solution1(暴力)

暴力蓝桥杯可以过，虽然理论上会超时~

#include<iostream>
using namespace std;
const int maxn = 2010;
int a[maxn][maxn] = {0};//0白棋，1黑棋 
int main(){
	int n, m, x1, x2, y1, y2;
	scanf("%d%d", &n, &m);
	while(m--){
		scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
		for(int i = x1; i <= x2; i++){
			for(int j = y1; j <= y2; j++){
				if(a[i][j]) a[i][j] = 0;
				else a[i][j] = 1;
			}
		}
	}
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			printf("%d", a[i][j]);
		}
		printf("\n");
	}
	return 0;
} 

solution2（差分）

记录操作次数，把对区间加一转化为差分序列的两点操作。
关于数值：偶数次取反为0，奇数次取反为1，可由求余2实现。

#include<iostream>
using namespace std;
const int maxn = 2010;
int a[maxn][maxn] = {0};
int main(){
	int n, m, x1, x2, y1, y2;
	scanf("%d%d", &n, &m);
	while(m--){
		scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
		for(int x = x1; x <= x2; x++){
			a[x][y1]++;
			a[x][y2 + 1]--;
		}
	}
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			a[i][j] += a[i][j - 1];
			printf("%d", a[i][j] % 2);
		}
		printf("\n");
	}
	return 0;
}