洛谷P4781 【模板】拉格朗日插值(拉格朗日插值)

题意

题目链接

Sol

记得NJU有个特别强的ACM队叫拉格朗，总感觉少了什么。。

不说了直接扔公式

\[f(x) = \sum_{i = 1}^n y_i \prod_{j \not = i} \frac{k - x[j]}{x[i] - x[j]}
\]

复杂度\(O(n^2)\)

如果\(x\)的取值是连续的话就前缀积安排一下，复杂度\(O(n)\)

#include<bits/stdc++.h>

using namespace std;

const int MAXN = 1e6 + 10, mod = 998244353;

inline int read() {

    char c = getchar(); int x = 0, f = 1;

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

int N, K, x[MAXN], y[MAXN];

int add(int x, int y) {

	if(x + y < 0) return x + y + mod;

	return x + y >= mod ? x + y - mod : x + y;

}

int mul(int x, int y) {

	return 1ll * x * y % mod;

}

int fp(int a, int p) {

	int base = 1;

	while(p) {

		if(p & 1) base = mul(base, a);

		a = mul(a, a); p >>= 1;

	}

	return base;

}

int Lagelangfuckchazhi(int k) {

	int ans = 0;

	for(int i = 1; i <= N; i++) {

		int up = 1, down = 1;

		for(int j = 1; j <= N; j++) {

			if(i == j) continue;

			up = mul(up, add(k, -x[j]));

			down = mul(down, add(x[i], -x[j]));

		}

		ans = add(ans, mul(y[i], mul(up, fp(down, mod - 2))));

	}

	return ans;

}

int main() {

	N = read(); K = read();

	for(int i = 1; i <= N; i++) x[i] = read(), y[i] = read();

	printf("%d", Lagelangfuckchazhi(K));

    return 0;

}

/*

*/