计算机数值计算与方法-二牛顿插值

最新推荐文章于 2024-07-06 09:35:21 发布

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本文介绍了一种使用C++实现的牛顿插值法算法。该算法通过输入一系列坐标点和目标点，找到最接近目标点的数据集，并计算目标点处的函数值。文章详细展示了如何构建差商表并利用其进行插值计算。

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代码

#include<iostream>


using namespace std;


double Newton(double x[], double y[], double t, int n)
{

    double *x2 = new double [n];
    double *y2 = new double [n];
    double *Separated = new double[((1 + n)*n) / 2];
    int dot=0;
    int i;
    for ( dot; t>x[dot]; dot++);//find t position
    cout << "find out dot:" << dot << endl;
    if (dot <= n / 2) {
        cout << "1.find dot should in";
        for (int j = 0; j<n; j++)
            cout << j << " ";
        cout << endl;
        i = 0;
        for (; i<n; i++)
        {
            x2[i] = x[i];
            y2[i] = y[i];
        }
    }
    else if (dot >= 20 - n / 2 + 1)
    {
        cout << "2.find dot should in";
        for (int j = 20 - n; j<20; j++)
            cout << j << " ";
        cout << endl;
        i = 20 - n ;
        int q = 0;
        for (i; i<20; i++)
        {
            x2[q] = x[i];
            y2[q] = y[i];
            q++;
        }
    }
    else {
        cout << "3.find dot should in";
        for (int j = dot - 4; j<dot+4; j++)
            cout << j << " ";
        cout << endl;
        i = dot - 4;
        int q = 0;
        for (i; i<dot+4; i++)
        {
            x2[q] = x[i];
            y2[q] = y[i];
            q++;
        }
    }
    static int num = 0;
    cout << x2[0] << "\t\t" << y2[0] << "\t\t" << endl;
    for (int i = 1; i < n; i++)
    {
        cout << x2[i] << "\t\t" << y2[i] << "\t\t";
        if(i >= 1)
        {
            Separated[num] = (y2[i]-y2[i-1])/(x2[i]-x2[i-1]);
            cout << Separated[num] << "["<<num<<"]\t\t";
            num++;
        }
        if (i >= 2)
        {
            Separated[num] = (Separated[num - 1] - Separated[num - i]) / (y2[i] - y2[i - 1]);
            cout << Separated[num] << "[" << num << "] \t\t";
            num++;
        }
        while (i >= 3)
        {
            for (int j = 1; j <= i - 2; j++)
            {
                Separated[num] = (Separated[num - j] - Separated[num - i]) / (Separated[num - j - 1] - Separated[num - i - 1]);
                cout << Separated[num] << "[" << num << "]\t\t";
                num++;
            }
            break;

        }
        cout << endl;
    }
    double fx = x2[0];
    double w;
    double q = 1;
    int n1=0;
    for (int i = 0; i < n-1; i++)
    {
        w = t - x[i];
        q = q*w;
        fx = fx + q*Separated[n1];
        cout <<"N1="<<n1<< "\t\toutput:fx=" << fx << endl;
        n1 = n1 + 2 + i;
    }
    return 0;
}           
int main()
{
    double x[] = { 0.10,0.15,0.20,0.25,0.30,0.35,0.40,0.45,0.50,0.55,0.60,0.65,0.70,0.75,0.80,0.85,0.90,0.95,1.00,1.05 };
    double y[] = { 0.1103329,0.1736223,0.2426552,0.3176729,0.3989105,0.4865951,0.5809439,0.6821617,0.7904390,0.9059492,
                   1.0288456,1.1592592,1.2972951,1.4430292,1.5965053,1.7577308,1.9266733,2.1032563,2.2873552,2.4787929 };
    int n = 8;
    double t;
    cout << "请输入待求点" << endl;
    cin >> t;
    Newton(x, y, t, n);
    system("pause");
        return 0;
}

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