算法问题.如何用c#实现多项式相加.
最近想学习下关于算法的知识.可是看到例子大部分是c程序,如果用c#实现的话比较困难.不安全代码不太会,
用静态链表也不是很懂.向高手求助..
[解决办法]
有固定输入模式没有?
[解决办法]
C#链表容易
[code=c#]
public class List <T>
{
class Node <T>
{
T data;
public Node <T> next;
public Node(){this.next=null;this.data=defatule(T);}
}
Node <T> root;
public void Add(Node <T> n)
{
n.next=root;
root=n;
}
public IEnumerable <T> GetNodes()
{
Node <T> current = root;
if(current!=null)
{
yield return current;
current=current.next;
}
}
}
[/code]
楼主继续写,如何用链表具体实现,大家一起来学习.
[解决办法]
要下班了,只能写个很简陋的,不过大概的思路不会错的
输出:2*x^0+4*x^1+2*x^2+2*x^3+2*x^5+
格式比较丑陋,楼主可以自己再控制一下输出
using System;using System.Collections.Generic;using System.Text;namespace ConsoleApplication3{ class Program { static void Main(string[] args) { //新建一个表达式和一些新项,并向表达式中插入这些项 MyExpression me = new MyExpression(); MyItem item1 = new MyItem(2, 0); MyItem item2 = new MyItem(2, 3); MyItem item3 = new MyItem(2, 1); MyItem item4 = new MyItem(2, 5); MyItem item5 = new MyItem(2, 2); me.InsertItem(item1); me.InsertItem(item2); me.InsertItem(item3); me.InsertItem(item3); me.InsertItem(item4); me.InsertItem(item5); foreach (MyItem mi in me.Expression) { Console.Write(mi.Num + "*x^" + mi.Exp + "+"); } Console.ReadKey(); } } /// <summary> /// 表达式类,用于存储所有的项 /// </summary> class MyExpression { private List<MyItem> _expression;//用于存储表达式的项 internal List<MyItem> Expression { get { return _expression; } set { _expression = value; } } /// <summary> /// 初始化一个表达式 /// </summary> public MyExpression() { Expression = new List<MyItem>(); MyItem item = new MyItem(0, 0); Expression.Add(item); } /// <summary> /// 插入一个项,根据指数来确定插入新项还是与原有项相加 /// </summary> /// <param name="item"></param> public void InsertItem(MyItem item) { int length = Expression.Count; bool hasInserted = false; for (int i = 0; i < length; i++) { if (Expression[i] > item) //当前项指数比插入项大时,插到当前项的位置 { Expression.Insert(i, item); hasInserted = true; break; } if (Expression[i].Exp == item.Exp)//指数相等时相加 { Expression[i].Num += item.Num; hasInserted = true; break; } } if (!hasInserted) { Expression.Add(item);//指数比已知项都要大,插到表达式尾部 } } protected void Sort() { this.Expression.Sort(); } } /// <summary> /// 结点类,存储每一项 /// </summary> class MyItem : IComparable<MyItem> { private int _num;//倍娄 public int Num { get { return _num; } set { _num = value; } } private int _exp;//指数 public int Exp { get { return _exp; } set { _exp = value; } } /// <summary> /// 给定值构造一个新项 /// </summary> /// <param name="num"></param> /// <param name="exp"></param> public MyItem(int num, int exp) { Num = num; Exp = exp; } public bool Compare(MyItem left, MyItem right) { if (left.Exp > right.Exp) { return true; } else return false; } public int CompareTo(MyItem item) { return this.Exp.CompareTo(item.Exp); } public static bool operator >(MyItem left, MyItem right) { return left.Exp.CompareTo(right.Exp) > 0; } public static bool operator <(MyItem left, MyItem right) { return left.Exp.CompareTo(right.Exp) > 0; } //public object Clone() //{ // return new MyItem(this.Num,this.Exp); //} }}
[解决办法]
这儿来个网上搜到的,已经成功通过本人的检验.(与原代码相比小有改变)
// using System; namespace Polyn { public class Polynominal { int HighPower;//多项式的次数 double[] CoeffArray = new double[100];//数组用来存储多项式的系数，数组的项次表示多项式的次数 //olynominal p; =new Polynominal(); public Polynominal() //构造函数，初始化。 { HighPower = 0; int i; for (i = 0; i < CoeffArray.Length; i++) CoeffArray[i] = 0; } public Polynominal(int Exponent) //为了在测试时减少每次输入的麻烦，故设此构造函数。 { HighPower = Exponent; int i; int t; for (i = 0, t = 1; i <= Exponent; i++, t++) { CoeffArray[i] = t * t; } } /// <summary> /// 读入多项式的系数和次数 /// </summary> public void ReadIn() //读入多项式的系数和次数 { int Exponent;//次数 double Coefficient; Console.WriteLine("当系数输入为0时表示多项式输入完毕."); while (true) //确保输入的数据有效 { Console.WriteLine("读入多项式的最高次数:"); if (!(int.TryParse(Console.ReadLine(), out HighPower))) { Console.WriteLine(" 输入的不是整数，请重新输入"); continue; } else break; } while (true) { Console.WriteLine("输入某一项的次数(int型)："); if (!(int.TryParse(Console.ReadLine(), out Exponent))) { Console.WriteLine("不是整数，请重新输入:"); continue; //确保输入的“次”符合要求 } else if (Exponent > HighPower) { Console.WriteLine("该项次数大于多项式的最高次数，请重新输入:"); continue; } Console.WriteLine("输入系数(浮点型)："); while (true) { if (!(double.TryParse(Console.ReadLine(), out Coefficient))) { Console.WriteLine("输入的不是实数，请重新输入"); continue; } break; } if (Coefficient == 0) //退出条件，当系数为零时退出 break; CoeffArray[Exponent] = Coefficient; } } /// <summary> /// 输出多项式,系数为零的不输出. /// </summary> public void PrintPoly() //输出多项式,系数为零的不输出. { int i = 0; Console.WriteLine(); Console.WriteLine("输出多项式："); Console.Write("Poly:="); while (CoeffArray[i] == 0) //寻找第一个系数不为零的项 { i++; } if (i != 0 && i != 1) //当第一个系数不为零的项是次数为零 或为1的项需要作特殊处理。不能在输出结果中出现形如12X^0+1X^1的形式，正确表达应为12+x { Console.Write("{0}X^{1}", CoeffArray[i], i); } else if (i == 0) { Console.Write("{0}", CoeffArray[i]); i++; if (CoeffArray[i] != 1) Console.Write("+{0}X", CoeffArray[i]); else Console.Write("+X"); } else if (CoeffArray[i] != 1) Console.Write("{0}X", CoeffArray[i]); else Console.Write("X"); i++; while (i < CoeffArray.Length) { if (CoeffArray[i] != 0) Console.Write("+{0}X^{1}", CoeffArray[i], i); i++;
[解决办法]
乘方运算没有添加进去,可参照下面的
using System;
using System.Collections.Generic;
using System.Text;
namespace String_c
{
class string_c
{
public string_c()
{
}
private string lfrq(string str)
{
str = str.Replace("-", "+-");
str = str.Replace("++", "+");
str = str.Replace("*+", "*");
str = str.Replace("/+", "/");
while (str[0] == '+')
{
str = str.Substring(1);
}
return str;
}
public string 值计算(string str)
{
str = lfrq(str);
string str1, str2, str3, str4;
int index1, index2;
char cha;
int p = 0;
//取出括号内字符串计算
index1 = str.IndexOf('(');
if (index1 != -1)
{
index2 = 0;
for (int i = index1 + 1; i < str.Length; i++)
{
if (str[i] == '(') p++;
if (str[i] == ')')
{
if (p == 0) index2 = i;
else p--;
}
}
str1 = str.Substring(0, index1);
str2 = str.Substring(index1 + 1, index2 - index1 - 1);
str3 = str.Substring(index2 + 1);
str = str1 + 值计算(str2) + str3;
}
//取得运算符位置，优先乘除运算符
index1 =
(str.IndexOf('*') > 0 ? str.IndexOf('*') : 9999) <
(str.IndexOf('/') > 0 ? str.IndexOf('/') : 9999) ?
(str.IndexOf('*') > 0 ? str.IndexOf('*') : 9999) :
(str.IndexOf('/') > 0 ? str.IndexOf('/') : 9999);
if (index1 == 9999) index1 = str.IndexOf('+');
if (index1 == -1) return (str);
//取得运算符
cha = str[index1];
//取得运算符前一个数和后一个数及余下的字符串
str1 = str.Substring(0, index1);
str2 = str.Substring(index1 + 1, str.Length - 1 - index1);
index1 =
str1.LastIndexOf('*') > str1.LastIndexOf('/') ?
str1.LastIndexOf('8') : str1.LastIndexOf('/');
index1 =
str1.LastIndexOf('+') > index1 ?
str1.LastIndexOf('+') : index1;
str3 = str1.Substring(0, index1 + 1);
str1 = str1.Substring(index1 + 1, str1.Length - 1 - index1);
index1 = str2.IndexOf('*') > -1 ? str2.IndexOf('*') : 9999;
index2 = str2.IndexOf('/') > -1 ? str2.IndexOf('/') : 9999;
index1 = index1 < index2 ? index1 : index2;
index2 = str2.IndexOf('+') > -1 ? str2.IndexOf('+') : 9999;
index1 = index1 < index2 ? index1 : index2;
if (index1 == 9999)
{
str4 = str2;
str2 = "";
}
else
{
str4 = str2.Substring(0, index1);
str2 = str2.Substring(index1, str2.Length - index1);
}
//运算
double d1, d2;
str1 = str1.Replace("E", "E+");
str4 = str4.Replace("E", "E+");
if (str1[0] == '-') d1 = 0 - double.Parse(str1.Substring(1, str1.Length - 1));
else d1 = double.Parse(str1);
if (str4[0] == '-') d2 = 0 - double.Parse(str4.Substring(1, str4.Length - 1));
else d2 = double.Parse(str4);
switch (cha)
{
case '+':
d1 = d1 + d2;
break;
case '*':
d1 = d1 * d2;
break;
case '/':
d1 = d1 / d2;
break;
default:
break;
}
str = str3 + d1.ToString() + str2;
str = str.Replace("E+", "E");
index1 = str.IndexOf('+');
index1 =
str.IndexOf('*') > index1 ?
str.IndexOf('*') : index1;
index1 =
str.IndexOf('/') > index1 ?
str.IndexOf('/') : index1;
if (index1 == -1)
return str;
else
str = 值计算(str);
return str;
}
}
}