[洛谷P4781]【模板】拉格朗日插值

题目大意：给你$n(n\leqslant2000)$个点，要你求$n-1$次经过这$n$个点的多项式在$k$处的值

题解：$Lagrange$插值：
$$
f_x=\sum\limits_{i=1}^ky_i\prod\limits_{j=1,j\not=i}^k\dfrac{x-x_j}{x_i-x_j}
$$
卡点：无

 

C++ Code：

#include <algorithm>
#include <cstdio>
#define maxn 2010
const int mod = 998244353;
namespace Math {
	inline int pw(int base, int p) {
		static int res;
		for (res = 1; p; p >>= 1, base = static_cast<long long> (base) * base % mod) if (p & 1) res = static_cast<long long> (res) * base % mod;
		return res;
	}
	inline int inv(int x) { return pw(x, mod - 2); }
}
inline void reduce(int &x) { x += x >> 31 & mod; }
inline int getreduce(int x) { return x + (x >> 31 & mod); }

int n, k, ans;
int x[maxn], y[maxn];
int main() {
	scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; ++i) scanf("%d%d", x + i, y + i);
	for (int i = 1; i <= n; ++i) {
		long long a = y[i], b = 1;
		for (int j = 1; j <= n; ++j) if (i != j) {
			a = a * getreduce(k - x[j]) % mod;
			b = b * getreduce(x[i] - x[j]) % mod;
		}
		reduce(ans += a * Math::inv(b) % mod - mod);
	}
	printf("%d\n", ans);
	return 0;
}