#define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ][]; ][]; ]; string t; int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); ;i<=;++i) ;j<=;++j) cin>>s[i][j]; cin.ignore(); getline(cin,t); ,cnt…

#define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; typedef struct{ int add,data,nex; }Node; Node node[],ans[],ans2[]; map<int,int>mp; int main(){ //ios::sync_with_stdio(false); //cin.tie(NULL); //cout.tie(NULL); int s,n,…

#define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; typedef struct node{ int data; node *lchild,*rchild; }tree; ]; tree *build(int l,int r){ if(l>r) return NULL; ,pos=; for(int i=l;i<=r;++i) if(a[i]<mn) mn=a[i],pos=i;…

#define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; ]; ]; int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n,m; cin>>n>>m; ;i<=m;++i){ int x,y; cin>>x>>y; v[x].insert(y);…

2019 秋季 PAT 甲级 备考总结 在 2019/9/8 的 PAT 甲级考试中拿到了满分,考试题目的C++题解记录在这里,此处对备考过程和考试情况做一个总结.如果我的方法能帮助到碰巧点进来的有缘人那就更好了Orz 目录 一.备考内容 1.  <算法笔记>(←阅读内容目录在这里) 2. PAT 题库刷题 3. 题型总结 + 刷题笔记 (←笔记目录在这里) 二.考试情况 1. 考前准备 2. 考试过程 3. 对这次考试的总结(碎碎念,可忽略) 一.备考内容 1.  <算法笔记>…

2019 秋季 PAT (Advanced Level) C++题解 考试拿到了满分但受考场状态和知识水平所限可能方法不够简洁,此处保留记录,仍需多加学习.备考总结(笔记目录)在这里 7-1 Forever (20 分) "Forever number" is a positive integer A with K digits, satisfying the following constrains: the sum of all the digits of A is m; the…

1080 Graduate Admission--PAT甲级练习题 It is said that in 2013, there were about 100 graduate schools ready to proceed over 40,000 applications in Zhejiang Province. It would help a lot if you could write a program to automate the admission procedure. Eac…

准备每天刷两题PAT真题.(一句话题解) 1001 A+B Format  模拟输出,注意格式 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; string ans = ""; int main() { ; cin >> a >> b; c = a + b; ) {…

链表处理 PAT (Advanced Level) Practice 链表题 目录 <算法笔记> 重点摘要:静态链表 1032 Sharing (25) 1052 Linked List Sorting (25) 1097 Deduplication on a Linked List (25) 1133 Splitting A Linked List (25) 附: 动态链表 <算法笔记> 7.3 链表 重点摘要 静态链表 ⭐ (1) 定义 结构体类型名和结构体变量名尽量不同 st…

排序题 PAT (Advanced Level) Practice 排序题 目录 <算法笔记> 6.9.6 sort()用法 <算法笔记> 4.1 排序题步骤 1012 The Best Rank (25) 1083 List Grades (25) 1137 Final Grading (25) 1141 PAT Ranking of Institutions (25) 1153 Decode Registration Card of PAT (25) <算法笔记>…

2019/4/3 1063 Set Similarity n个序列分别先放进集合里去重.在询问的时候,遍历A集合中每个数,判断下该数在B集合中是否存在,统计存在个数(分子),分母就是两个集合大小减去分子. // 1063 Set Similarity #include <set> #include <map> #include <cstdio> #include <iostream> #include <algorithm> using name…

1019. General Palindromic Number 题意:求数N在b进制下其序列是否为回文串,并输出其在b进制下的表示. 思路:模拟N在2进制下的表示求法,“除b倒取余”,之后判断是否回文. #include<iostream> #include<cstdio> using namespace std; ]; int main() { int n, b; scanf("%d%d", &n, &b); , tmp = n,yu=tmp…

正好这个"水水"的C4来了 先把甲级刷完吧.(开玩笑-2017.3.26) 这是一套"伪题解". wacao 刚才登出账号测试一下代码链接,原来是看不到..有空弄题(xia)解(che).. //"今天"的"收获": BST的中序遍历特性:栈那题利用树状数组+二分维护:哇咔咔,题意真难,根据案例猜题意.好吧,反正我又不考.不如睡觉? 出现了基础的二级最短路: 对DFS联通快愣了一下.. 哇塞呀!模拟题还是很劲的! 哇塞呀,手写…

模拟题 PAT (Advanced Level) Practice 模拟题 目录 1008 Elevator (20) 1042 Shuffling Machine (20) 1046 Shortest Distance (20) 1051 Pop Sequence (25) 1117 Eddington Number (25) 1128 N Queens Puzzle (20) 1008 Elevator (20) #include<cstdio> int main() { int N; s…

题目概述:Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now…

题目分析: 由于本题字符串长度有10^5所以直接暴力是不可取的,猜测最后的算法应该是先预处理一下再走一层循环就能得到答案,所以本题的关键就在于这个预处理的过程,由于本题字符串匹配的内容的固定的PAT,所以我们可以这样想,对于一个输入的串,我们找到每个A的位置,只要知道这个A的前面有几个P,这个A的后面有几个T,就可以得到以这个A为中心的所有种数,二者相乘即可,然后如果我们能得到0~s.size()-1范围内每个A的前面有多少个P,每个A后面有多少个T,只要从头遍历一遍并且求和就能得到最终答案,由…

前言: 本题是我在浏览了柳神的代码后,记下的一次半转载式笔记,不经感叹柳神的强大orz,这里给出柳神的题解地址:https://blog.csdn.net/liuchuo/article/details/52316405 题意: 有一个租借自行车的系统由n个点组成,用户可以在这n个点任意一个点借车或者还车,每个点都遵循一个共同的上限c,这里我们称一个点的状态时完美的如果这个点的车的数量为c/2,(每个点由1~n标号),而我们的出发点为0点,整个系统每次会有一个sp点发出警报,说明这个sp点的自行…

1022 Digital Library A Digital Library contains millions of books, stored according to their titles, authors, key words of their abstracts, publishers, and published years. Each book is assigned an unique 7-digit number as its ID. Given any query fro…

1053 Path of Equal Weight 给定一个非空的树,树根为 RR. 树中每个节点 TiTi 的权重为 WiWi. 从 RR 到 LL 的路径权重定义为从根节点 RR 到任何叶节点 LL 的路径中包含的所有节点的权重之和. 现在给定一个加权树以及一个给定权重数字,请你找出树中所有的权重等于该数字的路径(必须从根节点到叶节点). 例如,我们考虑下图的树,对于每个节点,上方的数字是节点 ID,它是两位数字,而下方的数字是该节点的权重. 假设给定数为 2424,则存在 44 个具有相同…

题目: Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Each input file contains one test case. Each case contains a pair of int…

题意:有n个人,每个人有k个爱好,如果两个人有某个爱好相同,他们就处于同一个集合.问总共有多少个集合,以及每个集合有多少人,并按从大到小输出. 很明显,采用并查集.vis[k]标记爱好k第一次出现的人的编号,如果为0则表示未出现. 当前第i个人若也存在爱好k,则只要将i与vis[k]两个人合并即可. 最后father[i]相同的即处在同一个集合中. #include <iostream> #include <cstdio> #include <algorithm> #i…

题意:给了图,以及s和t,让你求s到t花费的最短路程.最短时间,以及输出对应的路径.   对于最短路程,如果路程一样,输出时间最少的. 对于最短时间,如果时间一样,输出节点数最少的.   如果最短路程和最短时间路径一样,合并输出一次即可. 纯粹就是练习dijkstra,没什么难的. 第一次dijkstra求最短路程,记录下每个节点的路程和时间. 第二次dijkstra求最短时间,记录下每个节点的时间和经过的节点数. pre数组用来存储前驱节点,保存路径 #include <iostream>…

1121 Damn Single 模拟 // 1121 Damn Single #include <map> #include <vector> #include <cstdio> #include <iostream> #include <algorithm> using namespace std; map<int, int> m, vis; vector<int> p; ; int a[N]; int main()…

1071 Speech Patterns People often have a preference among synonyms of the same word. For example, some may prefer "the police", while others may prefer "the cops". Analyzing such patterns can help to narrow down a speaker's identity, w…

1100 Mars Numbers People on Mars count their numbers with base 13: Zero on Earth is called "tret" on Mars. The numbers 1 to 12 on Earch is called "jan, feb, mar, apr, may, jun, jly, aug, sep, oct, nov, dec" on Mars, respectively. For t…

1060 Are They Equal If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123*105 with simple chopping. Now given the number of significant digits on a machine and two…

1047 Student List for Course Zhejiang University has 40,000 students and provides 2,500 courses. Now given the registered course list of each student, you are supposed to output the student name lists of all the courses. Input Specification: Each inp…

1004 Counting Leaves A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Specification: Each input file contains one test case. Each case starts with a line containing 0<N<10…

1094 The Largest Generation A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level belong to the same generation. Your task is to find the generation with the largest population. Input Specification: Each inp…

1090 Highest Price in Supply Chain A supply chain is a network of retailers(零售商), distributors(经销商), and suppliers(供应商)-- everyone involved in moving a product from supplier to customer. Starting from one root supplier, everyone on the chain buys pro…