Hamming Codes(2진수 매거진)

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

 
제목:
N (1 ~ 64), B (1 ~ 8), D (1 ~ 7)를 제시하고 N 개의 B 비트 바이너리 수를 출력합니다. 이 수를 임의로 바이너리로 바꾸면 다른 수는 적어도 D 개가 있어야 합니다. 이 서열은 와 가장 작은 서열입니다.
 
생각:
일일이 열거하다.최대 8비트 2진법, 설명 최대 255, 과감한 2진법 매거, 1A.
 
    AC:

/*    
TASK:hamming    
LANG:C++    
ID:sum-g1    
*/
#include<stdio.h>
#include<math.h>
#include<string.h>

int n,b,d,ans;
int bin_a[10],bin_b[10];
int fin[70];

int test(int num)
{
    int k = 0;
    while(num)
    {
        k++;
        bin_a[k] = num % 2;
        num /= 2;
    }
    for(int i = 1;i <= ans;i++)
    {
        int change = fin[i];
        k = 0;
        memset(bin_b,0,sizeof(bin_b));
        while(change)
        {
            k++;
            bin_b[k] = change % 2;
            change /= 2;
        }

        int sum = 0,temp = 0;
        for(k = 1;k <= b;k++)
        {
            if(bin_a[k] != bin_b[k]) sum++;
            if(sum >= d)
            {
                temp = 1;
                break;
            }
        }
        if(!temp) return 0;
    }
    return 1;
}

int main()
{
    freopen("hamming.in","r",stdin);        
    freopen("hamming.out","w",stdout); 
    ans = 1;
    scanf("%d%d%d",&n,&b,&d);
    fin[1] = 0;
    printf("%d ",fin[1]);
    for(int i = 1;i <= pow(2,b) - 1;i++)
    {
        memset(bin_a,0,sizeof(bin_a));
        if(test(i))
        {
            ans++;
            fin[ans] = i;
            printf("%d",i);
            if(!(ans % 10) || ans == n) printf("
");
            else                        printf(" ");
        }
        if(ans == n) break;
    }
    return 0;
}