HDU 2700

Parity

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4334    Accepted Submission(s): 3264

Problem Description

A bit string hasodd parity if the number of 1's is odd. A bit string has even parity if thenumber of 1's is even.Zero is considered to be an even number, so a bit stringwith 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 consistsof one or more strings, each on a line by itself, followed by a line containingonly "#" that signals the end of the input. Each string contains 1–31bits followed by either a lowercase letter 'e' or a lowercase letter 'o'.

 

 

Output

Each line ofoutput must look just like the corresponding line of input, except that theletter at the end is replaced by the correct bit so that the entire bit stringhas 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

 

题意简述

看字符串中1的个数，e结尾变成偶数，o结尾变成单数

题意分析

明白了就是水题

 

代码总览

#include<stdio.h>
#include<string.h>
int main()
{
	//freopen("in.txt","r",stdin);
	int length,i,count;
	char str[101] ;
	while(scanf("%s",str) != EOF && str[0] != '#'){
		count = 0;
		length = strlen(str);
		for(i = 0; i<length-1;i++){
			if(str[i]=='1'){
				count++;
			} 
		}
		if(count%2==0 && str[length-1] == 'e'){
			str[length-1]='0';
		}else if(count%2!=0 && str[length-1] == 'e'){
			str[length-1]='1';
		}else if(count%2==0 && str[length-1] == 'o'){
			str[length-1]='1';
		}else if(count%2!=0 && str[length-1] == 'o'){
			str[length-1]='0';
		}
		for(i = 0; i<length;i++){
			printf("%c",str[i]);
			if(i == length-1){
				printf("\n");
			}
		}
	}
	return 0;
	//fclose(stdin);
}

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