#include <bits/stdc++.h>
using namespace std;
int a[110][110], n;

int main() {
	cin >> n;
	n = pow(2, n);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			if (i == 1 || j == i) {
				a[i][j] = 1;
			} else {
				a[i][j] = a[i - 1][j - 1] + a[i - 1][j];
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int k = n - i; k >= 1; k--) {
			cout << 0 << " ";
		}

		for (int j = 1; j <= i; j++) {
			if (a[i][j] % 2 == 0 ) {
				cout << 0 << " ";
			} else {
				cout << 1 << " ";
			}
		}
		cout << endl;
	}
	return 0;
}

这道题跟杨辉三角很像，用杨辉三角代码改的，对了一部分