Hamming Codes

链接

分析：码农模拟题，首先我们先求出对于0来说满足条件的数，然后在从集合当中筛选出任意两个都满足条件的数即可

/*
    PROB:hamming
    ID:wanghan
    LANG:C++
*/
#include "iostream"
#include "cstdio"
#include "cstring"
#include "string"
#include "vector"
#include "cmath"
#include "algorithm"
using namespace std;
const int maxn=1000+10;
int N,B,D;
vector<int> p;
int vis[maxn];
string transform(int x,int y,string s)  
{  
    string res="";  
    int sum=0;  
    for(int i=0;i<s.length();++i)  
    {  
        if(s[i]=='-')  continue;  
        if(s[i]>='0'&&s[i]<='9')  
        {  
            sum=sum*x+s[i]-'0';  
        }  
        else  
        {  
            sum=sum*x+s[i]-'A'+10;  
        }  
    }  
    while(sum)  
    {  
        char tmp=sum%y;  
        sum/=y;  
        if(tmp<=9)  
        {  
            tmp+='0';  
        }  
        else  
        {  
            tmp=tmp-10+'A';  
        }  
        res=tmp+res;  
    }  
    if(res.length()==0)  res="0";  
    if(s[0]=='-')  res='-'+res;  
    return res;  
}
string change(int x){
    string res="";
    while(x){
        int num=x%10;
        res+=num+'0';
        x/=10;
    }
    int i=0,j=res.length()-1;
    while(i<j){
        swap(res[i],res[j]);
        i++,j--;
    }
    return res;
}
bool solve(int a,int b){
    string s1=transform(10,2,change(a)),s2=transform(10,2,change(b));
    string res1="",res2="";
    int cha;
    if(s1.length()<B){
        cha=B-s1.length();
        for(int i=0;i<cha;i++)
            res1+='0';
    }
    res1+=s1;
    if(s2.length()<B){
        cha=B-s2.length();
        for(int i=0;i<cha;i++)
            res2+='0';
    }
    res2+=s2;
    int cnt=0;
    for(int i=0;i<B;i++){
        if(res1[i]!=res2[i]){
            cnt++;
        }
    }
    if(cnt>=D)
        return true;
    return false;
}
struct Node{
    int x;
    string res;
};
bool cmp(Node a,Node b){
    return a.res<b.res;
}
int main()
{
    freopen("hamming.in","r",stdin);
    freopen("hamming.out","w",stdout);
    cin>>N>>B>>D;
    p.push_back(0);
    int pos=0;
    int i=1;
    int num=pow(2,B+1)-1;
    while(i<num){
        if(solve(pos,i)){
            p.push_back(i);
        }
        i++;
    }
    memset(vis,0,sizeof(vis));
    int len=p.size();
    for(int i=0;i<len;i++){
        for(int j=i+1;j<len;j++){
            if(!vis[i]){
                if(vis[j]||!solve(p[i],p[j])){
                    vis[j]=1;
                }
            }
        }
    }
    vector<int> f;
    for(int i=0;i<len;i++){
        if(!vis[i]){
            int zz=p[i];
            f.push_back(zz);
        }
    }
    for(int i=0;i<N;i++){
        if(i%10==9){
            cout<<f[i]<<endl;
        }else{
            if(i==N-1)
                cout<<f[i]<<endl;
            else
                cout<<f[i]<<" ";
        }
    }
    return 0;
}