拉格朗日插值

 1 using System;
 2 using System.Collections.Generic;
 3 using System.Linq;
 4 using System.Text;
 5 namespace CsTest
 6 {   //可以确定的是n个插值点的拉格朗日多项式的最大次数为n
 7     public class Ploy   //多项式
 8     {
 9         public double[] a;//系数,指数即索引值
10         public Ploy(int n = 1)
11         {
12             a = new double[n]; //当未给数组中的每一项赋值时，其值为0
13         }
14         public static Ploy operator +(Ploy a1, Ploy b1)//多项式a1和b1的项数相同
15         {
16             Ploy p = new Ploy(a1.a.Length);
17             for (int i = 0; i < a1.a.Length; i++)
18             {
19                 p.a[i] = a1.a[i] + b1.a[i];//系数相加
20             }
21             return p;
22         }
23         public static Ploy operator *(Ploy a1, Ploy b1)
24         {
25             Ploy p = new Ploy(a1.a.Length);
26             for (int i = 0; i < a1.a.Length; i++)
27             {
28                 if (a1.a[i] == 0)
29                 {
30                     continue;  //此项系数为0
31                 }
32                 for (int j = 0; j < a1.a.Length; j++)
33                 {
34                     if (b1.a[j] == 0)//此项系数为0
35                     {
36                         continue;
37                     }
38                     if (i + j > a1.a.Length)
39                     {
40                         Console.WriteLine("运算出现错误");
41                         break;
42                     }
43                     p.a[i + j] += a1.a[i] * b1.a[j];
44                 }
45             }
46             return p;
47         }
48     }
49     class Program
50     {
51         static void Main(string[] args)
52         {
53             int n = 4;
54             //获取文件中的x值得数量并赋给n
55             //int[] X = new int[n];  //横坐标数组
56             //int[] Y = new int[n];  //纵坐标数组
57             int[] X = { 0, 1, 2, 3 };
58             int[] Y = { 2, -1, 4, 3 };
59             Ploy lag = new Ploy(n);//拉格朗日多项式
60             for (int i = 0; i < n; i++)
61             {
62                 Ploy py = new Ploy(X.Length);
63                 py.a[0] = Y[i];
64                 Ploy px = new Ploy(X.Length);
65                 px.a[0] = 1;
66                 for (int k = 0; k < X.Length; k++)
67                 {
68                     if (i != k)
69                     {
70                         px.a[0] *= X[i] - X[k];
71                     }
72                 }
73                 py.a[0] /= px.a[0];
74                 Ploy mul = new Ploy(n);
75                 mul.a[0] = 1;
76                 for (int j = 0; j < n; j++)
77                 {
78                     Ploy temp=new Ploy(n);
79                     temp.a[1]=1;
80                     Ploy temp2=new Ploy(n);
81                     if(i==j)
82                     {
83                         continue;
84                     }
85                     temp2.a[0]=-X[j];
86                     mul *= temp + temp2;
87                 }
88                 lag += py * mul;
89             }
90             //输出拉格朗日多项式
91             for(int i=n-1;i>=0;i--)
92             {
93                 if (lag.a[i] == 0)
94                     continue;
95                 Console.Write("{0}x^{1}+",lag.a[i],i);
96             }
97         }
98     }
99 }