【模拟】Parity

描述

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.

输入

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’.

输出

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’).

样例输入

101e
010010o
1e
000e
110100101o

样例输出

1010
0100101
11
0000
1101001010

题目来源

Mid-Central USA 2008

分析：
不算难的模拟，根据奇偶性来修改字符串。
代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    string s;
    int t;
    while(cin>>s)
    {
        t=0;
        if (s[0]=='#') break;
        int l=s.length();
        for (int i=0;i<l;i++)
        {
            if (s[i]=='1') t++;//统计1的个数来判断奇偶
        }
        if (t%2==0)//偶数
        {
            if (s[l-1]=='e')
                s[l-1]='0';
            if (s[l-1]=='o')
                s[l-1]='1';
        }
        else
        {
            if (s[l-1]=='e')
                s[l-1]='1';
            else
            if (s[l-1]=='o')
                s[l-1]='0';
        }
        cout<<s<<endl;
    }
    return 0;
}