西南科技大学OJ题 Hamming Distance 0056

最新推荐文章于 2021-11-08 21:53:06 发布

孤雪胜悲鸣

最新推荐文章于 2021-11-08 21:53:06 发布

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本文介绍了一种计算两个二进制字符串之间汉明距离的方法。汉明距离是指在相同位置上两个字符串对应位不同的数量，文章提供了一个通过逐位比较来计算这一距离的C语言程序示例。

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Hamming Distance

1000(ms)

65535(kb)

2037 / 6444

Have you ever heard of the Hamming distance? It is the number of positions for which the corresponding digits differ. Your task is to write a program that computes this distance for two binary strings.

输入

The input contains several test cases. Each test case consists of two lines. Each line contains one binary number. Any two numbers given in one test case have the same length, which is at most 100 binary digits. The last test case is followed by a line containing the uppercase letter “X”.

输出

Your program must output a single line for each test case. The line should contain the statement “Hamming distance is X.”, where X is the number of positions where the two numbers have different digits.

样例输入

0
1
000
000
1111111100000000
0000000011111111
101
000
X

样例输出

Hamming distance is 1.
Hamming distance is 0.
Hamming distance is 16.
Hamming distance is 2.

#include<stdio.h>
int main()
{
	char a[100],b[100];
	int c;
	for( ; ; )
	{
		scanf("%s",a);
		if(a[0]=='X')
		break;
		scanf("%s",b);
		c=0;
		for(int i=0;i<100;i++)
		{
			if(a[i]=='\0')
			break;
			else if(a[i]!=b[i])
			c++;
		}
		printf("Hamming distance is %d.\n",c);
	}
}