拉格朗日插值模板题 luoguP4871

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拉格朗日插值 (O(n^{2}))

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<queue>
#include<vector>
#include<string>
#include<fstream>

using namespace std;
typedef long long ll;
const int N = 4e6 + 105;
const int mod = 998244353;
const double Pi = acos(- 1.0);
const int INF = 0x3f3f3f3f;

int n, k;
int x[N], y[N];

int sub(const int& a, const int& b){
	int t = a - b;
	return t < 0 ? t + mod : t;
}


int inv(int x){
	int res = 1;
	int p = mod - 2;
	for(; p; p >>= 1, x = 1ll * x * x % mod)
		if(p & 1) res = 1ll * res * x % mod;
	return res;
}


int Lagrange(int n, int x[], int y[], int k){
	int res = 0;
	for(int i = 1; i <= n; ++ i){
		int s1 = 1, s2 = 1;	//s1:分子，s2:分母
		for(int j = 1; j <= n; ++ j){
			if(i != j){
				s1 = 1ll * s1 * sub(k, x[j]) % mod;
				s2 = 1ll * s2 * sub(x[i], x[j]) % mod;
			}
		}
		res = (res + 1ll * y[i] * s1 % mod * inv(s2) % mod) % mod; 
	}
	return res;
}

int main()
{
	scanf("%d%d",&n,&k);
	for(int i = 1; i <= n; ++ i) scanf("%d%d",&x[i],&y[i]);
	printf("%d
", Lagrange(n, x, y, k));
	return 0;
}

在 x 取值连续时的拉格朗日插值法 (O(nlog^{n}))

如果预处理逆元可以(O(n))

int n, k;
int pre[N], suf[N], fac[N];

int qpow(int x, int y){
	int res = 1;
	while(y){
		if(y & 1) res = 1ll * res * x % mod;
		x = 1ll * x * x % mod;
		y >>= 1;
	}
	return res;
}

int Lagrange(int n, int x[], int y[], int k){
	int res = 0;
	pre[0] = suf[n + 1] = fac[0] = 1;
	for(int i = 1; i <= n; ++ i) pre[i] = 1ll * pre[i - 1] * (k - i) % mod;
	for(int i = n; i >= 1; -- i) suf[i] = 1ll * suf[i + 1] * (k - i) % mod;
	for(int i = 1; i <= n; ++ i) fac[i] = 1ll * fac[i - 1] * i % mod;
	for(int i = 1; i <= n; ++ i){
		int s1 = (pre[i - 1] * suf[i + 1]) % mod;
		int s2 = (fac[i - 1] * fac[n - i]) % mod;
		if((n - i) & 1) s2 = (mod - s2) % mod;
		res = (1ll * res + 1ll * y[i] * s1 % mod * qpow(s2, mod - 2) % mod) % mod;
	}
	return res;
}