#include <iostream>
using namespace std;
#include <vector>
inline double newton_solution(double x[], double y[], int n, double num, int newton_time)
{
    vector<vector<double> > table(n + 1);
    for (int i = 0; i <= n; i++) {
        table[i].resize(n + 1);
    }
    for (int i = 0; i <= n; i++) table[0][i] = y[i];
    for (int i = 1; i <= n; i++) {
        for (int j = i; j <= n; j++) {
            table[i][j] = (table[i - 1][j] - table[i - 1][j - 1]) / (x[j] - x[j - i]);
        }
    }

    double res = 0.0;
    for (int i = 0; i <= newton_time; i++) {
        double temp = table[i][i];
        for (int j = 0; j < i; j++) {
            temp *= num - x[j];
        }
        res += temp;
    }

    return res;

}
int main(int argc, char const *argv[])
{
    int n = 0;
    cout << "插值节点个数-1:";
    cin >> n;
    double x[n + 1], y[n + 1];
    cout << "\n请输入x[i]:";
    for (int i = 0; i <= n; i++) {
        cin >> x[i];
    }
    cout << "\n请输入y[i]:";
    for (int i = 0; i <= n; i++) {
        cin >> y[i];
    }
    double num = 0;
    cout << "\n请输入要求的点的x:";
    cin >> num;
    cout << "\n请输入所求的插值多项式次数:";
    double newton_time = 0;
    cin >> newton_time;
    cout << newton_solution(x, y, n, num, newton_time) << endl;
    return 0;