[模板] 拉格朗日插值

[题目链接]

        https://www.luogu.org/problemnew/show/P4781

[算法]

       拉格朗日插值即可

       时间复杂度 : O(N ^ 2)

[代码]

        

#include<bits/stdc++.h>
using namespace std;
#define MAXN 2010
const int P = 998244353;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

int n , k;
int X[MAXN] , Y[MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline int exp_mod(int a , int n)
{
        int res = 1 , b = a;
        while (n > 0)
        {
                if (n & 1) res = 1ll * res * b % P;
                b = 1ll * b * b % P;
                n >>= 1;
        }
        return res;
}
inline int inv(int x)
{
        return exp_mod(x , P - 2);
}
inline void update(int &x , int y)
{
        x = (x + y) % P;
        if (x < 0) x += P;
}
inline int Lagrange(int x)
{
        int ret = 0;
        for (int i = 1; i <= n; i++)
        {
                int value = Y[i];        
                for (int j = 1; j <= n; j++)
                {
                        if (i != j)
                                value = 1ll * value * (x - X[j]) % P * inv(((X[i] - X[j]) % P + P) % P) % P;
                }
                update(ret , value);
        }        
        return ret;
}

int main()
{
        
        read(n); read(k);
        for (int i = 1; i <= n; i++)
        {
                read(X[i]);
                read(Y[i]);
        }
        printf("%d
" , Lagrange(k));
        
        return 0;
    
}