Question

这是一个程序c ++，它对多项式执行不同的操作。在此程序中，multiplication()函数存在逻辑问题。我无法计算得到的多项式中的项数，这是两个多项式相乘的结果。

该程序为大多数情况提供了正确的乘法输出，但很少有正确执行乘法的情况。以下是完整的计划。

#include<iostream>
using namespace std;

struct poly
{
    int coef, pow;
};

int k=0;
poly* getdata(poly *a,int n)
{
    for(int i=0; i<n; i++)
    {
        cout<<"\n Enter the coefficient and power of the term";
        cin>>a[i].coef>>a[i].pow;
    }
    return a;
}

poly* add(poly *c, poly *a, poly *b, int n1, int n2)
{
    int i=0, j=0;
    while(i<n1 && j<n2)
    {
        if(a[i].pow<b[j].pow)
        {
            c[k].pow = a[i].pow;
            c[k].coef = a[i].coef;
            i++;
            k++;
        }
        else if(a[i].pow == b[j].pow)
        {
            c[k].pow = a[i].pow;
            c[k].coef = a[i].coef+b[j].coef;
            i++; j++; k++;
        }
        else
            //(a[i].pow > b[j].pow)
        {
            c[k].pow = b[j].pow;
            c[k].coef = b[j].coef;
            j++; k++;
        }
    }
    while(i<n1)
    {
        c[k].pow = a[i].pow;
        c[k].coef = a[i].coef;
        i++;
        k++;
    }
    while(j<n2)
    {
        c[k].pow = b[j].pow;
        c[k].coef = b[j].coef;
        j++; k++;
    }
    return c;
}

void display(poly *p,int n)
{
    int i;
    cout<<"\n The polynomial is :-\n";
    for(i=0; i<n-1; i++)
    {
        cout<<p[i].coef<<"x^"<<p[i].pow<<"+";
    }
    cout<<p[i].coef<<"x^"<<p[i].pow;
    cout<<endl;
}

void multiplication(poly *c, poly *a, poly* b, int n1, int n2)
{
    int i, j;
    //int count = 0;

    for(i=0; i<n1*n2; i++)
    {
        c[i].coef = 0;
        c[i].pow = 0;
    }

    for(i=0; i<n1; i++)
    {
        for(j=0; j<n2; j++)
        {
            c[i+j].coef = a[i].coef*b[j].coef + c[i+j].coef;
            c[i+j].pow = a[i].pow + b[j].pow;
        }
    }
    //return count;
}

int main()
{
    poly a[10], b[10], c[10];
    poly *p, *q , *result;
    int n1, n2;
    cout<<"\n Enter the number of terms in polynomial 1 : ";
    cin>>n1;
    p = getdata(a,n1);
    display(p, n1);

    cout<<"\n Enter the number of terms in polynomial 2 : ";
    cin>>n2;
    q = getdata(b,n2);
    display(q, n2);

    result = add(c, a, b, n1, n2);
    cout<<"\n The Sum of the two polynomial is ";
    display(result, k);

    poly *mresult;
    mresult = new poly[n1+n2];
    multiplication(mresult, a, b, n1, n2);
    cout<<"\n The Multiplication of the two polynomials is :- ";
    display(mresult,n1+n2-1);

    return 0;
}

在少数情况下输出： -

 Enter the number of terms in polynomial 1 : 2

 Enter the coefficient and power of the term1
1

 Enter the coefficient and power of the term3
9

 The polynomial is :-
1x^1+3x^9

 Enter the number of terms in polynomial 2 : 2

 Enter the coefficient and power of the term1
3

 Enter the coefficient and power of the term2
6

 The polynomial is :-
1x^3+2x^6

 The Sum of the two polynomial is 
 The polynomial is :-
1x^1+1x^3+2x^6+3x^9

 The Multiplication of the two polynomials is :- 
 The polynomial is :-
1x^4+5x^12+6x^15

这是正确的。但请考虑这种情况： -

 Enter the number of terms in polynomial 1 : 2

 Enter the coefficient and power of the term1
1

 Enter the coefficient and power of the term3
9

 The polynomial is :-
1x^1+3x^9

 Enter the number of terms in polynomial 2 : 2

 Enter the coefficient and power of the term1
3

 Enter the coefficient and power of the term2
6

 The polynomial is :-
1x^3+2x^6

 The Sum of the two polynomial is 
 The polynomial is :-
1x^1+1x^3+2x^6+3x^9

 The Multiplication of the two polynomials is :- 
 The polynomial is :-
1x^4+5x^12+6x^15

这是错的。这里的乘法应该是： - x^4+2x^7+3x^12+6x^15

我尝试了很多逻辑，例如： -

int multiplication(poly *c, poly *a, poly* b, int n1, int n2)
{
    int i, j;
    int count = 0;

    for(i=0; i<n1*n2; i++)
    {
        c[i].coef = 0;
        c[i].pow = 0;
    }

    for(i=0; i<n1; i++)
    {
        for(j=0; j<n2; j++)
        {
            c[count].coef = a[i].coef*b[j].coef + c[count].coef;
            c[count].pow = a[i].pow + b[j].pow;
            if(c[count].pow != c[count-1].pow) count++;
        }
    }
    return count;
}

此函数返回多项式c中的项数。

注意： - c是两个多项式a和b相乘的结果。

Answer 1

你想要更像这样的逻辑：

for(i=0; i<n1; i++)
{
    for(j=0; j<n2; j++)
    {
        int coef = a[i].coef * b[j].coef;
        int pow = a[i].pow + b[j].pow;
        count = add_term_to(c,count,coef,pow);
    }
}

现在你只需要实现add_term_to()，以便它将术语与相同的力量结合起来。

Answer 2

当 f 和 g 是 x 的多项式时， f × g < / em>是 f 的程度加上 g 的程度。得到的多项式的系数将是几对乘积的总和; 例如当乘以（f_2 x ^ 2 + f_1 x + f_0）（g_2 x ^ 2 + g_1 x + g_0）时，x ^ 2的系数将为f_2 * g_0 + f_1 * g_1 + f_0 * G_2。您不会事先知道这些总和是否为零。二项式（x + 1）（x + 1）有三个非零项，但（x + 1）（x-1）有两个，而（x ^ 2 + 1）（x ^ 4 + x ^ 3）有四个

您可以通过将其索引为每个项的度数的向量中的系数求和来有效地计算此多项式。当您转换回来时，您可以删除与零相同的条款。

确定两个多项式相乘结果中的项数

2 个答案: