题目传送门
#include <bits/stdc++.h>
using namespace std;
//异或:不进位的加法
const int N = 110;
int n;
int a[N][N];
int c, r;
void gauss() {
for (c = 0, r = 0; c < n; c++) {
int t = r;
//找到第一个是1的行
for (int i = r; i < n; i++)
if (a[i][c]) {
t = i;
break;
}
//没有找到的话,下一列
if (!a[t][c]) continue;
//交换
for (int i = c; i <= n; i++) swap(a[r][i], a[t][i]);
for (int i = r + 1; i < n; i++)
//是1的才需要消元
if (a[i][c])
for (int j = n; j >= c; j--) a[i][j] ^= a[r][j];
//下一行
r++;
}
if (r < n) {
for (int i = r; i < n; i++)
if (a[i][n]) {
//无解
puts("No solution");
exit(0);
}
//无穷多组解
puts("Multiple sets of solutions");
exit(0);
}
//唯一解,还原成各个方程的解
for (int i = n - 1; i >= 0; i--)
for (int j = i + 1; j < n; j++)
a[i][n] ^= a[i][j] * a[j][n];
}
int main() {
//优化输入
ios::sync_with_stdio(false);
cin >> n;
for (int i = 0; i < n; i++)
for (int j = 0; j < n + 1; j++)
cin >> a[i][j];
//高斯消元
gauss();
//唯一解
for (int i = 0; i < n; i++) cout << a[i][n] << endl;
return 0;
}