HDU 2700 Parity

最新推荐文章于 2017-12-06 12:57:32 发布

原创最新推荐文章于 2017-12-06 12:57:32 发布 · 1.5k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#string #input #output #each #c

ACM题目解析专栏收录该内容

146 篇文章

订阅专栏

Parity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1389 Accepted Submission(s): 1088

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

Sample Output

思路：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。

AC代码：

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
	int al,i,s;
	char a[50];
	while(scanf("%s",a)!=EOF)
	{
		al=strlen(a);
		s=0;
		if(a[0]=='#')break;
		for(i=0;i<al-1;i++)
		{
           if(a[i]=='1')s++;
		   printf("%c",a[i]);
		}
		if(a[al-1]=='e')
		{
			if(s%2==0)printf("0\n");
			else printf("1\n");
		}
		else
		{
			if(s%2==0)printf("1\n");
			else printf("0\n");
		}
	}
	return 0;
}