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))
代码语言：javascript复制#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;
}
/*
*/
