有nn个未知数x1,x2,...,xnx_1,x_2,...,x_n，给出mm条消息，每条消息选出一些未知数并告诉你他们的和的奇偶性

你的目标是判断每个未知数的奇偶性

如果前kk条消息就可以确定所有未知数的奇偶性，输出kk以及所有未知数的奇偶性，否则输出存在多解

n≤1000,m≤2000n\le1000,m\le2000

Luogu P2447

Solution

可以直接高斯消元或者用线性基解异或方程组

线性基做的话，用bitset存，把常数项也压进去，当作一个新的变量跟着一起异或就行了

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 = 1000 + 100, Maxm = 2000 + 100;

int N, M;

namespace Basis
{
	bitset <Maxn> A[Maxm];

	inline int insert (bitset <Maxn> x)
	{
		for (int i = 0; i < N; ++i)
		{
			if (!x[i]) continue;
			if (!A[i].any())
			{
				for (int j = i + 1; j < N; ++j) if (x[j]) x ^= A[j];
				for (int j = 0; j < i; ++j) if (A[j][i]) A[j] ^= x;
				A[i] = x;
				return 1;
			}
			x ^= A[i];
		}
		return 0;
	}

	inline void query ()
	{
		for (int i = 0; i < N; ++i) A[i][N] ? puts("?y7M#") : puts("Earth");
	}
}

char S[Maxn];
bitset <Maxn> A;

inline void Solve ()
{
	int cnt = 0, pos = 0;
	for (int i = 1; i <= M; ++i)
	{
		scanf("%s", S);
		for (int j = 0; j < N; ++j) A[j] = S[j] - '0';
		int x = read<int>();
		A[N] = x;
		if (Basis :: insert (A)) ++cnt, pos = i;
	}

	if (cnt < N) { puts("Cannot Determine"); return ; }

	cout<<pos<<endl;
	Basis :: query ();
}

inline void Input ()
{
	N = read<int>(), M = read<int>();
}

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