USACO SECTION 2.1 Hamming Codes

最新推荐文章于 2024-05-04 16:56:52 发布

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最新推荐文章于 2024-05-04 16:56:52 发布

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分类专栏： USACO 文章标签： distance output each integer file pair

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本文介绍了一种基于Hamming距离的编码算法实现方法，该算法能够生成满足特定长度及距离要求的码字集合。通过使用C语言编程，文章提供了一个具体的实现方案，包括输入输出格式及示例。

Hamming Codes
Rob Kolstad

Given N, B, and D: Find a set of N codewords (1 <= N <= 64), each of length B bits (1 <= B <= 8), such that each of the codewords is at least Hamming distance of D (1 <= D <= 7) away from each of the other codewords. The Hamming distance between a pair of codewords is the number of binary bits that differ in their binary notation. Consider the two codewords 0x554 and 0x234 and their differences (0x554 means the hexadecimal number with hex digits 5, 5, and 4):

        0x554 = 0101 0101 0100
        0x234 = 0010 0011 0100
Bit differences: xxx  xx

Since five bits were different, the Hamming distance is 5.

PROGRAM NAME: hamming

INPUT FORMAT

N, B, D on a single line

SAMPLE INPUT (file hamming.in)

16 7 3

OUTPUT FORMAT

N codewords, sorted, in decimal, ten per line. In the case of multiple solutions, your program should output the solution which, if interpreted as a base 2^B integer, would have the least value.

SAMPLE OUTPUT (file hamming.out)

0 7 25 30 42 45 51 52 75 76
82 85 97 102 120 127

#include<stdlib.h>
#include<string.h>

int CodeWords[64];
int WordsNumber,Length,Distance;

int IsTrue(int i,int NextWords)
{
	int j,k;
	int a,b,count;
	for(j=0;j<i;j++)
	{
		a=CodeWords[j];
		b=NextWords;
		count=0;
		for(k=0;k<Length;k++)
		{
			if((a&1)!=(b&1))
				count++;
			a>>=1;
			b>>=1;
		}
		if(count<Distance)
			return 0;
	}
	return 1;
}

int main()
{
	FILE *fin,*fout;
	fin=fopen("hamming.in","r");
	fout=fopen("hamming.out","w");
	
	fscanf(fin,"%d %d %d",&WordsNumber,&Length,&Distance);
	
	int i,j,k;
	int NextWord;
	
	CodeWords[0]=0;
	for(i=1;i<WordsNumber;i++)
	{
		NextWord=CodeWords[i-1]+1;
		while(!IsTrue(i,NextWord))
		{
			NextWord++;
		}	
		CodeWords[i]=NextWord;
	}
	
	for(i=0;i<WordsNumber;i++)
	{
		fprintf(fout,"%d",CodeWords[i]);
		if((i+1)%10==0)
			fprintf(fout,"\n");
		else
		if(i!=WordsNumber-1)
			fprintf(fout," ");
	}
	if(i%10!=0)
		fprintf(fout,"\n");
	
	return 0;
}

感觉题好水。。没有以前花好长时间做出来的那种喜悦的感觉了= =|||

多去NOCOW上面看看别人的做法。看看能不能多点收获