HDU 2700 Parity（找规律）

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

Parity

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

Problem Description

A bit string has odd parity if the number of 1's is odd. A bitstring has even parity if the number of 1's is even.Zero isconsidered to be an even number, so a bit string with no 1's haseven 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 byitself, followed by a line containing only "#" that signals the endof the input. Each string contains 1–31 bits followed by either alowercase letter 'e' or a lowercase letter 'o'.

 

 

Output

Each line of output must look just like the corresponding lineof input, except that the letter at the end is replaced by thecorrect bit so that the entire bit string has even parity (if theletter was 'e') or odd parity (if the letter was 'o').

 

 

Sample Input

101e010010o 1e 000e 110100101o #

 

 

Sample Output

10100100101 11 0000 1101001010

 

 

Source

2008 Mid-Central USA

 

 

Recommend

zty

 

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
    elsec[len-1]='1';
      puts(c);
  }
  return 0;
}