C语言实现一元多项式加法运算 - 代码详解

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

struct node { int coefficient; // 系数 int exponent; // 指数 struct node *next; // 指向下一个节点 };

struct node *create_poly(); // 创建多项式 void print_poly(struct node *poly); // 输出多项式 struct node *add_poly(struct node *poly1, struct node *poly2); // 多项式加法

int main() { struct node *poly1, *poly2, *sum;

printf("输入第一个多项式：\n");
poly1 = create_poly();

printf("\n输入第二个多项式：\n");
poly2 = create_poly();

sum = add_poly(poly1, poly2);

printf("\n第一个多项式：\n");
print_poly(poly1);

printf("\n第二个多项式：\n");
print_poly(poly2);

printf("\n它们的和：\n");
print_poly(sum);

return 0;

}

struct node *create_poly() { struct node *head, *p, *q; int coefficient, exponent;

head = (struct node *)malloc(sizeof(struct node));

p = head;

printf("请输入系数和指数，输入-1表示结束：\n");
while (1) {
    scanf("%d%d", &coefficient, &exponent);
    if (coefficient == -1 && exponent == -1) {
        break;
    }

    q = (struct node *)malloc(sizeof(struct node));
    q->coefficient = coefficient;
    q->exponent = exponent;
    p->next = q;
    p = q;
}

p->next = NULL;

return head->next;

}

void print_poly(struct node *poly) { while (poly != NULL) { printf("%dX^%d", poly->coefficient, poly->exponent); poly = poly->next; if (poly != NULL) { printf("+"); } } }

struct node *add_poly(struct node *poly1, struct node *poly2) { struct node *head, *p, *q, *sum;

head = (struct node *)malloc(sizeof(struct node));

p = poly1;
q = poly2;
sum = head;

while (p != NULL && q != NULL) {
    if (p->exponent < q->exponent) {
        sum->next = p;
        p = p->next;
    } else if (p->exponent > q->exponent) {
        sum->next = q;
        q = q->next;
    } else {
        int coefficient = p->coefficient + q->coefficient;
        if (coefficient != 0) {
            struct node *node = (struct node *)malloc(sizeof(struct node));
            node->coefficient = coefficient;
            node->exponent = p->exponent;
            sum->next = node;
            p = p->next;
            q = q->next;
            sum = sum->next;
        } else {
            p = p->next;
            q = q->next;
        }
    }

    if (p != NULL) {
        sum->next = p;
    } else {
        sum->next = q;
    }
}

return head->next;

}