http://codeforces.com/problemset/problem/405/C

Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.

The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.

Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:

The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.

However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:

  1. given a row index i, flip all the values in the i-th row in A;
  2. given a column index i, flip all the values in the i-th column in A;
  3. find the unusual square of A.

To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.

Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?

Input

The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij(0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.

The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:

  • 1 i — flip the values of the i-th row;
  • 2 i — flip the values of the i-th column;
  • 3 — output the unusual square of A.

Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.

Output

Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.

 

题意：给定一个01方阵，有修改操作：第i行/列全部元素反转，询问Σ  第i行向量与第i列向量的点积（数量积），i∈[1,n]，答案模2。

思路：答案要模2，那么答案只有0，1两种。其实答案就相当于Σa(i,j)*a(j,i),i,j∈[1,n]。

假如要反转第i行，考虑元素a(i,j)，i和j不同，a(i,j)*a(j,i)要么是1，要么是0，是确定的值，这一对向量分别在作为第i/j组向量时答案相同，模2后都是0。

也就是说，如果i，j不同，那么反转a(i,j)对答案的贡献是0。

如果i和j相同，反转了a(i,i)，对答案的贡献是1，相当于答案取反。

那么，方法就是：先求初始ans，然后每反转一次，ans=!ans。

#include<bits/stdc++.h>
using namespace std;
#define maxn 1005

int n,m,ans;
int a[maxn][maxn];

int main()
{
	//freopen("input.in","r",stdin);
	cin>>n;
	for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)scanf("%d",&a[i][j]);
	for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)ans=(ans+a[i][j]*a[j][i])&1;
	cin>>m;
	while(m--)
	{
		int x;
		scanf("%d",&x);
		if(x==3)printf("%d",ans);
		else scanf("%d",&x),ans=!ans;
	}
	return 0;
}