帮我写多项式乘法的代码

当然可以，以下是多项式乘法的代码：

#include <stdio.h>
#include <stdlib.h>

#define MAXN 100

int main()
{
    int n, m, a[MAXN], b[MAXN], c[MAXN*2];

    // 读入多项式 A
    scanf("%d", &n);
    for (int i = ; i <= n; i++)
        scanf("%d", &a[i]);

    // 读入多项式 B
    scanf("%d", &m);
    for (int i = ; i <= m; i++)
        scanf("%d", &b[i]);

    // 多项式乘法
    for (int i = ; i <= n+m; i++)
        c[i] = ;
    for (int i = ; i <= n; i++)
        for (int j = ; j <= m; j++)
            c[i+j] += a[i] * b[j];

    // 输出结果
    printf("%d", c[]);
    for (int i = 1; i <= n+m; i++)
        printf(" %d", c[i]);
    printf("\n");

    return ;
}

希望能够帮到你！

相关问题

两个一元多项式乘法代码

下面是Python实现的两个一元多项式乘法代码示例：

1. 使用列表存储一元多项式

def poly_mult(poly1, poly2):
    """
    一元多项式乘法，使用列表存储多项式系数
    
    :param poly1: 多项式1，如 [3, 2, 1] 表示 3x^2 + 2x + 1
    :param poly2: 多项式2，如 [4, 2] 表示 4x + 2
    :return: 乘积多项式，如 [12, 14, 4, 1] 表示 12x^3 + 14x^2 + 4x + 1
    """
    m, n = len(poly1), len(poly2)
    res = [0] * (m + n - 1)
    for i in range(m):
        for j in range(n):
            res[i+j] += poly1[i] * poly2[j]
    return res

2. 使用字典存储一元多项式

def poly_mult(poly1, poly2):
    """
    一元多项式乘法，使用字典存储多项式系数
    
    :param poly1: 多项式1，如 {2: 3, 1: 2, 0: 1} 表示 3x^2 + 2x + 1
    :param poly2: 多项式2，如 {1: 4, 0: 2} 表示 4x + 2
    :return: 乘积多项式，如 {3: 12, 2: 14, 1: 4, 0: 1} 表示 12x^3 + 14x^2 + 4x + 1
    """
    res = {}
    for i, a in poly1.items():
        for j, b in poly2.items():
            res[i+j] = res.get(i+j, 0) + a * b
    return res

这两个函数的输入参数都是两个多项式，其中`poly1`和`poly2`可以用列表或字典表示。输出结果也是一个多项式，同样用列表或字典表示。其中，`poly_mult`函数使用的是暴力算法，时间复杂度为$O(mn)$，其中$m$和$n$分别为两个多项式的项数。如果使用更高效的Karatsuba算法可以将时间复杂度降为$O(n^{\log_2 3})$。

C语言实现两个一元多项式乘法代码

#include <stdio.h>
#include <stdlib.h>

typedef struct PolyNode *Polynomial;
struct PolyNode{
    int coef;
    int expon;
    Polynomial link;
};

Polynomial ReadPoly();
Polynomial Add(Polynomial P1, Polynomial P2);
Polynomial Multiply(Polynomial P1, Polynomial P2);
void PrintPoly(Polynomial P);

int main()
{
    Polynomial P1, P2, PP, PS;

    P1 = ReadPoly();
    P2 = ReadPoly();
    PP = Multiply(P1, P2);
    PrintPoly(PP);

    return 0;
}

Polynomial ReadPoly()
{
    int N, c, e;
    Polynomial P, Rear, Tmp;

    scanf("%d", &N);
    P = (Polynomial)malloc(sizeof(struct PolyNode));
    P->link = NULL;
    Rear = P;
    while (N--) {
        scanf("%d %d", &c, &e);
        Tmp = (Polynomial)malloc(sizeof(struct PolyNode));
        Tmp->coef = c;
        Tmp->expon = e;
        Tmp->link = NULL;
        Rear->link = Tmp;
        Rear = Tmp;
    }
    Tmp = P; /* 删除多余的头结点 */
    P = P->link;
    free(Tmp);
    return P;
}

Polynomial Add(Polynomial P1, Polynomial P2)
{
    Polynomial Front, Rear, Temp;
    int sum;

    Rear = (Polynomial)malloc(sizeof(struct PolyNode));
    Front = Rear;
    while (P1 && P2) {
        if (P1->expon == P2->expon) {
            sum = P1->coef + P2->coef;
            if (sum) {
                Temp = (Polynomial)malloc(sizeof(struct PolyNode));
                Temp->coef = sum;
                Temp->expon = P1->expon;
                Rear->link = Temp;
                Rear = Rear->link;
            }
            P1 = P1->link;
            P2 = P2->link;
        } else if (P1->expon > P2->expon) {
            Temp = (Polynomial)malloc(sizeof(struct PolyNode));
            Temp->coef = P1->coef;
            Temp->expon = P1->expon;
            Rear->link = Temp;
            Rear = Rear->link;
            P1 = P1->link;
        } else {
            Temp = (Polynomial)malloc(sizeof(struct PolyNode));
            Temp->coef = P2->coef;
            Temp->expon = P2->expon;
            Rear->link = Temp;
            Rear = Rear->link;
            P2 = P2->link;
        }
    }
    while (P1) {
        Temp = (Polynomial)malloc(sizeof(struct PolyNode));
        Temp->coef = P1->coef;
        Temp->expon = P1->expon;
        Rear->link = Temp;
        Rear = Rear->link;
        P1 = P1->link;
    }
    while (P2) {
        Temp = (Polynomial)malloc(sizeof(struct PolyNode));
        Temp->coef = P2->coef;
        Temp->expon = P2->expon;
        Rear->link = Temp;
        Rear = Rear->link;
        P2 = P2->link;
    }
    Rear->link = NULL;
    Temp = Front;
    Front = Front->link;
    free(Temp);
    return Front;
}

Polynomial Multiply(Polynomial P1, Polynomial P2)
{
    Polynomial P, Rear, Tmp, t1, t2;
    int c, e;

    if (!P1 || !P2) return NULL;
    t1 = P1; t2 = P2;
    P = (Polynomial)malloc(sizeof(struct PolyNode));
    Rear = P;
    while (t2) { /* 先用P1的第1项乘以P2，得到P */
        c = t1->coef * t2->coef;
        e = t1->expon + t2->expon;
        if (c) {
            Tmp = (Polynomial)malloc(sizeof(struct PolyNode));
            Tmp->coef = c;
            Tmp->expon = e;
            Rear->link = Tmp;
            Rear = Rear->link;
        }
        t2 = t2->link;
    }
    t1 = t1->link;
    while (t1) { /* P1的第2项开始分别乘以P2，得到各项插入P中 */
        t2 = P2; Rear = P;
        while (t2) {
            c = t1->coef * t2->coef;
            e = t1->expon + t2->expon;
            while (Rear->link && Rear->link->expon > e)
                Rear = Rear->link;
            if (Rear->link && Rear->link->expon == e) {
                if (Rear->link->coef + c)
                    Rear->link->coef += c;
                else {
                    Tmp = Rear->link;
                    Rear->link = Tmp->link;
                    free(Tmp);
                }
            } else {
                Tmp = (Polynomial)malloc(sizeof(struct PolyNode));
                Tmp->coef = c;
                Tmp->expon = e;
                Tmp->link = Rear->link;
                Rear->link = Tmp;
                Rear = Rear->link;
            }
            t2 = t2->link;
        }
        t1 = t1->link;
    }
    Tmp = P; P = P->link; free(Tmp);
    return P;
}

void PrintPoly(Polynomial P)
{
    int flag = 0;

    if (!P) {
        printf("0 0\n");
        return;
    }
    while (P) {
        if (!flag)
            flag = 1;
        else
            printf(" ");
        printf("%d %d", P->coef, P->expon);
        P = P->link;
    }
    printf("\n");
}