Parity HDU 2700



Parity

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2770    Accepted Submission(s): 2139

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
   
   
   
   

 

Source

这个题卡了半天。。。。原因就是循环里面下表上限搞错了。。。。

#include<stdio.h>
#include<string.h>

char st[50];
int len,k,i;

int main()
{
   while(gets(st) != NULL && st[0]!='#')
   {
       len = strlen(st);
       k = 0;
       for(i = 0 ;i < len -1;i++)
           if (st[i] == '1') k++;

        if(st[len-1] == 'e' )
        st[len-1] = (k%2!=0) ? '1':'0';

        else if(st[len -1] =='o')
       st[len-1] = (k%2!=0) ? '0':'1';
        puts(st);
   }
    return 0;
}