两个大数相乘

本文介绍如何使用C++编程语言实现两个大数的乘法运算,通过字符串表示大数并利用数组存储乘积结果。
两个大数相乘 char* multiply(char*,char*);

给了两个字符串,每个都是代表了一个很长的10进制表示的数

比如 char str1[] = "23456789009877666555544444"
char str2[] = "346587436598437594375943875943875"
最后求出他们的乘积


string multiply(string num1, string num2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> res(num1.size () + num2.size (), 0);
        for (int i = 0; i < num1.size (); i++) {
            int carry = 0;
            for (int j = 0; j < num2.size (); j++) {
                int d1 = num1[num1.size () - 1 - i] - '0';
                int d2 = num2[num2.size () - 1 - j] - '0';
                carry = d1 * d2 + res[i + j] + carry;
                res[i + j] = carry % 10;
                carry = carry / 10;
            }
            res[i + num2.size ()] = carry;
        }
        
        int i = num1.size () + num2.size () - 1;
        while (i > 0 && res[i] == 0) i--;
        string result;
        while (i >= 0) result += char (res[i --] + '0');
        
        return result;
    }



以下是一个基于链表的 C 语言代码,用于计算两个大数的乘积。 ```c #include <stdio.h> #include <stdlib.h> struct node { int data; struct node *next; }; typedef struct node Node; Node *create_node(int num) { Node *new_node = (Node *) malloc(sizeof(Node)); new_node->data = num; new_node->next = NULL; return new_node; } Node *insert_at_end(Node *head, int num) { Node *new_node = create_node(num); if (head == NULL) { head = new_node; return head; } Node *temp = head; while (temp->next != NULL) { temp = temp->next; } temp->next = new_node; return head; } Node *reverse_list(Node *head) { Node *prev = NULL, *current = head, *next = NULL; while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; } return prev; } Node *multiply_lists(Node *head1, Node *head2) { Node *result = NULL; Node *temp1 = head1, *temp2; int carry = 0, sum; while (temp1 != NULL) { temp2 = head2; Node *new_node = NULL; carry = 0; while (temp2 != NULL) { sum = (temp1->data * temp2->data) + carry; carry = sum / 10; new_node = insert_at_end(new_node, sum % 10); temp2 = temp2->next; } if (carry > 0) { new_node = insert_at_end(new_node, carry); } new_node = reverse_list(new_node); for (int i = 0; i < temp1->data; i++) { new_node = insert_at_end(new_node, 0); } result = add_lists(result, new_node); temp1 = temp1->next; } return result; } Node *add_lists(Node *head1, Node *head2) { Node *result = NULL; Node *temp1 = head1, *temp2 = head2; int carry = 0, sum; while (temp1 != NULL || temp2 != NULL) { sum = carry + (temp1 != NULL ? temp1->data : 0) + (temp2 != NULL ? temp2->data : 0); carry = sum / 10; result = insert_at_end(result, sum % 10); if (temp1 != NULL) { temp1 = temp1->next; } if (temp2 != NULL) { temp2 = temp2->next; } } if (carry > 0) { result = insert_at_end(result, carry); } return result; } void display_list(Node *head) { Node *temp = head; while (temp != NULL) { printf("%d", temp->data); temp = temp->next; } printf("\n"); } int main() { Node *num1 = NULL, *num2 = NULL, *result = NULL; char c; int num; printf("Enter first number:\n"); while ((c = getchar()) != '\n') { num = c - '0'; num1 = insert_at_end(num1, num); } printf("Enter second number:\n"); while ((c = getchar()) != '\n') { num = c - '0'; num2 = insert_at_end(num2, num); } num1 = reverse_list(num1); num2 = reverse_list(num2); result = multiply_lists(num1, num2); printf("Result:\n"); display_list(result); return 0; } ``` 该代码将两个大数存储为链表,并使用一个辅助函数来反转链表。然后,它将两个链表相乘,产生一个新的链表,该链表包含结果。最后,它将结果链表打印到控制台上。
评论
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值