HDU 2700 Parity

Parity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3354 Accepted Submission(s): 2529



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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'.

[align=left]Output[/align]
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').

[align=left]Sample Input[/align]

101e
010010o
1e
000e
110100101o
#

[align=left]Sample Output[/align]

1010
0100101
11
0000
1101001010

[align=left]Source[/align]
2008 Mid-Central USA

题意：
统计每组数中的1的个数

e是指判断1的个数为偶数 为真 变成0
为假 变成1

o是指判断1的个数为奇数 为真 变成0
为假 变成1

注意：

1：水题

#include<stdio.h>
#include<math.h>
#include<string.h>
int main (void)
{
int i,sum=0;
char a[34];
int leng;
while(~scanf("%s",a))
{
sum=0;
if(a[0]=='#')
return 0;
leng=strlen(a);
for(i=0;i<strlen(a)-1;i++)
{
sum+=a[i]-'0';printf("%c",a[i]);
}
if(((sum%2==0)&&(a[leng-1]=='e'))||((sum%2==1)&&(a[leng-1]=='o')))
{
printf("0\n");
}
else
{
printf("1\n");
}
}
return 0;
}