考虑差分，使得 $b_i = a_i - a_{i - 1}$，$b_1 = a_1 - a_n$。

但此时 $\displaystyle \sum _{i = 1} ^n b_i = 0$ 导致丢失了一部分信息，于是我们再给 $b_1$ 减去 $a_1$，此时 $b_1 = -a_n$。

有了 $a_i$ 和 $b_i = a_i - a_{i - 1}$，我们就可以递推得到 $a_{i - 1} = a_i - b_i$，至此问题得到解决。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin >> n;
    cout << 1 << ' ' << 1 << endl;
    for (int i = 1; i <= n; i++)
        cout << i << ' ';
    cout << n + 1 << endl;
    for (int i = 1; i <= n - 1; i++)
        cout << i + 1 << ' ';
    cout << 1 << ' ' << n + 1 << endl;
    return 0;
}

/* ATTENTION!!! THIS IS THE BARRIER!!! */

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 12;
long long a[N], b[N];
int main()
{
    int n, X0 = 1, Y0 = 1;
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> b[i];
    a[n] = -b[1];
    for (int i = n - 1; i >= 1; i--)
        a[i] = a[i + 1] - b[i + 1];
    for (int i = 1; i <= n - 1; i++)
        cout << a[i] << ' ';
    cout << a[n] << endl;
    return 0;
}