我想从用户那里得到一个数学（单变量代数）方程，例如 x ^ 3-4x -9 = 0 ，并希望针对不同的 x 。
我尝试创建两个数组；一个用于获取 x 的幂的输入，另一个用于系数。该程序向用户询问方程式中存在的项数，然后使该数为1的数组。 但是，该算法仅有助于打印方程式，而不能操纵或评估方程式。
/* this code can take some relevant input from the user and form an equation on the screen to show it to the user. */
#include<bits/stdc++.h>
#include<conio.h>
using namespace std;
class Bisection
{
int noofcaparr, *coeffarr, *powerarr, eqn;
char *eqnprnt;
public:
void geteqn();
void showeqn();
void setupeqn();
};
void Bisection::geteqn()
{
int c, i, n;
system("cls");
cout<<"\n\n\t\t How many terms do you have in your equation? ";
cout<<"\n\t For Example: x^3 - 4x - 9 = 0 , has '3' terms and ";
cout<<"\n\t x^4 + x^3 - 7x^2 - x + 5 = 0 , has '5' terms";
cout<<"\n\t Enter the number of terms present in your equation: ";
cin>>this->noofcaparr;
n = this->noofcaparr-1;
this->coeffarr = new int[n];
this->powerarr = new int[n];
for(i=0, c=1; i<=n; i++, c++ )
{
cout<<endl<<endl<<"\t\t Please enter the "<<c<<" th/st/nd/rd highest degree of x: ";
cin>>this->powerarr[i];
cout<<endl<<endl<<"\t\t Please enter the coefficient of "<<c<<" th/st/nd/rd highest degree of x (with sign -/+): ";
cin>>this->coeffarr[i];
}
cout<<endl<<endl<<"\n\n\t Values Set!";
getch();
}
void Bisection::showeqn()
{
int i, n;
n = this->noofcaparr-1;
system("cls");
cout<<endl<<endl<<"\t\t Your equation is: ";
for(i=0; i<=n; i++ )
{
if(this->powerarr[i]==0)
{
if(i==0)
{
if(this->coeffarr[i]>= 0)
{
if(this->coeffarr[i]==1)
{
cout<<" ";
}
else
{
cout<<" "<<(this->coeffarr[i])<<" ";
}
}
else
{
if(this->coeffarr[i]== -1)
{
cout<<" -"<<" ";
}
else
{
cout<<" "<<(this->coeffarr[i])<<" ";
}
}
}
else
{
if(this->coeffarr[i]>= 0)
{
cout<<" +"<<(this->coeffarr[i])<<" ";
}
else
{
cout<<" "<<(this->coeffarr[i])<<" ";
}
}
}
else
{
if(this->powerarr[i]==1)
{
if(i==0)
{
if(this->coeffarr[i]>= 0)
{
if(this->coeffarr[i]==1)
{
cout<<"x";
}
else
{
cout<<(this->coeffarr[i])<<"x";
}
}
else
{
if(this->coeffarr[i]== -1)
{
cout<<" -"<<"x";
}
else
{
cout<<(this->coeffarr[i])<<"x";
}
}
}
else
{
if(this->coeffarr[i]>= 0)
{
cout<<"+"<<(this->coeffarr[i])<<"x";
}
else
{
cout<<(this->coeffarr[i])<<"x";
}
}
}
else
{
if(i==0)
{
if(this->coeffarr[i]>= 0)
{
if(this->coeffarr[i]==1)
{
cout<<"x^"<<this->powerarr[i]<<" ";
}
else
{
cout<<" "<<(this->coeffarr[i])<<" "<<"x^"<<this->powerarr[i]<<" ";
}
}
else
{
if(this->coeffarr[i]== -1)
{
cout<<" -"<<"x^"<<this->powerarr[i]<<" ";
}
else
{
cout<<" "<<(this->coeffarr[i])<<" "<<"x^"<<this->powerarr[i]<<" ";
}
}
}
else
{
if(this->coeffarr[i]>= 0)
{
cout<<" +"<<(this->coeffarr[i])<<" "<<"x^"<<this->powerarr[i]<<" ";
}
else
{
cout<<" "<<(this->coeffarr[i])<<" "<<"x^"<<this->powerarr[i]<<" ";
}
}
}
}
}
cout<<" = 0";
getch();
}
int main()
{
Bisection a;
a.geteqn();
a.showeqn();
getch();
return(0);
}
尝试检查此代码。如果要求输入，请尝试一个示例：在第一个输入中，键入 3 ，在第二个输入 3 ，然后输入 1 ，然后输入 1 ，然后是 -4 ，然后是 0 ，然后是 -9 。这将在屏幕上打印以下方程式： x ^ 3-4x-9 = 0
但是，我无法操纵或评估这个方程。如果我想通过使方程等于 fx 并采用不同的 x 值来计算方程，然后评估 fx 的值，我不能这样做。
我已经尝试过在Internet上搜索它，但是所有解决方案都没有帮助或太复杂以至于无法理解。
我是一个非常新的程序员，对数据结构，野牛或任何类似的解析器一无所知。请以一种尽可能简单的方式来说明/帮助我。
请不要投票，如果您在问题中发现任何错误，请在评论中告诉我；我会问我的问题。 预先感谢！
答案 0 :(得分：0)
就像NO_NAME认为的那样，评估功能并不困难，尽管在当前数据布局下，我们无法通过其建议的迭代来计算power ，因为没有每个指数的术语。但是此变体有效：
double Bisection::evalfun(int x)
{
double f = 0;
for (int i = 0; i < this->noofcaparr; ++i)
f += coeffarr[i] * pow(x, powerarr[i]);
return f;
}
示例要求 x的不同值：
cout <<endl;
for (int x = -5; x <= 5; ++x) cout <<a.evalfun(x) <<'\t';
cout <<endl;
也许您想使用double x而不是int x。