possible(num_str):
   sub1 = 0

   # Find total number of odd digits
   for digit in num_str:
      if int(digit) % 2 != 0:
         sub1 += 1

   # Return false if sub1 is odd or 0
   if sub1 % 2 != 0 or sub1 == 0:
      return False
   return True

# Main function
if __name__ == "__main__":
   num_str = "3232"

   if is_split_possible(num_str):
      print("Yes, it is possible to split the string into two substrings.")
   else:
      print("No, it is not possible to split the string into two substrings.")