拉格朗日插值

luogu 4781

金牌导航 拉格朗日插值-1

题目大意

给出n个点，让你确定经过这n个点且不超过n-1次的方程，现在给出k，让你求出其函数值

样例#1

输入样例#1

3 100
1 4
2 9
3 16

输出样例#1

10201

样例#2

输入样例#2

3 100
1 1
2 2
3 3

输出样例#2

100

数据范围

$1⩽n⩽2\times 10^3,1⩽x_i,y_i,k⩽998244353$

解题思路

直接代入拉格朗日插值公式即可（如下）
$$f(k)=\sum _{i=0}^n{y}_i\prod _{i\ne j}\frac{k-x_j}{x_i-x_j}$$

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 2010
#define wyc 998244353
using namespace std;
ll n, m, k, g1, g2, ans, x[N], y[N];
ll Counting(ll x, ll y)
{ll s = 1;while(y){if (y&1) s = s * x % wyc;x = x * x % wyc;y >>= 1;}return s;
}
int main()
{scanf("%lld%lld", &n, &k);for (ll i = 1; i <= n; ++i)scanf("%lld%lld", &x[i], &y[i]);for (ll i = 1; i <= n; ++i){g1 = 1;g2 = 1;for (ll j = 1; j <= n; ++j)//计算公式if (i != j){g1 = g1 * (k - x[j]) % wyc;g2 = g2 * (x[i] - x[j]) % wyc;}ans = (ans + y[i] * g1 % wyc * Counting(g2, wyc - 2) % wyc + wyc) % wyc;}printf("%lld", ans);return 0;
}