add a1022, a1032

2026-10-03 00:23:15 +08:00 · 2020-04-26 16:46:48 +08:00
parent 89028a2524
commit 856567f179
6 changed files with 581 additions and 0 deletions
@@ -0,0 +1,120 @@
+---
+layout: post
+date: 2020-04-26 15:32:53 +0800
+title:  "PAT Advanced 1022. Digital Library (30) (C语言实现)"
+categories: Advanced
+tags: []
+permalink: Advanced/1022.html
+---
+
+## 题目
+
+{% include_relative html/a1022.md %}
+
+## 思路
+
+{% include_relative analysis/a1022.md %}
+
+## 代码
+
+[Github最新代码](https://github.com/OliverLew/PAT/blob/master/PATAdvanced/1022.c)，欢迎交流
+
+```c
+{% raw %}#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+typedef struct book {
+    int ID;
+    char title[81];
+    char author[81];
+    char keywords[5][11];
+    char publisher[81];
+    int year;
+} Book;
+
+int cmpbyid(const void *a, const void *b)
+{
+    return (*(Book**)a)->ID - (*(Book**)b)->ID;
+}
+
+int main()
+{
+    int N, M, count, query_type, query_year;
+    char space, query_str[81];
+    Book books[10000] = {0}, *p, *booksbyid[10000];
+
+    scanf("%d", &N);
+    for(int i = 0; i < N; i++)
+    {
+        p = booksbyid[i] = books + i;
+        scanf("%d%c", &p->ID, &space);
+        scanf("%[^\n]%c", p->title, &space);
+        scanf("%[^\n]%c", p->author, &space);
+        space = '\0';
+        for(int k = 0; space != '\n'; k++)
+            scanf("%s%c", p->keywords[k], &space);
+        scanf("%[^\n]%c", p->publisher, &space);
+        scanf("%d", &p->year);
+    }
+
+    qsort(booksbyid, N, sizeof(Book*), cmpbyid);
+
+    scanf("%d", &M);
+    for(int i = 0; i < M; i ++)
+    {
+        count = 0;
+        scanf("%d: %[^\n]%c", &query_type, query_str, &space);
+        printf("%d: %s\n", query_type, query_str);
+        for(int j = 0; j < N; j++)
+        {
+            p = booksbyid[j];
+            switch(query_type)
+            {
+                case 1:
+                    if(strcmp(query_str, p->title) == 0)
+                    {
+                        printf("%07d\n", p->ID);
+                        count ++;
+                    }
+                    break;
+                case 2:
+                    if(strcmp(query_str, p->author) == 0)
+                    {
+                        printf("%07d\n", p->ID);
+                        count ++;
+                    }
+                    break;
+                case 3:
+                    for(int k = 0; k < 5; k ++)
+                        if(strcmp(query_str, p->keywords[k]) == 0)
+                        {
+                            printf("%07d\n", p->ID);
+                            count ++;
+                        }
+                    break;
+                case 4:
+                    if(strcmp(query_str, p->publisher) == 0)
+                    {
+                        printf("%07d\n", p->ID);
+                        count ++;
+                    }
+                    break;
+                case 5:
+                    sscanf(query_str, "%d", &query_year);
+                    if(query_year == p->year)
+                    {
+                        printf("%07d\n", p->ID);
+                        count ++;
+                    }
+                    break;
+                default:
+                    break;
+            }
+        }
+        if(count == 0) puts("Not Found");
+    }
+
+    return 0;
+}
+{% endraw %}```
@@ -0,0 +1,64 @@
+---
+layout: post
+date: 2020-04-26 16:28:50 +0800
+title:  "PAT Advanced 1032. Sharing (25) (C语言实现)"
+categories: Advanced
+tags: [linked list]
+permalink: Advanced/1032.html
+---
+
+## 题目
+
+{% include_relative html/a1032.md %}
+
+## 思路
+
+{% include_relative analysis/a1032.md %}
+
+## 代码
+
+[Github最新代码](https://github.com/OliverLew/PAT/blob/master/PATAdvanced/1032.c)，欢迎交流
+
+```c
+{% raw %}#include <stdio.h>
+
+typedef struct Node Node;
+struct Node {
+    int checked;
+    struct Node *next;
+};
+
+int main()
+{
+    char data;
+    int start1, start2, address, next, N;
+    Node list[100000] = {0}, *p;
+
+    scanf("%d %d %d", &start1, &start2, &N);
+    /* record a linked list */
+    for(int i = 0; i < N; i++)
+    {
+        scanf("%d %c %d", &address, &data, &next);
+        list[address].next = next == -1 ? NULL : &list[next];
+    }
+
+    if(start1 == -1 || start2 == -1)
+    {
+        printf("-1");
+        return 0;
+    }
+    /* First traverse the first string */
+    for(p = list + start1; p; p = p->next)
+        p->checked = 1;
+    /* Then traverse the second looking for checked node */
+    for(p = list + start2; p && !p->checked; p = p->next)
+        ;
+
+    if(p)
+        printf("%05ld", p - list);
+    else
+        printf("-1");
+
+    return 0;
+}
+{% endraw %}```
@@ -0,0 +1,12 @@
+<!--
+2020-04-26 15:32:53 +0800
+
+-->
+
+数据的存储与查询。
+
+使用了最简单暴力的结构数组来存储，然后简单地按照ID排下序，遍历数组核对就行了。
+
+这样写是很简单的，如果要使用复杂的数据结构肯定会更快。
+
+作为参考：最后一个测试点用了660毫秒，上限为1200。
@@ -0,0 +1,22 @@
+<!--
+2020-04-26 16:28:50 +0800
+linked list
+-->
+
+一个链表相关的题目。
+
+依然暴力地（再次说明，很不实用。不过最坏情况就需要这么多，所以对于OJ来说，不算浪费）申请一个100000长度的数组，直接用数组索引作为链表地址。
+
+思路很简单：
+```
+记录链表：
+    结构数组[Address] => 结构数组[Next]
+遍历第一个单词：
+    标记所有遍历的结点
+遍历第二个单词：
+    输出第一个被标记的结点
+```
+
+我定义的结构只有下一个结点的地址和一个标记，没有记录字符，因为你会发现它没用。。。
+
+**盲点**：单词开头地址可能为`NULL`。盲生，你是否发现这个了华点？
@@ -0,0 +1,150 @@
+### 我的PAT系列文章更新重心已移至Github，欢迎来看PAT题解的小伙伴请到[Github Pages](https://oliverlew.github.io/PAT)浏览最新内容([本篇文章链接](https://oliverlew.github.io/PAT/Advanced/1032.html))。此处文章目前已更新至与Github Pages同步。欢迎star我的[repo](https://github.com/OliverLew/PAT)。
+
+## 题目
+
+To store English words, one method is to use linked lists and store a word
+letter by letter. To save some space, we may let the words share the same
+sublist if they share the same suffix. For example, `loading` and `being` are
+stored as showed in Figure 1.
+
+![fig.jpg](https://images.ptausercontent.com/ef0a1fdf-3d9f-46dc-9a27-21f989270fd4.jpg)
+
+Figure 1
+
+You are supposed to find the starting position of the common suffix (e.g. the
+position of `i` in Figure 1).
+
+### Input Specification:
+
+Each input file contains one test case. For each case, the first line contains
+two addresses of nodes and a positive $N$ ( $\le 10^5$ ), where the two
+addresses are the addresses of the first nodes of the two words, and $N$ is
+the total number of nodes. The address of a node is a 5-digit positive
+integer, and NULL is represented by $-1$ .
+
+Then $N$ lines follow, each describes a node in the format:
+
+    
+    
+    Address Data Next
+    
+
+where`Address` is the position of the node, `Data` is the letter contained by
+this node which is an English letter chosen from { a-z, A-Z }, and `Next` is
+the position of the next node.
+
+### Output Specification:
+
+For each case, simply output the 5-digit starting position of the common
+suffix. If the two words have no common suffix, output `-1` instead.
+
+### Sample Input 1:
+
+    
+    
+    11111 22222 9
+    67890 i 00002
+    00010 a 12345
+    00003 g -1
+    12345 D 67890
+    00002 n 00003
+    22222 B 23456
+    11111 L 00001
+    23456 e 67890
+    00001 o 00010
+    
+
+### Sample Output 1:
+
+    
+    
+    67890
+    
+
+### Sample Input 2:
+
+    
+    
+    00001 00002 4
+    00001 a 10001
+    10001 s -1
+    00002 a 10002
+    10002 t -1
+    
+
+### Sample Output 2:
+
+    
+    
+    -1
+    
+
+
+
+## 思路
+
+
+一个链表相关的题目。
+
+依然暴力地（再次说明，很不实用。不过最坏情况就需要这么多，所以对于OJ来说，不算浪费）申请一个100000长度的数组，直接用数组索引作为链表地址。
+
+思路很简单：
+```
+记录链表：
+    结构数组[Address] => 结构数组[Next]
+遍历第一个单词：
+    标记所有遍历的结点
+遍历第二个单词：
+    输出第一个被标记的结点
+```
+
+我定义的结构只有下一个结点的地址和一个标记，没有记录字符，因为你会发现它没用。。。
+
+**盲点**：单词开头地址可能为`NULL`。盲生，你是否发现这个了华点？
+
+## 代码
+
+[最新代码@github](https://github.com/OliverLew/PAT/blob/master/PATAdvanced/1032.c)，欢迎交流
+```c
+#include <stdio.h>
+
+typedef struct Node Node;
+struct Node {
+    int checked;
+    struct Node *next;
+};
+
+int main()
+{
+    char data;
+    int start1, start2, address, next, N;
+    Node list[100000] = {0}, *p;
+
+    scanf("%d %d %d", &start1, &start2, &N);
+    /* record a linked list */
+    for(int i = 0; i < N; i++)
+    {
+        scanf("%d %c %d", &address, &data, &next);
+        list[address].next = next == -1 ? NULL : &list[next];
+    }
+
+    if(start1 == -1 || start2 == -1)
+    {
+        printf("-1");
+        return 0;
+    }
+    /* First traverse the first string */
+    for(p = list + start1; p; p = p->next)
+        p->checked = 1;
+    /* Then traverse the second looking for checked node */
+    for(p = list + start2; p && !p->checked; p = p->next)
+        ;
+
+    if(p)
+        printf("%05ld", p - list);
+    else
+        printf("-1");
+
+    return 0;
+}
+```