P4781 【模板】拉格朗日插值

传送门

把公式实现一下即可：
当$x$连续的时候可以优化为$O(N)$。

// Problem: P4781 【模板】拉格朗日插值
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P4781
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;

//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;

const int N=1000010,mod=998244353,INF=0x3f3f3f3f;
const double eps=1e-6;

LL n,k;
pair<LL,LL> p[N];

LL qmi(LL a,LL b) {
	LL ans=1;
	a%=mod; a+=mod; a%=mod;
	while(b) {
		if(b&1) ans=ans*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ans%mod;
}

int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);
	
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++) scanf("%lld%lld",&p[i].X,&p[i].Y);
	LL ans=0;
	for(int i=1;i<=n;i++) {
		LL now1=p[i].Y,now2=1;
		for(int j=1;j<=n;j++) {
			if(i==j) continue;
			now1*=((k-p[j].X+mod)%mod); now1%=mod;
			now2*=p[i].X-p[j].X; now2%=mod;
		}
		now2=qmi(now2,mod-2);
		ans+=now1*now2%mod; ans%=mod;
	}
	printf("%lld\n",ans%mod);
	
		
	



	return 0;
}
/*

*/