考场上想到了没打完，细节思路还是不是很优，我原先的想法是每一次找完后标记那个点，下次再继续找（并不是这个意思，说不清楚）但实际上和平衡树一样加个大小就很好写了
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=500005;
const int DEP=31;
int N,M;
ll ans,s[MAXN];
struct node{
ll val;
int x,k;
bool operator <(const node& A) const {
return val<A.val;
}
};
priority_queue<node> q;
struct Node{
int ch[2],sum;
}trie[MAXN*(DEP+2)];
int head[MAXN],cnt;
void insert(Node c,Node& u,ll val,int d){
u.sum=c.sum+1;
if(d<0) return;
int x=(val>>d)&1ll;
u.ch[!x]=c.ch[!x];
insert(trie[c.ch[x]],trie[u.ch[x]=++cnt],val,d-1);
}
ll query(Node u,ll val,int d,int k){
if(d<0) return 0;
int x=(val>>d)&1;
int lsum=trie[u.ch[!x]].sum;
if(lsum>=k)
return ((ll)1<<d)+(ll)query(trie[u.ch[!x]],val,d-1,k);
return (ll)query(trie[u.ch[x]],val,d-1,k-lsum);
}
int main(){
trie[0].ch[0]=trie[0].ch[1]=trie[0].sum=0;
insert(trie[0],trie[head[0]=++cnt],0,DEP);
scanf("%d%d",&N,&M);
for(int i=1;i<=N;++i){
ll a;
scanf("%lld",&a);
s[i]=s[i-1]^a;
insert(trie[head[i-1]],trie[head[i]=++cnt],s[i],DEP);
q.push((node){query(trie[head[i-1]],s[i],DEP,1),i,1});
}
for(int i=1;i<=M;++i){
ans+=q.top().val;
int x=q.top().x,k=q.top().k;
q.pop();
if(k==x) continue;
q.push((node){query(trie[head[x-1]],s[x],DEP,k+1),x,k+1});
}
printf("%lld",ans);
return 0;
}