• 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
     
     
     
    /*
    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("
    ");
        }
    }
    

      

  • 相关阅读:
    javascript DOM 操作
    遍历Map集合四中方法
    遍历List集合的三种方法
    Java中static关键字用法总结
    java中this关键字的作用
    深入理解Java的接口和抽象类
    MySQL查看所有用户及拥有权限
    MySQL新建用户,授权
    VMware Ubuntu如何连接互联网
    控件不接收用户交互的情况以及事件响应顺序
  • 原文地址:https://www.cnblogs.com/Roni-i/p/7206115.html
Copyright © 2020-2023  润新知