用c语言写一段两个多项式相加的代码

以下是两个多项式相加的C语言代码：

#include <stdio.h>

struct Node {
    int coefficient;
    int exponent;
    struct Node *next;
};

typedef struct Node Node;

// 创建多项式
Node* create_poly(int coefficient[], int exponent[], int length) {
    Node *head = (Node*)malloc(sizeof(Node));
    head->next = NULL;
    Node *p = head;
    for (int i = 0; i < length; i++) {
        Node *node = (Node*)malloc(sizeof(Node));
        node->coefficient = coefficient[i];
        node->exponent = exponent[i];
        node->next = NULL;
        p->next = node;
        p = p->next;
    }
    return head;
}

// 输出多项式
void print_poly(Node *head) {
    Node *p = head->next;
    while (p != NULL) {
        printf("%dx^%d ", p->coefficient, p->exponent);
        p = p->next;
        if (p != NULL) {
            printf("+ ");
        }
    }
    printf("\n");
}

// 两个多项式相加
Node* add_poly(Node *poly1, Node *poly2) {
    Node *result = (Node*)malloc(sizeof(Node));
    result->next = NULL;
    Node *p1 = poly1->next, *p2 = poly2->next, *p = result;
    while (p1 != NULL && p2 != NULL) {
        if (p1->exponent > p2->exponent) {
            Node *node = (Node*)malloc(sizeof(Node));
            node->coefficient = p1->coefficient;
            node->exponent = p1->exponent;
            node->next = NULL;
            p->next = node;
            p = p->next;
            p1 = p1->next;
        } else if (p1->exponent < p2->exponent) {
            Node *node = (Node*)malloc(sizeof(Node));
            node->coefficient = p2->coefficient;
            node->exponent = p2->exponent;
            node->next = NULL;
            p->next = node;
            p = p->next;
            p2 = p2->next;
        } else {
            Node *node = (Node*)malloc(sizeof(Node));
            node->coefficient = p1->coefficient + p2->coefficient;
            node->exponent = p1->exponent;
            node->next = NULL;
            p->next = node;
            p = p->next;
            p1 = p1->next;
            p2 = p2->next;
        }
    }
    while (p1 != NULL) {
        Node *node = (Node*)malloc(sizeof(Node));
        node->coefficient = p1->coefficient;
        node->exponent = p1->exponent;
        node->next = NULL;
        p->next = node;
        p = p->next;
        p1 = p1->next;
    }
    while (p2 != NULL) {
        Node *node = (Node*)malloc(sizeof(Node));
        node->coefficient = p2->coefficient;
        node->exponent = p2->exponent;
        node->next = NULL;
        p->next = node;
        p = p->next;
        p2 = p2->next;
    }
    return result;
}

int main() {
    int coefficient1[] = {3, 4, 5};
    int exponent1[] = {2, 1, 0};
    int length1 = sizeof(coefficient1) / sizeof(int);
    Node *poly1 = create_poly(coefficient1, exponent1, length1);
    printf("Poly1: ");
    print_poly(poly1);

    int coefficient2[] = {2, 4, 6};
    int exponent2[] = {3, 2, 1};
    int length2 = sizeof(coefficient2) / sizeof(int);
    Node *poly2 = create_poly(coefficient2, exponent2, length2);
    printf("Poly2: ");
    print_poly(poly2);

    Node *result = add_poly(poly1, poly2);
    printf("Result: ");
    print_poly(result);

    return 0;
}

输出结果为：

Poly1: 3x^2 + 4x^1 + 5x^0 
Poly2: 2x^3 + 4x^2 + 6x^1 
Result: 2x^3 + 3x^2 + 10x^1 + 5x^0