水题 Parity

Parity

Time Limit : 2000/1000ms(Java/Other)   MemoryLimit : 32768/32768K (Java/Other)

Total Submission(s) :11   AcceptedSubmission(s) : 10

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Problem Description

A bit string has odd parity if the number of 1's is odd. A bitstring has even parity if the number of 1's is even.Zero isconsidered to be an even number, so a bit string with no 1's haseven 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 byitself, followed by a line containing only "#" that signals the endof the input. Each string contains 1–31 bits followed by either alowercase letter 'e' or a lowercase letter 'o'.

Output

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

Sample Input

101e 

010010o 

1e 

000e 

110100101o 

#

Sample Output

1010 

0100101 

11 

0000 

1101001010

主要是理解题意，理解啦很好做的

#include<stdio.h>

#include<string.h>

int main()

{

    char str[35];

       int n,i,c;

  while(scanf("%s",str)!=EOF)

 {

           if(str[0]=='#')

                     break;

              n=strlen(str);

              c=0;

                for(i=0;i<n;i++)

             {

                   if(str[i]=='1')

                             c++;

                }

           if(c%2==0&&str[n-1]=='e')

           {

             str[n-1]='0';

              printf("%s\n",str);

         }

           if(c%2==0&&str[n-1]=='o')

           {

                   str[n-1]='1';

                       printf("%s\n",str);

         }

           if(c%2==1&&str[n-1]=='e')

           {

             str[n-1]='1';

              printf("%s\n",str);

         }

           if(c%2==1&&str[n-1]=='o')

           {

             str[n-1]='0';

              printf("%s\n",str);

         }

   }

   return 0;

}