【线性基】Codeforces-1101G. (Zero XOR Subset)-less

题目链接

time limit per test 2 seconds

memory limit per test 256 megabytes

input standard input

output standard output

Description

You are given an array $a_1,a_2,…,a_n$ of integer numbers.

Your task is to divide the array into the maximum number of segments in such a way that:

• each element is contained in exactly one segment;
• each segment contains at least one element;
• there doesn't exist a non-empty subset of segments such that bitwise XOR of the numbers from them is equal to $0$.

Print the maximum number of segments the array can be divided into. Print $-1$ if no suitable division exists.

Input

The first line contains a single integer $n(1≤n≤2⋅10^5)$ — the size of the array.

The second line contains nn integers$a_1,a_2,…,a_n (0≤a_i≤10^9)$.

Output

Print the maximum number of segments the array can be divided into while following the given constraints. Print $-1$ if no suitable division exists.

Examples

input

4
5 5 7 2

output

2

input

3
1 2 3

output

-1

input

3
3 1 10

output

3

Note

In the first example $2$ is the maximum number. If you divide the array into$\{[5],[5,7,2]\}$, the XOR value of the subset of only the second segment is $5⊕7⊕2=0$,$\{[5,5],[7,2]\}$ has the value of the subset of only the first segment being $5⊕5=0$. However, $\{[5,5,7],[2]\}$ will lead to subsets $\{[5,5,7]\}$ of XOR $7$, $\{[2]\}$ of XOR $2$ and $\{[5,5,7],[2]\}$ of XOR $5⊕5⊕7⊕2=5$.

Let's take a look at some division on $3$ segments — $\{[5],[5,7],[2]\}$. It will produce subsets:

• $\{[5]\}$, XOR $5$;
• $\{[5,7]\}$, XOR $2$;
• $\{[5],[5,7]\}$, XOR $7$;
• $\{[2]\}$, XOR $2$;
• $\{[5],[2]\}$, XOR $7$;
• $\{[5,7],[2]\}$, XOR $0$;
• $\{[5],[5,7],[2]\}$, XOR $5$;

As you can see, subset $\{[5,7],[2]\}$ has its XOR equal to $0$, which is unacceptable. You can check that for other divisions of size $3$ or $4$, non-empty subset with 00 XOR always exists.

The second example has no suitable divisions.

The third example array can be divided into $\{[3],[1],[10]\}$. No subset of these segments has its XOR equal to $0$.

题意

给出一个长度为$n$的序列$\{a_i\}$，试将其划分为尽可能多的非空子段，满足

每一个元素出现且仅出现在其中一个子段中，每个段至少包含一个元素，不存在段组成的非空子集使得这些段的数的异或和为$0$．$n(1≤n≤2∗10^5),a_i(0≤a_i≤109)$

思路

把这个数组分段为 $(a_1...a_{x1})(a_{x1+1}...a_{x2})...(...a_n)$

每个段都有一个异或值，每一段异或起来又可以看成是它的两个前缀异或和的异或值

即第一段$S(x_1)$^$S(0)$ ，第二段$S(x_2)$^$S(x_1)$...

又有要求：不存在段组成的非空子集使得这些段的数的异或和为$0$

即$S(0)$~$S(n)$里面取偶数个异或出来的值不能为$0$，也就意味着取出来的数必然是线性无关的，实际上就是求一个极大线性无关组，做前缀和的线性基。

这个线性基的大小就是最多的段数。

有一个特殊情况，$S(n) = 0$时，不管怎么做都不可能取到满足条件的段，所有的异或起来就会是$0$。

AC代码

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define inf 0x3f3f3f3f
#define PII pair<int,int>
#define endl '\n'
const int N = 2e5 + 10;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);
typedef long long ll;
int t, n, m;
int arr[N], base[65];//最高位为第i位的基
int ans;
void insert(int x) {
	for (int i = 60; i >= 0; i--) {
		if (!(x >> i)) {
			continue;
		}
		if (base[i] == 0) {
			base[i] = x;
			break;
		}
		else x ^= base[i];
	}
	if (x) ans++;
	return;
}
void init() {
	for (int i = 0; i <= 64; i++) {
		base[i] = 0;
	}
	for (int i = 0; i <= n; i++) {
		arr[i] = 0;
	}
	ans = 0;
	return;
}
void solve() {
	init();
	cin >> arr[1];
	insert(arr[1]);
	int spj = arr[1];
	for (int i = 2; i <= n; i++) {
		cin >> arr[i];
		spj ^= arr[i];
		insert(arr[i]);
	}
	if (spj == 0) cout << -1 << endl;
	else cout << ans << endl;
	return;
}
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	while (cin >> n) {
		solve();
	}
	return 0;
}