长整数加法运算

Posted 2020-12-14 lancelee98

问题描述 :

假设2个任意长度的整数x、y分别由双向链表A和B存储，现要求设计一个算法，实现x+y。计算结果存储在链表C中。

 

说明：

由于A和B输出时需要从头至尾遍历，而做加法时需要从尾至头遍历，因此使用双向链表存储。

可以从长整数的低位开始拆分（4位为一组，即不超过9999的非负整数），依次存放在链表的每个结点的数据域中；头结点的数据域存放正负数标志（正数或0：1，负数：-1）。

 

输入说明 :

第一行：长整数x

第二行：长整数y 

输出说明 :

第一行：格式化后的长整数x（从低位到高位每4位用","分开）

第二行：格式化后的长整数y（从低位到高位每4位用","分开）

第三行：空行

第四行：单链表C的遍历结果

第五行：格式化后的计算结果（从低位到高位每4位用","分开）

 

输入范例 :

-53456467576846547658679870988098
435643754856985679

输出范例 :

-5345,6467,5768,4654,7658,6798,7098,8098
43,5643,7548,5698,5679

5345->6467->5768->4611->2014->9250->1400->2419
-5345,6467,5768,4611,2014,9250,1400,2419

  1 #define _CRT_SECURE_NO_WARNINGS
  2 #include <stdio.h>
  3 #include <stdlib.h>
  4 #include <string.h>
  5 #include <iostream>
  6 #include <algorithm>
  7 #include<iomanip>
  8 #include <vector>
  9 #define MY_MAX_LEN 100000
 10 using namespace std;
 11 typedef struct node
 12 {
 13     int  val;//做头结点时 -1为负数 0为正数
 14     struct node* next;
 15     struct node* pre;
 16     node() {}
 17     node(int v) :val(v), next(0), pre(0) {}
 18 }Node;
 19 void display(Node* head)
 20 {
 21     Node* p = head->next;
 22     while (p && p->val == 0)p = p->next;
 23     if (!p) { cout << 0 << endl; return; }
 24     if (p)
 25     {
 26         cout << p->val;
 27         p = p->next;
 28     }
 29     while (p)
 30     {
 31         cout << "->";
 32         cout << setw(4) << setfill(‘0‘) << p->val;
 33         p = p->next;
 34     }
 35     cout << endl;
 36 }
 37 void my_print(Node* head)
 38 {
 39     Node* p = head->next;
 40     while (p && p->val == 0)p = p->next;
 41     if (!p) { cout << 0 << endl; return; }
 42     if (head->val == -1)cout << "-";
 43     if (p)
 44     {
 45         cout << p->val;
 46         p = p->next;
 47     }
 48     while (p)
 49     {
 50         cout << ",";
 51         cout << setw(4) << setfill(‘0‘) << p->val;
 52         p = p->next;
 53     }
 54     cout << endl;
 55 }
 56 Node* creatByVector(vector<Node*>& node_vec)
 57 {
 58     int i;
 59     if (node_vec.size() >= 2)
 60     {
 61         node_vec[0]->next = node_vec[1];
 62         node_vec[node_vec.size() - 1]->pre = node_vec[node_vec.size() - 2];
 63         node_vec[node_vec.size() - 1]->next = NULL;
 64     }
 65     for (i = 1; i < node_vec.size() - 1; i++)
 66     {
 67         node_vec[i]->next = node_vec[i + 1];
 68         node_vec[i]->pre = node_vec[i - 1];
 69     }
 70     return node_vec[0];
 71 }
 72 //使用atoi时一定要记得将字符串末尾设为
 73 Node* creatByStr(char* str)
 74 {
 75     Node* head = new Node();
 76     if (str[0] == ‘-‘)
 77     {
 78         head->val = -1;
 79         strcpy(str, &str[1]);
 80     }
 81     else
 82         head->val = 0;
 83     vector<Node*> node_vec;
 84     node_vec.push_back(head);
 85     char temp[5];
 86     int len = strlen(str);
 87     int x = len % 4;
 88     if (x)
 89     {
 90         strncpy(temp, str, x);
 91         temp[x] = ‘‘;
 92         node_vec.push_back(new Node(atoi(temp)));
 93         len -= x;
 94         strcpy(str, &str[x]);
 95     }
 96     while (len)
 97     {
 98         strncpy(temp, str, 4);
 99         temp[4] = ‘‘;
100         node_vec.push_back(new Node(atoi(temp)));
101         strcpy(str, &str[4]);
102         len -= 4;
103     }
104     return creatByVector(node_vec);
105 }
106 //符号相同时 直接算加法
107 Node* add_fun(Node* head_a, Node* head_b)
108 {
109     //* ta = NULL, * tb = NULL,
110     Node* pa = head_a->next, * pb = head_b->next;
111     while (pa->next)pa = pa->next; //ta = pa;
112     while (pb->next)pb = pb->next; //tb = pb;
113     vector<Node*> node_vec;
114     int carry = 0, cur;
115     while (pa != head_a && pb != head_b)
116     {
117         cur = pa->val + pb->val + carry;
118         //cout << pa->val << " " << pb->val << " " << carry << " " << cur % 10000 << " " << endl;
119         carry = cur / 10000;
120         node_vec.push_back(new Node(cur % 10000));
121         pa = pa->pre;
122         pb = pb->pre;
123     }
124     while (pa != head_a)
125     {
126         cur = pa->val + carry;
127         carry = cur / 10000;
128         node_vec.push_back(new Node(cur % 10000));
129         pa = pa->pre;
130     }
131     while (pb != head_b)
132     {
133         cur = pb->val + carry;
134         carry = cur / 10000;
135         node_vec.push_back(new Node(cur % 10000));
136         pb = pb->pre;
137     }
138     if (carry)
139         node_vec.push_back(new Node(carry));
140     node_vec.push_back(new Node(head_a->val));
141     //for (auto i : node_vec)cout << i->val << ‘ ‘;
142     //cout << endl;
143     reverse(node_vec.begin(), node_vec.end());
144     //for (auto i : node_vec)cout << i->val<<‘ ‘;
145     //cout << endl;
146     return creatByVector(node_vec);
147 }
148 //符号相反时 比较绝对值的大小 最后的符号与绝对值大的相同 head_a为绝对值大的数
149 Node* sub_fun(Node* head_a, Node* head_b)
150 {
151     //* ta = NULL, * tb = NULL,
152     Node* pa = head_a->next, * pb = head_b->next;
153     while (pa->next)pa = pa->next; //ta = pa;
154     while (pb->next)pb = pb->next; //tb = pb;
155     vector<Node*> node_vec;
156     int carry = 0, cur;
157     while (pa != head_a && pb != head_b)
158     {
159         //cout << pa->val << " " << pb->val << " " << carry << " " << pa->val - carry - pb->val 
160         //    << " " << pa->val + 10000 - carry - pb->val<< endl;
161         if (pa->val - carry >= pb->val)//若可以减
162         {
163             node_vec.push_back(new Node(pa->val - carry - pb->val));
164             //cout << "1 "<<pa->val - carry - pb->val << endl;
165             carry = 0;
166         }
167         else//若不够减
168         {
169             node_vec.push_back(new Node(pa->val + 10000 - carry - pb->val));
170             //cout << "2 " << pa->val + 10000 - carry - pb->val << endl;
171             carry = 1;
172         }
173         pa = pa->pre;
174         pb = pb->pre;
175     }
176     while (pa != head_a)//被减数还有剩余
177     {
178         cur = pa->val - carry;
179         if (cur > 0)
180         {
181             node_vec.push_back(new Node(cur));
182             //cout << "3 " << cur << endl;
183             carry = 0;
184         }
185         else if (cur < 0)
186         {
187             node_vec.push_back(new Node(cur + 1000));
188             //cout << "4 " << cur+10000 << endl;
189             carry = 1;
190         }
191         pa = pa->pre;
192     }
193     node_vec.push_back(new Node(head_a->val));
194     //for (auto i : node_vec)cout << i->val << ‘ ‘;
195     //cout << endl;
196     reverse(node_vec.begin(), node_vec.end());
197     //for (auto i : node_vec)cout << i->val<<‘ ‘;
198     //cout << endl;
199     return creatByVector(node_vec);
200 }
201 Node* my_fun(Node* ha, Node* hb, int flag)
202 {
203     if (ha->val == hb->val) return add_fun(ha, hb);
204     else if (flag == -1) return sub_fun(hb, ha);
205     else if (flag == 1)return sub_fun(ha, hb);
206     else return new Node(0);
207 }
208 //比较a与b的绝对值 谁大
209 int judge(char* a, char* b)
210 {
211     char ta[1000], tb[1000];
212     if (a[0] == ‘-‘)strcpy(ta, &a[1]);
213     else strcpy(ta, a);
214     if (b[0] == ‘-‘)strcpy(tb, &b[1]);
215     else strcpy(tb, b);
216     int lena = strlen(ta), lenb = strlen(tb);
217     if (lena > lenb) return 1;
218     else if (lenb > lena) return -1;
219     else
220     {
221         if (!strcmp(ta, tb))return 0;
222         int i = 0;
223         while (i != lena)
224         {
225             if (ta[i] > tb[i])return 1;
226             else if (ta[i] < tb[i])return -1;
227             i++;
228         }
229     }
230 }
231 int main()
232 {
233     char str_a[MY_MAX_LEN];
234     char str_b[MY_MAX_LEN];
235     scanf("%s", str_a); getchar();
236     scanf("%s", str_b);
237     int flag = judge(&str_a[0], &str_b[0]);
238     Node* head_a = creatByStr(str_a);
239     Node* head_b = creatByStr(str_b);
240     my_print(head_a);
241     my_print(head_b);
242     cout << endl;
243     Node* res = my_fun(head_a, head_b, flag);
244     display(res);
245     my_print(res);
246     return 0;
247 }