USACO2.1.5Hamming Codes

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
题解：赤果果的枚举+位运算啊。枚举范围为 [0,2^B-1],先对两个编码进行异或运算得到一个新数字，然后对数字进行与运算和移位运算把其中的1的个数计算出来，再判断Hamming距离是否大于D，如果是则是符合要求的（必须与其他所有的数相比，Hamming距离都符合要求,这个数才正确）。

View Code

 1 /*
 2 ID:spcjv51
 3 PROG:hamming
 4 LANG:C
 5 */
 6 #include<stdio.h>
 7 int n,a[500];
 8 int distance(int x)
 9 {
10     int ans;
11     ans=0;
12     while(x)
13     {
14         ans+=x&1;
15         x>>=1;
16     }
17     return ans;
18 }
19 void print()
20 {
21     int i;
22     for(i=1;i<n;i++)
23     {
24         if(i%10==0) printf("%d\n",a[i]);
25         else
26         printf("%d ",a[i]);
27     }
28     printf("%d\n",a[i]);
29 }
30 int main(void)
31 {
32     freopen("hamming.in","r",stdin);
33     freopen("hamming.out","w",stdout);
34     int b,d,ans,i,j,k,x,y;
35     scanf("%d%d%d",&n,&b,&d);
36     for(i=0; i<(1<<b)-1; i++)
37     {
38         ans=1;
39         a[ans]=i;
40         for(j=i+1; j<(1<<b); j++)
41         {
42             x=i^j;
43             if(distance(x)>=d)
44             {
45                 for(k=1; k<=ans; k++)
46                 {
47                     y=j^a[k];
48                     if(distance(y)<d)
49                         break;
50                 }
51                 if(k==ans+1)
52                 {
53                     ans++;
54                     a[ans]=j;
55                 }
56             }
57             if(ans==n) break;
58 
59         }
60         if(ans==n)
61         {
62             print();
63             break;
64         }
65     }
66     return 0;
67 }