题意
Sol
记(s_i)表示前(i)个数的前缀异或和，我们每次相当于要找一个(j)满足(0 < j < i)且((s_i oplus s_j) + s_j)最大
然后下面的就和标算相差十万八千里了。
[egin{aligned}
&(s_i oplus s_j) + s_j\
=&(s_i oplus s_j oplus s_j) + ((s_i oplus s_j) & s_j )\
=&(s_i + ( ext{~}s_i & s_j))
end{aligned}
]
也就是对于每个(i)，我们要在前面找一个(j)使得( ext{~}s[i] & s[j])最大
然后这里暴力处理子集就行了(一开始还想了半天trie树)。
加一个记忆化可以保证复杂度
最后复杂度为(O(2^{20} + n log{a_i}))
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define ull unsigned long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 3e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '
';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
template <typename A, typename B> inline LL fp(A a, B p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;}
template <typename A> A inv(A x) {return fp(x, mod - 2);}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, a[MAXN], s[MAXN];
bool mark[MAXN];
void insert(int x) {
//if(mark[x]) return ;
mark[x] = 1;
for(int i = 0; i < 20; i++)
if((x >> i & 1) && (!mark[x ^ (1 << i)]))
insert(x ^ (1 << i));
}
int Query(int x) {
int ans = 0;
for(int i = 19; ~i; i--)
if((x >> i & 1) && mark[ans | 1 << i])
ans |= 1 << i;
return ans;
}
signed main() {
//freopen("ex_childhood2.in", "r", stdin);
N = read();
for(int i = 1; i <= N; i++) a[i] = read(), s[i] = s[i - 1] ^ a[i];
for(int i = 1; i <= N; i++) {
// for(int j = i - 1; j >= 0; j--) chmax(ans, (s[i] ^ s[j]) + s[j]);
//for(int j = i - 1; j >= 0; j--) chmax(ans, (~s[i]) & s[j]);
int ans = Query(~s[i]);
cout << s[i] + ans * 2 << ' ';
insert(s[i]);
}
puts("");
return 0;
}