HDU2700:Parity

Problem DescriptionA 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. 

InputThe 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'. 

OutputEach 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

101e010010o1e000e110100101o#

 

Sample Output

101001001011100001101001010

 

 

 

//水題，就是以e結尾就把1不成偶數，以o結尾則補成奇數

 

#include <stdio.h>#include <string.h>int main(){    char str[10000];    int len,i,cnt;    while(gets(str))    {        cnt = 0;        if(!strcmp(str,"#"))        break;        len = strlen(str);        for(i = 0;i<len;i++)        {            if(str[i] == '1')            cnt++;        }        if(str[len-1] == 'e')        {            if(cnt%2 == 0)            str[len-1] = '0';            else            str[len-1] = '1';        }        else if(str[len-1] == 'o')        {            if(cnt%2 == 0)            str[len-1] = '1';            else            str[len-1] = '0';        }        str[len] = '\0';        printf("%s\n",str);    }    return 0;}