Atcoder ABC-F Xor Sum 3 (线性基)

Xor Sum 3

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define sc scanf
#define itn int
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=1e6+5;
const long long mod=1e9+7;
const long long mod2=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned long long ull;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
inline ll mul_64(ll x,ll y,ll c){return (x*y-(ll)((long double)x/c*y)*c+c)%c;}
inline ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=a*a%c)if(b&1)ans=ans*a%c;return ans;}
inline void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=x*(a/b);}
int n;
ll a[N];
ll dp[65];
bool vis[65]={0};
void insert(ll x){
  for(int i=60;i>=0;i--){
    if(!vis[i])continue; 
    if(x>>i&1){
      if(!dp[i]){
        dp[i]=x;
        break;
      }
      x^=dp[i];
    }
  }
}
ll Q(){
  ll ans=0;
  for(int i=60;i>=0;i--)if(vis[i]&&(ans>>i&1)==0)ans^=dp[i];// ans的i位为0则异或该位的值
  return ans;
}
int main(){
  cin>>n;
  ll sum=0;
  for(int i=1;i<=n;i++){
    sc("%lld",&a[i]);
    sum^=a[i];
  }
  for(int i=0;i<=60;i++){
    if(sum>>i&1)continue;//该位有奇数个1跳过
    vis[i]=1;//将偶数个1的位插入到线性基里
  }
  for(int i=1;i<=n;i++)insert(a[i]);
  ll g=Q();
  o((sum^g)+g);
}