牛客地址

思路：全部组合异或，很容易想到使用线性基，正好线性基中有一个求第k小的用法，那我们可以二分来找 K是第几小的数，然后用总数减去。

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <cstdlib>
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f3f3f3f
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define re register
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(a) ((a)&-(a))
#define ios std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);
#define fi first
#define rep(i,n) for(int i=0;(i)<(n);i++)
#define rep1(i,n) for(int i=1;(i)<=(n);i++)
#define se second
#define scd(a) scanf("%d",&a)
#define scdd(a,b) scanf("%d%d",&a,&b)
#define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define ac cout<<ans<<"\n"
using namespace std;
typedef long long  ll;
typedef unsigned long long  ull;
typedef pair<ll,ll> pii;
int dx[4]= {-1,1,0,0},dy[4]= {0,0,1,-1};
const ll mod=1e9+7;
const ll N =1e6+10;
const double eps = 1e-4;
//const double pi=acos(-1);
ll gcd(ll a,ll b){return !b?a:gcd(b,a%b);}
ll n,k,x,tot;
ll d[70];
int cnt;
void ins(ll x){for(int i=60;i>=0;i--){if(x&(1ll<<i)){if(d[i]) x^=d[i];else{d[i]=x;return;}}}tot=1;
}
void rebuild()
{for(int i=1;i<=60;i++)for(int j=1;j<=i;j++)if(d[i]&(1ll<<(j-1)))d[i]^=d[j-1];for(int i=0;i<=60;i++) if(d[i]) cnt++;
}
ll check(ll x){//x-=tot;if(!x) return 0;if(x>=(1ll<<cnt)) return INF;ll ans=0;for(int i=0;i<=60;i++){if(d[i]){if(x&1) ans^=d[i];x/=2;}}return ans;
}
void sovle(){cin>>n>>k;for(int i=1;i<=n;i++){cin>>x;ins(x);}rebuild();ll s=(1ll<<cnt)-1;ll l=0,r=s;while(l<r){ll mid=(l+r+1)>>1;if(check(mid)<=k) l=mid;else r=mid-1;}cout<<s-l<<endl;
}
int main()
{iosint t=1;// cin>>t;while(t--){sovle();}return 0;
}