从0-2^B枚举即可,将i和前面存下的所有答案对比,满足条件就存进去,这里我们判不同位的个数可以直接用异或(相同为0不同为1),然后用函数求出异或后'1'的个数即可

#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define lowbit(a) (a&(-a))
#define _mid(a,b) ((a+b)/2)
#define _mem(a,b) memset(a,0,(b+3)<<2)
#define fori(a) for(int i=0;i<a;i++)
#define forj(a) for(int j=0;j<a;j++)
#define ifor(a) for(int i=1;i<=a;i++)
#define jfor(a) for(int j=1;j<=a;j++)
#define mem(a,b) memset(a,b,sizeof(a))
#define IN freopen("in.txt","r",stdin)
#define OUT freopen("out.txt","w",stdout)
#define IO do{\
    ios::sync_with_stdio(false);\
    cin.tie(0);\
    cout.tie(0);}while(0)
using namespace std;
typedef long long ll;
const int maxn =  150+10;
const int INF = 0x3f3f3f3f;
const int inf = 0x3f;
const double EPS = 1e-7;
const double Pi = acos(-1);
const int MOD = 1e9+7;
int Minbit;
int a[maxn];
int cnt;
bool juge(int x){
    fori(cnt)
        if(__builtin_popcount(x^a[i])<Minbit)    //编译器内置函数,求出x^a[i]的二进制数中'1'的个数
            return false;
    return true;
}
int main(){
    int n,b;
    cin >> n >>b >>Minbit;
    fori(1<<b){
        if(juge(i))
            a[cnt++] = i;
        if(cnt == n)
            break;
    }
    int cn = 1;
    fori(cnt){
        cout <<a[i];
        if(cn%10==0)
            cout << endl;
        else if(i!=cnt-1)
            cout <<" ";
        cn++;
    }
    return 0;
}