Biegral Blog

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;
namespace CsTest
{   //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
    public class Ploy   //多项式
    {
        public double[] a;//系数,指数即索引值
        public Ploy(int n = 1)
        {
            a = new double[n]; //当未给数组中的每一项赋值时，其值为0
        }
        public static Ploy operator +(Ploy a1, Ploy b1)//多项式a1和b1的项数相同
        {
            Ploy p = new Ploy(a1.a.Length);
            for (int i = 0; i < a1.a.Length; i++)
            {
                p.a[i] = a1.a[i] + b1.a[i];//系数相加
            }
            return p;
        }
        public static Ploy operator *(Ploy a1, Ploy b1)
        {
            Ploy p = new Ploy(a1.a.Length);
            for (int i = 0; i < a1.a.Length; i++)
            {
                if (a1.a[i] == 0)
                {
                    continue;  //此项系数为0
                }
                for (int j = 0; j < a1.a.Length; j++)
                {
                    if (b1.a[j] == 0)//此项系数为0
                    {
                        continue;
                    }
                    if (i + j > a1.a.Length)
                    {
                        Console.WriteLine("运算出现错误");
                        break;
                    }
                    p.a[i + j] += a1.a[i] * b1.a[j];
                }
            }
            return p;
        }
    }
    class Program
    {
        static void Lagrange(int[] X,int[]Y,out Ploy lag)
        {
            int n = 0;
            n = X.Length;
            lag = new Ploy(n);//拉格朗日多项式
            for (int i = 0; i < n; i++)
            {
                Ploy py = new Ploy(X.Length);
                py.a[0] = Y[i];
                Ploy px = new Ploy(X.Length);
                px.a[0] = 1;
                for (int k = 0; k < X.Length; k++)
                {
                    if (i != k)
                    {
                        px.a[0] *= X[i] - X[k];
                    }
                }
                py.a[0] /= px.a[0];
                Ploy mul = new Ploy(n);
                mul.a[0] = 1;
                for (int j = 0; j < n; j++)
                {
                    Ploy temp = new Ploy(n);
                    temp.a[1] = 1;
                    Ploy temp2 = new Ploy(n);
                    if (i == j)
                    {
                        continue;
                    }
                    temp2.a[0] = -X[j];
                    mul *= temp + temp2;
                }
                lag += py * mul;
            }
        }
        static void Main(string[] args)
        {
            int n =4;//没选取n个节点进行一次拉格朗日运算
            string Inpath = @"E://360壁纸//color.txt";//原数据文件路径
            string Outpath = @"F://Rout.txt";//输出数据文件
            StreamReader sr = new StreamReader(Inpath, Encoding.Default);
            StreamWriter sw = new StreamWriter(Outpath, true);
            string temps;
           // ArrayList aL = new ArrayList();//将节点存入数组
            while ((temps = sr.ReadLine()) != null)
            {
                int Length=0;
                string[] s = temps.Split(new char[] { ' ' }, StringSplitOptions.RemoveEmptyEntries);//根据空格区分数据
                Length = s.Length; //获取一行有多少节点
                int tempInt;
                for (int i = 0; i < Length; i++)
                {
                    tempInt = Convert.ToInt32(s[i]);  //string转换成int,基本数据类型的转换
                 //   aL.Add(tempInt);
                }
                int[] Z = new int[Length * 2 - 1];//存放当前行所做拉格朗日分段运算的值
                for (int i = 0; i < Length-n+1; i++)//对当前行进行分段拉格朗日运算
                {
                    int[] X = new int[n];
                    for (int j = 0; j < n; j++)
                    {
                        X[j] = j;
                    }
                    int[] Y = new int[n];
                    for (int k = 0; k < n; k++)
                    {
                        Y[k] = Convert.ToInt32(s[i + k]);
                    }
                    Ploy lag;
                    Program.Lagrange(X, Y, out lag);
                    int[] r = new int[n * 2 -1]; 
                    for (int m = 0; m < n; m++)//带入横坐标求值
                    {
                        double xhalf=Convert.ToDouble(m)+0.5;
                        double result=0;
                        for(int t=n-1;t>=0;t--)//求对应于result的拉格朗日值
                        {
                            result+=lag.a[t]*Math.Pow(xhalf,t);
                        }
                        r[m * 2] = Y[m];
                        if (result > 255)
                            result = 255;
                        if (m != n - 1)
                        {
                            r[m * 2 + 1] = Convert.ToInt32(result);
                        }
                        
                    }
                    if (i == 0) //说明是第一个指定n个节点所做的拉格朗日运算
                    {
                        for (int m = 0; m < r.Length; m++)
                        {
                            Z[i + m] = r[m];
                        }
                    }
                    else    //倒数第2个之前的值与Z中对应的值求平均
                    {
                        for (int m = 0; m < r.Length - 2; m++)
                        {
                            Z[i * 2+m] = (Z[i * 2+m] + r[m]) / 2;
                        }
                        Z[i*2+r.Length - 2] = r[r.Length - 2];
                        Z[i*2+r.Length - 1] = r[r.Length - 1];
                    }
                }
                //将Z中的值输出到文件中
                for (int i = 0; i < Z.Length; i++)
                {
                    sw.Write(Z[i].ToString()+" ");
                }
                sw.WriteLine();
            }
            sw.Flush();
            sw.Close();
            sr.Close();
            //获取文件中的x值得数量并赋给n
            //int[] X = new int[n];  //横坐标数组
            //int[] Y = new int[n];  //纵坐标数组
            //int[] X = { 0, 1, 2, 3 };
            //int[] Y = { 2, -1, 4, 3 };
            
            //输出拉格朗日多项式
          /* for (int i = n - 1; i >= 0; i--)
            {
                if (lag.a[i] == 0)
                    continue;
                Console.Write("{0}x^{1}+", lag.a[i], i);
            }*/
        }
    }
}