HDU 2700 Parity（找规律）

位串奇偶校验

最新推荐文章于 2017-12-02 16:11:44 发布

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本文介绍了一道关于位串奇偶校验的算法题目，包括题目的要求、输入输出样例及解析，并提供了一个简单的C语言解决方案。

http://acm.hdu.edu.cn/showproblem.php?pid=2700

Parity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1137 Accepted Submission(s): 891

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

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寻找规律

#include<stdio.h>
#include<string.h>
int main()
{
char c[35];
char *p=c;
int len,i,count;
while(scanf("%s",c)!=EOF)
{
    if(c[0]=='#') break;
len=strlen(c);
count=0;
    for(i=0;i<len-1;i++)
    {
    if(*(p+i)=='1')
         count++;
    }
    if((count%2==0&&c[len-1]=='e')||(count%2==1&&c[len-1]=='o'))//跟字符串里面的字符1个数有关
       c[len-1]='0';//注意加‘’符号，否则会输出很奇怪的东西
    else c[len-1]='1';
       puts(c);
}
return 0;
}