bzoj 4269 再见Xor 线性基

题面

题目传送门

解法

第一问就是线性基的裸题

第二问也很类似,从低位向高位枚举,如果线性基上这一位有数,那么直接异或后返回

时间复杂度:\(O(n\ log\ a_i)\)

代码

#include <bits/stdc++.h>
#define int long long
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Linear_Base {
	int a[40];
	void ins(int x) {
		for (int i = 31; ~i; i--)
			if ((x >> i) & 1) {
				if (!a[i]) {a[i] = x; break;}
				x ^= a[i];
			}
	}
} b;
main() {
	int n; read(n);
	for (int i = 1; i <= n; i++) {
		int x; read(x);
		b.ins(x);
	}
	int mx = 0;
	for (int i = 31; ~i; i--)
		if ((mx ^ b.a[i]) > mx) mx ^= b.a[i];
	int mx2 = mx;
	for (int i = 0; i <= 31; i++)
		if (b.a[i]) {mx2 ^= b.a[i]; break;}
	cout << mx << ' ' << mx2 << "\n";
	return 0;
}

posted @ 2018-08-14 18:49  谜のNOIP  阅读(139)  评论(0编辑  收藏  举报