任意长的整数加法运算



【问题描述】

      设计一个实现任意长的整数进行加法运算的演示程序。

【基本要求】

1. 系统以菜单提示方式工作；

2. 基本功能包括大整数输入、加法运算、大整数输出；

3. 利用双向循环链表实现长整数的存储，每个结点含一个整型变量。任何整型变量的范围是-（2¹⁵-1）～（2¹⁵-1）；

4. 输入和输出形式：按中国对于长整数的表示习惯，每四位一组，组间用逗号隔开。如：1，0000，0000，0000，0000

   【测试数据】

1. 0；0；应输入“0”。

2. -2345，6789；-7654，3211；应输出“-1，0000，0000”。

3. -9999，9999；1，0000，0000，0000；应输出“9999，0000，0001”。

4. 1，0001，0001；-1，0001，0001；应输出“0”。

5. 1，0001，0001；-1，0001，0000；应输出“1”。

6. -9999，9999，9999；-9999，9999，9999；应输出“-1，9999，9999，9998”。

7. 1，0000，9999，9999；1；应输出“1，0001，0000，0000”。

   //设计两个函数   一个纯加法计算   另一个减法运算#include "stdafx.h"#include <iostream>#include <string.h>#include <stdlib.h>#include <cstdlib>#include <string>using namespace std;//加数和被加数的存储结构struct Cltype{int num;Cltype *nextNode;Cltype *preNode;};//结果的存储结构struct Answer{int num = 0;int caorbo = 0;Answer *nextNode;Answer *preNode;};Cltype*FindEnd(Cltype*head)  //查找尾部节点{Cltype*htemp;htemp = head;while (htemp->nextNode != NULL){htemp = htemp->nextNode;}return htemp;}Answer*FindEnd(Answer*head)  //查找结果链表的尾部节点{Answer*htemp;htemp = head;while (htemp->nextNode != NULL){htemp = htemp->nextNode;}return htemp;}Cltype*AddnodeNum(int num, Cltype*head)  //添加节点（数据）{Cltype *htemp, *node;if (!(node = new Cltype)){cout << "分配内存失败";return NULL;}else{node->num = num;node->nextNode = NULL;if (head == NULL){head = node;return head;}htemp = head;while (htemp->nextNode != NULL){htemp = htemp->nextNode;}htemp->nextNode = node;node->preNode = htemp;return head;}}Answer*AddnodeAns(int num, Answer*head)  //添加存储结果的链表的节点{Answer *htemp, *node;if (!(node = new Answer)){cout << "分配内存失败";return NULL;}else{node->num = num;node->nextNode = NULL;if (head == NULL){head = node;return head;}htemp = head;while (htemp->nextNode != NULL){htemp = htemp->nextNode;}htemp->nextNode = node;node->preNode = htemp;return head;}}Answer*add(Cltype*head1, Cltype*head2, Answer*head3)  //调用这个函数之前应该把各个节点的数值存放进链表中{                                                      //相加函数Cltype*end1 = FindEnd(head1);Cltype*end2 = FindEnd(head2);Answer*End = FindEnd(head3);while (end2 != head2){End->num = end1->num + end2->num + End->caorbo;if (End->num > 9){End->num = End->num % 10;End->preNode->caorbo = 1;}end1 = end1->preNode;end2 = end2->preNode;End = End->preNode;}End->num = head2->num + end1->num + End->caorbo;  //head2此时就是end2，从尾遍历至头部if (End->num > 9)            //如果遍历至头部的话还是大于等于10 的，就进位{End->num = End->num % 10;End->preNode->caorbo = 1;}while (end1 != head1){end1 = end1->preNode;End = End->preNode;End->num = end1->num + End->caorbo;if (End->num > 9)            //如果遍历至头部的话还是大于等于10 的，就进位{End->num = End->num % 10;End->preNode->caorbo = 1;}}End = End->preNode;End->num = End->caorbo;return head3;}//******************************************//减法函数  其中head1是大数链表的头指针 head2是小数链表的头指针//******************************************Answer*subs(Cltype*head1, Cltype*head2, Answer*head3){Cltype*end1 = FindEnd(head1);Cltype*end2 = FindEnd(head2);Answer*End = FindEnd(head3);while (end2 != head2){End->num = end1->num - end2->num + End->caorbo;if (End->num < 0){End->num += 10;End->preNode->caorbo -= 1;}end1 = end1->preNode;end2 = end2->preNode;End = End->preNode;}End->num = end1->num - head2->num + End->caorbo;if (End->num < 0)          //如果最后一次还是小于0，那么就继续加一个条件，让借位等于-1{End->num += 10;End->preNode->caorbo -= 1;}while (end1 != head1){end1 = end1->preNode;End = End->preNode;End->num = end1->num + End->caorbo;if (End->num < 0)          //如果最后一次还是小于0，那么就继续加一个条件，让借位等于-1{End->num += 10;End->preNode->caorbo -= 1;}}return head3;}void AllNode(Answer*head)   //需要修改{int a[100];memset(a, 0, 100);int i = 0;Answer*temp, *thend;thend = FindEnd(head);temp = head;while (temp){if (temp->num == -1 || temp->num == 0)   //判断结果是不是都是0如果是则输出一个0{if (temp == thend)cout << "0";}if (temp->num != -1 && temp->num != 0)break;temp = temp->nextNode;}Answer*htemp;int flag = 0;htemp = head;while (htemp){if (htemp->num != -1 && htemp->num != 0 || flag == 1){a[i++] = htemp->num;flag = 1;}htemp = htemp->nextNode;}if (i > 4){int k, w = i % 4;for (k = 0; k < w; k++)cout << a[k];if (w != 0)        //通过这个判断语句让对四如果为空的后面不输出逗号cout << ",";for (; k < i; k++){for (int d = 0; d < 4; d++)    //四位四位一输出{cout << a[k];k++;}k--;if (k != i - 1)    //作为一个判断标志，让最后一位数的后面的标点符号不输出cout << ",";}}else{for (int g = 0; g < i; g++)cout << a[g];}}int count(string s){int cnt = 0;for (auto c : s)if (isdigit(c))++cnt;return cnt;}//****************************************//将最初的串中的字符剔除掉,并将符号存在下标为0的位置//****************************************void send(string org, char*&num, int n){int k = 1;if (isdigit(org[0]))num[0] = '1';    //1代表正数else num[0] = '0';   //0代表负数for (int i = 0; k < n, i < org.length(); i++){if (isdigit(org[i])){num[k] = org[i];k++;}}}//******************************************//return 1  num1的绝对值大于num2//******************************************int compare(string s1, string s2){int a = count(s1) + 1;   //除掉字符时候数字的个数+1是存储符号int b = count(s2) + 1;if (a > b) return 1;if (a < b) return 0;char*num1 = new char[a];char*num2 = new char[b];memset(num1, 0, a);        //初始化存储加数和被加数的字符型数组memset(num2, 0, b);send(s1, num1, a);         //调用函数将字符串中的符号去掉并将数字存在num1 num2中send(s2, num2, b);for (int i = 1; i < a; i++){if (num1[i]>num2[i])return 1;if (num1[i] < num2[i])return 0;}return 1;}Cltype*ass(Cltype*head, char*num, int n){for (int i = 1; i <= n - 1; i++){int su = num[i] - '0';head = AddnodeNum(su, head);}return head;}int _tmain(int argc, _TCHAR* argv[]){Cltype*head1 = NULL, *head2 = NULL;/***************我们要让Num1是大数*/Answer*head3 = NULL;//int length1, length2;string s1, s2;cin >> s1 >> s2;int a = count(s1) + 1;   //除掉字符时候数字的个数+1是存储符号int b = count(s2) + 1;char*num1 = new char[a];char*num2 = new char[b];memset(num1, 0, a);        //初始化存储加数和被加数的字符型数组memset(num2, 0, b);send(s1, num1, a);         //调用函数将字符串中的符号去掉send(s2, num2, b);head1 = ass(head1, num1, a);  //传入链表head2 = ass(head2, num2, b);int q = a > b ? a + 1 : b + 1;for (int j = 1; j < q; j++)head3 = AddnodeAns(-1, head3);if (compare(s1, s2)){if (num1[0] == '0' && num2[0] == '0'){add(head1, head2, head3);cout << "-";AllNode(head3);}if (num1[0] == '1' && num2[0] == '0'){subs(head1, head2, head3);AllNode(head3);}if (num1[0] == '0'&& num2[0] == '1'){subs(head1, head2, head3);cout << "-";AllNode(head3);}if (num1[0] == '1' && num2[0] == '1'){add(head1, head2, head3);AllNode(head3);}}else{if (num1[0] == '0' && num2[0] == '0'){add(head2, head1, head3);cout << "-";AllNode(head3);}if (num1[0] == '1' && num2[0] == '0'){subs(head2, head1, head3);cout << "-";AllNode(head3);}if (num1[0] == '0' && num2[0] == '1'){subs(head2, head1, head3);AllNode(head3);}if (num1[0] == '1' && num2[0] == '1'){add(head2, head1, head3);AllNode(head3);}}return 0;}