注意到完全 DAG 中 $c_0 = 2^{n - 2}$，考虑造出 $n = 33$ 的完全 DAG 后二进制拆分。

给 $c$ 数组排序，让 $i + 1$ 号点出发的路径有 $c_i$ 条（$c_{k+1} = 0$），连边 $(1, 2), (2, 3), \ldots, (k, k + 1)$，一共需要 $32 + 1 + 33 = 66$ 个点，至此问题得到解决。

#include <bits/stdc++.h>
using namespace std;
deque < pair <int,int> > q;
void add(int x, int y)
{
	q.push_back(make_pair(x, y));
}
void insert(int x, int y)
{
	q.push_front(make_pair(x ,y));
}
void print()
{
	cout << 66 << ' ' << q.size() << endl;
	while (!q.empty())
	{
		cout << q.front().first << ' ' << q.front().second << endl;
		q.pop_front();
	}
}
long long c[42];
int main()
{
	int k, n;
	cin >> k >> n;
	for (int i = 0; i <= k; i++)
		cin >> c[i];
	for (int i = k + 1; i <= 32; i++)
		c[i] = c[0];
	for (int i = 0; i <= 32; i++)
		c[i] = c[i] * 1000 + i;
	c[0] += 100;
	for (int i = 34; i <= 65; i++)
		for (int j = i + 1; j <= 66; j++)
			add(i, j);
	sort(c, c + 33);
	for (int i = 0; i <= 32; i++)
	{
		long long d=c[i] / 1000;
		if (i) d -= c[i - 1] / 1000;
		for (int j = 0; j <= 31; j++)
			if(d & (1ll << j)) add(i + 1, 65 - j);
	}
	for (int i = 32; i >= 1; i--)
		for (int j = 0; j <= 32; j++)
			if (c[j] % 1000 == i) insert(j + 1, j + 2);
	print();
	return 0;
}