P4781 【模板】拉格朗日插值

Note$\text{Note}$$\text{Note}$#

考虑构造函数$f$

$$f(k)=\sum_{i=1}^{n} y_i\prod_{j\neq i }\frac{k-x_j}{x_i-x_j}$$

求多项式在某一位置的取值#

对于求$f(k)$，将$k$带入即可，时间复杂度为$O(n^2)$

求多项式每一项的系数#

考虑将$f$展开，时间复杂度的瓶颈在于求$\prod_{j\neq i }(x-x_j)$
先预处理出$S = \prod (x-x_j)$
所以$\prod_{j\neq i }(x-x_j) = \frac{S}{x - x_i}$，此时用朴素的多项式除法即可，时间复杂度仍为$O(n^2)$

Code$\text{Code}$$\text{Code}$#

求多项式在某一位置的取值#

Copy
#include<cstdio>
#define LL long long
using namespace std;
const int P = 998244353;
int n; LL m,x[2005],y[2005];

LL fpow(int x,LL y)
{
	LL res = 1;
	for (; x; x >>= 1,y = y * y % P)
		if (x & 1) res = res * y % P;
	return res;
}
int main()
{
	scanf("%d%lld",&n,&m);
	for (int i = 1; i <= n; i++) scanf("%lld%lld",&x[i],&y[i]);
	LL ans = 0;
	for (int i = 1; i <= n; i++)
	{
		LL p = y[i],q = 1;
		for (int j = 1; j <= n; j++) 
			if (j != i) p = p * (m - x[j]) % P,q = q * (x[i] - x[j]) % P;
		p = (p + P) % P,q = (q + P) % P;
		ans = (ans + p * fpow(P - 2,q) % P) % P;
	}
	printf("%lld\n",ans);
 }