[线性基] BZOJ 4269: 再见Xor

$Solution$

划水中。。迟早要完
显然要线性基。。
考虑求$k$大。
对于基的每一个位置，因为其最高非零位就是位置标号，那看一下$k$的这一位是否为，看是否要异或上（或者不异或）使得这一位得到的答案较大。

#include <bits/stdc++.h>
#define show(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long ll;
typedef pair<int, int> Pairs;

const int N = 101010;

inline char get(void) {
    static char buf[100000], *S = buf, *T = buf;
    if (S == T) {
        T = (S = buf) + fread(buf, 1, 100000, stdin);
        if (S == T) return EOF;
    }
    return *S++;
}
template<typename T>
inline void read(T &x) {
    static char c; x = 0; int sgn = 0;
    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
    if (sgn) x = -x;
}

struct linearBasis {
    int a[32];
    int siz;
    inline int size(void) {
        return siz;
    }
    inline int insert(int x) {
        for (int i = 30; ~i; i--)
            if (x >> i & 1) {
                if (!a[i]) {
                    a[i] = x; return ++siz;
                } else x ^= a[i];
            }
        return 0;
    }
    inline int query(int k) {
        int res = 0, cnt = siz;
        for (int i = 30; ~i; i--) {
            if (!a[i]) continue;
            --cnt;
            int tmp = max(res, res ^ a[i]);
            if (k >> cnt & 1) res = tmp;
            else res = tmp ^ a[i];
        }
        return res;
    }
} LB;
int n, x, mx;

int main(void) {
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
    read(n);
    while (n--) {
        read(x); LB.insert(x);
    }
    int L = 0, R = (1 << LB.size()) - 1, Mid, Ans;
    printf("%d ", mx = LB.query(R));
    while (L <= R) {
        int Mid = (L + R) >> 1;
        if (LB.query(Mid) < mx) L = (Ans = Mid) + 1;
        else R = Mid - 1;
    }
    printf("%d\n", LB.query(Ans));
    return 0;
}