在 c++ 中对两个多项式的加法进行编码时,我可以在 no 和项的顺序相同时计算加法。 diff时显示错误,为什么?
While coding addition of two polynomials in c++ , I could compute the addition when no and order of terms are same . It shows error when its diff,why?
在这里,我尝试用 c++ 编写两个单变量多项式的加法代码。当两个输入与另一个输入同步时,代码运行良好。但是当它不是时,它会显示一些疯狂的事情发生并打印出完整的胡言乱语。我将我的代码与输出一起粘贴到此处。任何建议都会有所帮助。
我的代码
#include <iostream>
using namespace std;
class Term
{
public :
int coeff;
int exp;
};
class Poly
{
private:
int n;
Term *terms;
public:
Poly(int n)
{
this->n = n;
terms = new Term[n];
}
void create();
void Display();
Poly* add(Poly &p2);
~Poly()
{
delete []terms;
}
};
void Poly::create()
{
cout << "Enter Terms " << endl;
for(int i = 0; i<n;i++)
{
cin >> terms[i].coeff >> terms[i].exp ;
}
}
void Poly::Display()
{
for(int i = 0;i<n;i++)
{
cout <<terms[i].coeff<<"x"<<terms[i].exp <<" + ";
}
}
Poly* Poly :: add(Poly &p2)
{
int i,j,k;
Poly* sum = new Poly(n+p2.n);
i = j = k = 0 ;
while(i<n && j<p2.n)
{
if(terms[i].exp>p2.terms[j].exp)
{
sum->terms[k++] = terms[i++];
}
else if(terms[i].exp<p2.terms[j].exp)
sum->terms[k++] = terms[j++];
else
sum->terms[k].exp = terms[i].exp;
sum->terms[k++].coeff = terms[i++].coeff +p2.terms[j++].coeff ;
}
for(;i<n;i++)
{
sum->terms[k++] = terms[i];
}
for(;j<p2.n;j++)
{
sum->terms[k++] = p2.terms[j];
}
sum->n = k;
return sum ;
}
int main()
{
int n,m;
cout << "Enter no.of terms for 1st Polynomial" << endl ;
cin >> n;
Poly p1(n);
p1.create();
cout << "Enter no.of terms for 2nd Polynomial" << endl ;
cin >> m ;
Poly p2(m) ;
p2.create();
Poly *p3 ;
p3 = p1.add(p2);
cout << endl;
p1.Display();
cout << endl;
cout << "-------" << endl ;
p2.Display();
cout << endl;
cout << " ------------" << endl ;
p3->Display();
return 0;
}
输出 1:(通过第一个测试用例)
Enter no.of terms for 1st Polynomial
5
Enter Terms
2 5
3 4
6 3
7 2
8 1
Enter no.of terms for 2nd Polynomial
5
Enter Terms
3 5
8 4
1 3
5 2
6 1
2x5 + 3x4 + 6x3 + 7x2 + 8x1 +
-------
3x5 + 8x4 + 1x3 + 5x2 + 6x1 +
------------
5x5 + 11x4 + 7x3 + 12x2 + 14x1 +
Process returned 0 (0x0) execution time : 48.853 s
Press any key to continue.
输出 II(测试用例 2 失败)
Enter no.of terms for 1st Polynomial
5
Enter Terms
3 5
2 4
8 3
4 2
6 1
Enter no.of terms for 2nd Polynomial
4
Enter Terms
7 5
4 3
8 1
2 0
3x5 + 2x4 + 8x3 + 4x2 + 6x1 +
-------
7x5 + 4x3 + 8x1 + 2x0 +
------------
10x5 + 2x4 + 12x-1 + 4x2 + 14x0 + 2x0 +
Process returned 0 (0x0) execution time : 56.271 s
Press any key to continue.
任何解决问题的帮助建议将不胜感激。谢谢。
您代码中的主要错误是这一行:
sum->terms[k++].coeff = terms[i++].coeff +p2.terms[j++].coeff ;
更准确地说,它在整个例程中的位置。
也就是说,它位于过去 if - else if - else
构造。因此,尽管指数满足 'less-than'、'greater-than' 或 'equal' 条件,它仍会执行;并且应该只在最后一种情况下完成。
代码的相关部分应该用另一对大括号括起来:
while(i<n && j<p2.n)
{
if(terms[i].exp>p2.terms[j].exp)
sum->terms[k++] = terms[i++];
else if(terms[i].exp<p2.terms[j].exp)
sum->terms[k++] = terms[j++];
else
{
sum->terms[k].exp = terms[i].exp;
sum->terms[k++].coeff = terms[i++].coeff +p2.terms[j++].coeff ;
}
}
另一个错误在 else if
分支中:您使用 j
索引到 this->terms
数组:
sum->terms[k++] = terms[j++];
而你应该使用 p2.terms
:
sum->terms[k++] = p2.terms[j++];
在这里,我尝试用 c++ 编写两个单变量多项式的加法代码。当两个输入与另一个输入同步时,代码运行良好。但是当它不是时,它会显示一些疯狂的事情发生并打印出完整的胡言乱语。我将我的代码与输出一起粘贴到此处。任何建议都会有所帮助。
我的代码
#include <iostream>
using namespace std;
class Term
{
public :
int coeff;
int exp;
};
class Poly
{
private:
int n;
Term *terms;
public:
Poly(int n)
{
this->n = n;
terms = new Term[n];
}
void create();
void Display();
Poly* add(Poly &p2);
~Poly()
{
delete []terms;
}
};
void Poly::create()
{
cout << "Enter Terms " << endl;
for(int i = 0; i<n;i++)
{
cin >> terms[i].coeff >> terms[i].exp ;
}
}
void Poly::Display()
{
for(int i = 0;i<n;i++)
{
cout <<terms[i].coeff<<"x"<<terms[i].exp <<" + ";
}
}
Poly* Poly :: add(Poly &p2)
{
int i,j,k;
Poly* sum = new Poly(n+p2.n);
i = j = k = 0 ;
while(i<n && j<p2.n)
{
if(terms[i].exp>p2.terms[j].exp)
{
sum->terms[k++] = terms[i++];
}
else if(terms[i].exp<p2.terms[j].exp)
sum->terms[k++] = terms[j++];
else
sum->terms[k].exp = terms[i].exp;
sum->terms[k++].coeff = terms[i++].coeff +p2.terms[j++].coeff ;
}
for(;i<n;i++)
{
sum->terms[k++] = terms[i];
}
for(;j<p2.n;j++)
{
sum->terms[k++] = p2.terms[j];
}
sum->n = k;
return sum ;
}
int main()
{
int n,m;
cout << "Enter no.of terms for 1st Polynomial" << endl ;
cin >> n;
Poly p1(n);
p1.create();
cout << "Enter no.of terms for 2nd Polynomial" << endl ;
cin >> m ;
Poly p2(m) ;
p2.create();
Poly *p3 ;
p3 = p1.add(p2);
cout << endl;
p1.Display();
cout << endl;
cout << "-------" << endl ;
p2.Display();
cout << endl;
cout << " ------------" << endl ;
p3->Display();
return 0;
}
输出 1:(通过第一个测试用例)
Enter no.of terms for 1st Polynomial
5
Enter Terms
2 5
3 4
6 3
7 2
8 1
Enter no.of terms for 2nd Polynomial
5
Enter Terms
3 5
8 4
1 3
5 2
6 1
2x5 + 3x4 + 6x3 + 7x2 + 8x1 +
-------
3x5 + 8x4 + 1x3 + 5x2 + 6x1 +
------------
5x5 + 11x4 + 7x3 + 12x2 + 14x1 +
Process returned 0 (0x0) execution time : 48.853 s
Press any key to continue.
输出 II(测试用例 2 失败)
Enter no.of terms for 1st Polynomial
5
Enter Terms
3 5
2 4
8 3
4 2
6 1
Enter no.of terms for 2nd Polynomial
4
Enter Terms
7 5
4 3
8 1
2 0
3x5 + 2x4 + 8x3 + 4x2 + 6x1 +
-------
7x5 + 4x3 + 8x1 + 2x0 +
------------
10x5 + 2x4 + 12x-1 + 4x2 + 14x0 + 2x0 +
Process returned 0 (0x0) execution time : 56.271 s
Press any key to continue.
任何解决问题的帮助建议将不胜感激。谢谢。
您代码中的主要错误是这一行:
sum->terms[k++].coeff = terms[i++].coeff +p2.terms[j++].coeff ;
更准确地说,它在整个例程中的位置。
也就是说,它位于过去 if - else if - else
构造。因此,尽管指数满足 'less-than'、'greater-than' 或 'equal' 条件,它仍会执行;并且应该只在最后一种情况下完成。
代码的相关部分应该用另一对大括号括起来:
while(i<n && j<p2.n)
{
if(terms[i].exp>p2.terms[j].exp)
sum->terms[k++] = terms[i++];
else if(terms[i].exp<p2.terms[j].exp)
sum->terms[k++] = terms[j++];
else
{
sum->terms[k].exp = terms[i].exp;
sum->terms[k++].coeff = terms[i++].coeff +p2.terms[j++].coeff ;
}
}
另一个错误在 else if
分支中:您使用 j
索引到 this->terms
数组:
sum->terms[k++] = terms[j++];
而你应该使用 p2.terms
:
sum->terms[k++] = p2.terms[j++];