HDU 2700 Parity（字符串，奇偶性）

         Parity

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1855    Accepted Submission(s): 1447

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

 

 

 

/*
e：是要使字符串中1的个数变成偶数
o：是要使字符串中1的个数变成奇数
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
using namespace std;
int main()
{
    char s[100];
    int cnt;
    while(~scanf("%s",s))
    {
        cnt=0;
        if(s[0]=='#')
            break;
        int len=strlen(s);
        for(int i=0;i<len;i++)
        {
            if(s[i]=='1')
                cnt++;
        }
        if(cnt%2==0)
            {
                if(s[len-1]=='e')//最后一位是len-1 ！
                    s[len-1]='0';
                if(s[len-1]=='o')
                    s[len-1]='1';
            }
            else
            {
                if(s[len-1]=='o')
                    s[len-1]='0';
                if(s[len-1]=='e')
                    s[len-1]='1';
            }
        for(int i=0;i<len;i++)
        {
            printf("%c",s[i]);
        }
        printf("\n");
    }
}