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#include <bits/stdc++.h>
using namespace std;
const int N = 11;
int a[N] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
void print() {
int s = 0;
for (int i = 1; i < N; i++) {
s += a[i];
cout << s % 2 << " ";
}
cout << "\n";
int main() {
/*
01矩阵+区间修改+单点查询 <=> 非01矩阵+单点修改+区间前缀和查询
办法：
① 区间[l,r]取反 => 单点修改：a[l]++,a[r+1]++
② 单点查询a[k] => 区间前缀和查询： sum(a[1~k])%2
*/
//将2-4取反
a[2]++, a[4 + 1]--;
print();
//将2-4再取一次反
//将3-4再取一次反
a[3]++, a[4 + 1]--;
return 0;
输出：
0 1 1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 1 1 0 0 0 0 0 0
