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多思考少编码

PAT甲级真题1001 - 1005题详细题解（C++）（个人题解）本文为PAT甲级真题中的1001 - 1005题的详细题解，本文的解法仅个人的思路和解法，有什么更优解欢迎讨论（AI一键生成答案的年代还能坚持写题解吗？）

【PAT甲级真题】- Recover the Smallest Number (30)Recover the Smallest Number (30) 这是牛客上面最后的一个题了，之后会去写 PTA 上面的题

【PAT甲级真题】- Table Tennis (30)Table Tennis (30) - PTATable Tennis (30) - 牛客注意点：A table tennis club has N tables available to the public. The tables are numbered from 1 to N. For any pair of players, if there are some tables open when they arrive, they will be assigned to the available

【PAT甲级真题】- Favorite Color Stripe (30)Favorite Color Stripe (30)注意点Eva is trying to make her own color stripe out of a given one. She would like to keep only her favorite colors in her favorite order by cutting off those unwanted pieces and sewing the remaining parts together to form her favo

【PAT甲级真题】- Stack (30)Stack (30)注意点Stack is one of the most fundamental data structures, which is based on the principle of Last In First Out (LIFO). The basic operations include Push (inserting an element onto the top position) and Pop (deleting the top element). Now you are

【PAT甲级真题】- Cars on Campus (30)Cars on Campus (30)-牛客 Cars on Campus (30)-PTA注意点：Zhejiang University has 6 campuses and a lot of gates. From each gate we can collect the in/out times and the plate numbers of the cars crossing the gate. Now with all the information available, you are su

【PAT甲级真题】- All Roads Lead to Rome (30)All Roads Lead to Rome (30)注意点：Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness.

【PAT甲级真题】- Count PAT‘s (25)Count PAT’s (25)注意点The string APPAPT contains two PAT’s as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters.

【PAT甲级真题】- Insert or Merge (25)Insert or Merge (25)与Insertion or Heap Sort (25)这题几乎一致，只是一个是单步堆排，一个是单步归并

【PAT甲级真题】- Reversing Linked List (25)Reversing Linked List (25)题目描述点击链接自行查看注意点：Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you must

【PAT甲级真题】- Talent and Virtue (25)Talent and Virtue (25)About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people’s talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a “sage（圣人）”; being le

【PAT甲级真题】- Linked List Sorting (25)Linked List Sorting (25)题目描述点击链接自行查看注意点：一开始还在想写一个静态链表的归并排序然后发现合并的时候不知道怎么复制。。。后面反应过来发现好像根本不用自己写

2024 秋季PAT认证甲级(题解A1-A4)这一次PAT甲级应该是最近几次最简单的一次了，3个小时的比赛差不多30分钟就ak了（也是拿下了整场比赛的rk1），下面是题解报告，每个题目差不多都是20-30行代码，难度在洛谷普及组左右（cf 1000-1200分）

PAT 1097 Deduplication on a Linked List（25分）原题链接：PAT 1097 Deduplication on a Linked List（25分）Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolut

PTA题解 --- 静静的推荐（C语言）今天是PTA题库解法讲解的第七天，今天我们要讲解静静的推荐，题目如下：解题思路：这个问题的核心在于如何在满足给定条件的情况下，最大化推荐学生的数量。首先，我们需要过滤出所有天梯赛成绩不低于175分的学生。然后，我们要按天梯赛成绩排序，如果天梯赛成绩相同，再根据PAT成绩排序。在推荐学生时，我们需要按批次进行，确保每一批的成绩严格递增，同时如果同一天梯赛成绩的学生PAT成绩达到了企业的面试分数线，也可以被接受。

PAT (Basic Level) Practice | 判断题判断题的评判很简单，本题就要求你写个简单的程序帮助老师判题并统计学生们判断题的得分。输入在第一行给出两个不超过 100 的正整数 N 和 M，分别是学生人数和判断题数量。第二行给出 M 个不超过 5 的正整数，是每道题的满分值。第三行给出每道题对应的正确答案，0 代表“非”，1 代表“是”。随后 N 行，每行给出一个学生的解答。数字间均以空格分隔。

天梯赛L2-030 冰岛人https://pintia.cn/problem-sets/994805046380707840/exam/problems/1111914599412858888?type=7&page=1

PAT 1114 Family Property个人学习记录，代码难免不尽人意Sample Input: 10 6666 5551 5552 1 7777 1 100 1234 5678 9012 1 0002 2 300 8888 -1 -1 0 1 1000 2468 0001 0004 1 2222 1 500 7777 6666 -1 0 2 300 3721 -1 -1 1 2333 2 150 9012 -1 -1 3 1236 1235 1234 1 100 1235 5678 9012 0 1 50 2222 1236 2468 2 6

PAT 1097 Deduplication on a Linked List个人学习记录，代码难免不尽人意 Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolute value of its key equals K will

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