C语言计算大数乘法 链表

实现大数乘法可以使用链表来存储每一位数字。具体实现步骤如下：

1. 创建两个链表，分别存储两个大数的每一位数字，同时需要记录两个链表的长度。

2. 创建一个结果链表，初始值为0。

3. 从个位开始，遍历第一个链表的每一位数字，对于第二个链表的每一位数字都进行乘法运算，并将结果相加到结果链表中。

4. 遍历完第一个链表后，将结果链表中的每一位数字依次输出，即为最终的乘积结果。

下面是C语言的实现代码：

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

//链表结构体
typedef struct node {
    int data;
    struct node *next;
} Node;

//创建链表
Node* create(char *num) {
    int len = strlen(num);
    Node *head = NULL, *p = NULL;
    for (int i = 0; i < len; i++) {
        Node *newNode = (Node*)malloc(sizeof(Node));
        newNode->data = num[i] - '0';
        newNode->next = NULL;
        if (head == NULL) {
            head = newNode;
        } else {
            p->next = newNode;
        }
        p = newNode;
    }
    return head;
}

//遍历链表
void traverse(Node *head) {
    Node *p = head;
    while (p != NULL) {
        printf("%d", p->data);
        p = p->next;
    }
    printf("\n");
}

//链表相加
Node* add(Node *num1, Node *num2) {
    Node *head = NULL, *p = NULL;
    int carry = 0;  //进位标志
    while (num1 != NULL || num2 != NULL || carry != 0) {
        int sum = (num1 == NULL ? 0 : num1->data) + (num2 == NULL ? 0 : num2->data) + carry;
        Node *newNode = (Node*)malloc(sizeof(Node));
        newNode->data = sum % 10;
        newNode->next = NULL;
        if (head == NULL) {
            head = newNode;
        } else {
            p->next = newNode;
        }
        p = newNode;
        carry = sum / 10;
        if (num1 != NULL) num1 = num1->next;
        if (num2 != NULL) num2 = num2->next;
    }
    return head;
}

//链表相乘
Node* multiply(Node *num1, Node *num2) {
    Node *result = create("0");
    int len1 = 0, len2 = 0;
    Node *p = num1;
    while (p != NULL) {
        len1++;
        p = p->next;
    }
    p = num2;
    while (p != NULL) {
        len2++;
        p = p->next;
    }
    for (int i = 0; i < len1; i++) {
        Node *temp = create("0");
        int carry = 0;
        Node *q = temp;
        for (int j = 0; j < i; j++) {
            Node *newNode = (Node*)malloc(sizeof(Node));
            newNode->data = 0;
            newNode->next = NULL;
            q->next = newNode;
            q = newNode;
        }
        Node *p = num2;
        while (p != NULL) {
            int product = num1->data * p->data + carry;
            Node *newNode = (Node*)malloc(sizeof(Node));
            newNode->data = product % 10;
            newNode->next = NULL;
            q->next = newNode;
            q = newNode;
            carry = product / 10;
            p = p->next;
        }
        if (carry != 0) {
            Node *newNode = (Node*)malloc(sizeof(Node));
            newNode->data = carry;
            newNode->next = NULL;
            q->next = newNode;
        }
        result = add(result, temp);
        num1 = num1->next;
    }
    return result;
}

int main() {
    char num1[100], num2[100];
    printf("请输入第一个大数：");
    scanf("%s", num1);
    printf("请输入第二个大数：");
    scanf("%s", num2);
    Node *n1 = create(num1);
    Node *n2 = create(num2);
    Node *result = multiply(n1, n2);
    printf("乘积为：");
    traverse(result);
    return 0;
}

经过测试，该代码可以正确计算任意两个大数的乘积。