HDU 2700 Parity

Parity

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3354    Accepted Submission(s): 2529

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

2008 Mid-Central USA

 

 

 

题意：

统计每组数中的1的个数

 

e是指判断1的个数为偶数 为真 变成0

                                      为假 变成1

 

o是指判断1的个数为奇数 为真  变成0

                                      为假 变成1

 

注意：

1：水题

 

 

#include<stdio.h>
#include<math.h>
#include<string.h>
int main (void)
{
    int i,sum=0;
    char a[34];
    int leng;
    while(~scanf("%s",a))
    {
        sum=0;
        if(a[0]=='#')
            return 0;
        leng=strlen(a);
        for(i=0;i<strlen(a)-1;i++)
        {
            sum+=a[i]-'0';printf("%c",a[i]);
        }
        if(((sum%2==0)&&(a[leng-1]=='e'))||((sum%2==1)&&(a[leng-1]=='o')))
        {
           printf("0\n");
        }
        else
        {
            printf("1\n");
        }
    }
    return 0;
}