noi.ac#309 Mas的童年(子集乱搞)

时间：2021-06-06 23:15:45

题意

题目链接

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。

\[\begin{aligned}
&(s_i \oplus s_j) + s_j\\
=&(s_i \oplus s_j \oplus s_j) + ((s_i \oplus s_j) \& s_j )\\
=&(s_i + (\text{~}s_i \& s_j))
\end{aligned}
\]

也就是对于每个\(i\)，我们要在前面找一个\(j\)使得\(\text{~}s[i] \& s[j]\)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 << '\n';}

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;

}

相关文章