HDOJ--2700 Parity

Problem Description
  
  
   A bit string has odd parity if the number of 1's is odd. A bit string has even parity if the number of 1's is even.Zero is considered to be an even number, so a bit string with no 1's has even parity. Note that the number of
0's does not affect the parity of a bit string.
  

  
   Input
  
  
   The input consists of one or more strings, each on a line by itself, followed by a line containing only "#" that signals the end of the input. Each string contains 1–31 bits followed by either a lowercase letter 'e' or a lowercase letter 'o'.

  

  
   Output
  
  
   Each line of output must look just like the corresponding line of input, except that the letter at the end is replaced by the correct bit so that the entire bit string has even parity (if the letter was 'e') or odd parity (if the letter was 'o').
  
  

  
   Sample Input
  
 
     101e
010010o
1e
000e
110100101o
#
    
   
   


    Sample Output
   

   
      1010
0100101
11
0000
1101001010
     

   
     
   


import java.util.Scanner;

public class Main {

   public static void main(String[] args) {
      Scanner sc=new Scanner(System.in);
      while(sc.hasNext()){
         String str=sc.nextLine();
         if(str.equals("#")){
            return;
         }
         int num=0;
         for(int i=0;i<str.length()-1;i++){
            if(str.charAt(i)=='1'){
               num++;
            }
         }
            if(num%2==0){//1 出现的次数为偶数
               if(str.charAt((str.length()-1))=='e'){ //偶数奇偶矫正
                  System.out.println(str.substring(0, str.length()-1)+'0');
                  
               }else{
                  System.out.println(str.substring(0, str.length()-1)+'1');
               }
            }else{
               if(str.charAt((str.length()-1))=='o'){
                  System.out.println(str.substring(0, str.length()-1)+'0');
               }else{
                  System.out.println(str.substring(0, str.length()-1)+'1');
               }
            }
         }
      }
}