\(\text{Note}\)
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\(O(n^2)\)
\(\text{Code}\)
求多项式在某一位置的取值
#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);
}