mt19937 umi(8300901);

int main() {
	int n = 300000, m = 200 * (__lg(n) + 1000);
	out(n), outln(m);
	for(int i = 1; i <= n; ++i) printf("%d%c", 0721, " \n"[i == n]);
	for(int i = 1; i <= 200; ++i) {
		for(int j = 1; j * 2 <= n; j <<= 1) 
			printf("4 1 %d %d %d\n", j, j + 1, j << 1);
		for(int j = 1; j <= 1000; ++j) {
			int a = umi() % n + 1, b = umi() % n + 1;
			if(a > b) swap(a, b);
			printf("2 %d %d %d\n", a, b, int(umi() % 0721) + 1);
		}
	}
	return 0;
}

击落题解：

• https://www.luogu.com.cn/article/0wj7y2vr

• https://www.luogu.com.cn/article/s5p57vt4

• https://www.luogu.com.cn/article/7w9is0o5

@Mooncrying @123456xwd @cheng2010

请求添加数据并撤下题解。