拉格朗日插值

拉格朗日插值

简要阐述

结论给定$n+1$个点最多可以得到一个$n$次多项式的表达式，并且$f(x)={\sum }_{i=1}^n{y}_i\prod _{j\nmid i}\frac{x-x_j}{x_i-y_j}$我们随便往里带入一个$x_i,y_j$可以得到等式是成立的，所以我们只要套模板即可求得某个函数的值。

特殊情况

当$x$的取值是连续的，我们还可以得到一个更优的算法。

$pre[i]=(x-x_1)(x-x_2)(x-x_3)\dots (x-x_{i-1})(x-x_i),suc[i]=(x-x_i)(x-x_{i+1})(x-x_{i+2})\dots (x-x_{n-1})(x-x_n)$

所以上面式子的分母可以写成$pre[i-1]\times suc[i+1]$，同样的分子写成$(-1{)}^{n-i}(i-1)!(n-i)!$，这个时候提前处理好阶乘逆元，前缀积，后缀积即可达到$O(n)$来求解了。

两个模板题

P4781 【模板】拉格朗日插值

/* Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h> using namespace std; typedef long long ll; const int inf = 0x3f3f3f3f; const double eps = 1e-7; const int N = 2e3 + 10, mod = 998244353; ll x[N], y[N], ans, s1, s2, n, k; ll quick_pow(ll a, int n) { ll ans = 1; while(n) { if(n & 1) ans = ans * a % mod; a = a * a % mod; n >>= 1; } return ans; } ll inv(ll x) { return quick_pow(x, mod - 2); } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); scanf("%lld %lld", &n, &k); for(int i = 1; i <= n; i++) { scanf("%lld %lld", &x[i], &y[i]); } for(int i = 1; i <= n; i++) { s1 = y[i], s2 = 1; for(int j = 1; j <= n; j++) { if(i == j) continue; s1 = (s1 * (k - x[j]) % mod + mod) % mod, s2 = (s2 * (x[i] - x[j]) % mod + mod) % mod; } ans = (ans + s1 * inv(s2) % mod) % mod; } printf("%lld\n", ans); return 0; }

$\sum _{i=1}^n{i}^k$

/* Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h> using namespace std; typedef long long ll; const int inf = 0x3f3f3f3f; const double eps = 1e-7; const int N = 1e6 + 10, mod = 1e9 + 7; ll fac[N], pre[N], suc[N], inv[N], prime[N], sum[N], n, k, cnt; bool st[N]; ll quick_pow(ll a, int n) { ll ans = 1; while(n) { if(n & 1) ans = ans * a % mod; a = a * a % mod; n >>= 1; } return ans; } void init() { sum[1] = 1; for(int i = 2; i < N; i++) { if(!st[i]) { prime[cnt++] = i; sum[i] = quick_pow(i, k); } for(int j = 0; j < cnt && i * prime[j] < N; j++) { st[i * prime[j]] = 1; sum[i * prime[j]] = 1ll * sum[i] * sum[prime[j]] % mod; if(i % prime[j] == 0) break; } } fac[0] = inv[0] = 1; for(int i = 1; i < N; i++) { sum[i] = (sum[i] + sum[i - 1]) % mod; fac[i] = 1ll * fac[i - 1] * i % mod; } inv[N - 1] = quick_pow(fac[N - 1], mod - 2); for(int i = N - 2; i >= 1; i--) { inv[i] = 1ll * inv[i + 1] * (i + 1) % mod; } } ll solve(ll n, int k) { ll ans = 0; init(); pre[0] = suc[k + 3] = 1; for(int i = 1; i <= k + 2; i++) pre[i] = 1ll * pre[i - 1] * (n - i) % mod; for(int i = k + 2; i >= 1; i--) suc[i] = 1ll * suc[i + 1] * (n - i) % mod; for(int i = 1; i <= k + 2; i++) { ll a = 1ll * pre[i - 1] * suc[i + 1] % mod, b = 1ll * inv[i - 1] * inv[k + 2 - i] % mod; if((k + 2 - i) & 1) b *= -1; ans = ((ans + 1ll * sum[i] * a % mod * b % mod) % mod + mod) % mod; } return ans; } int main() { // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); scanf("%lld %lld", &n, &k); printf("%lld\n", solve(n, k)); return 0; }