给出一个长度为nn的序列{ai}\{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，且在这些子段中任取若干子段，它们包含的所有数的异或和不能为00．

n≤105,ai≤109n\le10^5, a_i\le 10^9

CF1101G

Solution

一看到“异或”、“子段”的字眼就不难想到与前缀异或和有关

再仔细想想就会发现题目就是要求最多选出多少个前缀异或和，使得它们线性无关，且sumnsum_n必选，即sumn≠0sum_n\ne 0

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 2e5 + 100;

int N, A[Maxn], Sum[Maxn];

namespace Basis
{
	const int Maxlen = 30;

	int A[Maxlen + 10];

	inline int insert (int x)
	{
		for (int i = Maxlen; ~i; --i)
		{
			if (!(x & (1 << i))) continue;
			if (!A[i])
			{
				for (int j = 0; j < i; ++j) if (x & (1 << j)) x ^= A[j];
				for (int j = i + 1; j <= Maxlen; ++j) if (A[j] & (1 << i)) A[j] ^= x;
				A[i] = x;
				return 1;
			}
			x ^= A[i];
		}
		return 0;
	}
}

inline void Solve ()
{
	for (int i = 1; i <= N; ++i) Sum[i] = Sum[i - 1] ^ A[i];
	if (!Sum[N]) { puts("-1"); return ; }
	
	Basis :: insert(Sum[N]);
	int ans = 1;
	for (int i = 1; i < N; ++i) ans += Basis :: insert (Sum[i]);
	cout<<ans<<endl;
}

inline void Input ()
{
	N = read<int>();
	for (int i = 1; i <= N; ++i) A[i] = read<int>();
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("G.in", "r", stdin);
	freopen("G.out", "w", stdout);
#endif
	Input();
	Solve();
	return 0;
}