题目给出了明确的提示“把 \(x_i,y_i\) 看成一个整体组来考虑”，于是容易想到 Lagrange 插值插出 \(f\) 函数，根据经典结论可知插出的函数恰有 \(n\) 项，然后再编一个大于 \(\max y_i\) 的模数（例如 \(10^9+7\)）在模意义下插值。此时 \(f\) 函数每一项的系数都不超过 \(2^{30}\)，直接转成二进制扔到 \(s\) 串里传输即可。
而对于逆变换，直接套 Lagrange 插值板子即可。
namespace Loyalty
{inline void init() { }int x[N], y[N], a[N];char s[N];inline void dasongsong(){int n;cin >> n;for (int i = 1; i <= n; ++i)cin >> x[i] >> y[i];for (int i = 1; i <= n; ++i)for (int j = 1; j <= n; ++j){int numerator = y[j], denominator = 1;for (int k = 1; k <= n; ++k)if (k != j){numerator = numerator * (i - x[k] + mod) % mod;denominator = denominator * (x[j] - x[k] + mod) % mod;}a[i] = (a[i] + numerator * inversion(denominator) % mod) % mod;}cout << n * 30 << '\n';for (int i = 1; i <= n; ++i)for (int j = 0; j < 30; ++j)cout << (a[i] >> j & 1);cout << '\n';}inline void ruahao(){int n, q, k;cin >> n >> q >> k;scanf("%s", s + 1);for (int i = 1; i <= n; ++i){int l = i * 30 - 29, r = i * 30;a[i] = 0;for (int j = r; j >= l; --j)a[i] = a[i] << 1 | (s[j] & 1);}while (q--){int k;cin >> k;int sum = 0;for (int i = 1; i <= n; ++i){int numerator = a[i], denominator = 1;for (int j = 1; j <= n; ++j)if (j != i){numerator = numerator * (k - j + mod) % mod;denominator = denominator * (i - j + mod) % mod;}sum = (sum + numerator * inversion(denominator) % mod) % mod;}cout << sum << '\n';}}inline void main([[maybe_unused]] int _ca, [[maybe_unused]] int atc){int ty;cin >> ty;if (ty == 1)dasongsong();elseruahao();}
}