[2700]：Parity

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’).[even:偶数的, odd:奇数的]

Sample Input
101e
010010o
1e
000e
110100101o
#

Sample Output
1010
0100101
11
0000
1101001010

简单题目，看懂上面标黑部分即可

#include<stdio.h>#include<string.h>#define N 100int main(){    char str[N];    while(scanf("%s", str)!=EOF){        if(str[0] == '#'){            break;        }        int len = strlen(str);        int i = 0, count = 0;        for(; i<len - 1; i++){            if(str[i] == '1'){                count++;            }        }        if(str[len-1] == 'e' && count%2 == 0){            str[len-1] = '0';        }        if(str[len-1] == 'e' && count%2 == 1){            str[len-1] = '1';        }        if(str[len-1] == 'o' && count%2 == 1){            str[len-1] = '0';        }        if(str[len-1] == 'o' && count%2 == 0){            str[len-1] = '1';        }        printf("%s\n", str);    }    return 0;}