1006 Sign In and Sign Out (25)(25 分) At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed…

https://pintia.cn/problem-sets/994805342720868352/problems/994805516654460928 At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of…

PATA1006 Sign In and Sign Out At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to fin…

题目分析:由于不存在相同的两个时间(24:00:00和00:00:00不会同时存在),则我们假设两个全局变量存放到达的最早的时间和达到的最晚的时间,设置最早的初值为“23:59:59”,设置最晚的初值为“00:00:00”,只要一个人到达的时间比最早的早则更新最早时间同时将id记录下来,最晚的情况也是同样的 #include<iostream> #include<string> using namespace std; int main(){ int n; while(scanf(…

给定一个链表,你需要删除那些绝对值相同的节点,对于每个绝对值K,仅保留第一个出现的节点.删除的节点会保留在另一条链表上.简单来说就是去重,去掉绝对值相同的那些.先输出删除后的链表,再输出删除了的链表. 建立结构体节点,包括起始地址addr,下一个地址to,值value.链表数组索引为地址,接下来就是模拟链表的操作了,并且建立一个flag数组标记对应值K是否出现,若出现则flag[k]=addr,未出现则为-1,注意这里不能为0因为地址值存在为0的情况.最后的输出地址前面要补0,如果是链尾的话,-…

我先在CSDN上面发表了同样的文章,见https://blog.csdn.net/weixin_44385565/article/details/88863693 排版比博客园要好一些.. 1135 Is It A Red-Black Tree (30 分) There is a kind of balanced binary search tree named red-black tree in the data structure. It has the following 5 proper…

1043 Is It a Binary Search Tree (25 分)   A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than the node's key. The right subtree of a…

1033 To Fill or Not to Fill (25 分)   With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may…

题意: 输入一个正整数N(<=10000),接着依次输入N个学生的ID.输入一个正整数Q,接着询问Q次,每次输入一个学生的ID,如果这个学生的ID不出现在之前的排行榜上输出Are you kidding,否则如果已经询问过输出Checked,否则如果这位学生排名第一输出Mystery Award,否则如果这位学生的排名是质数输出Minion,否则输出Chocolate. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/…

题意: 输入一个正整数N(<=1000),接着输入N个整数([-1000,1000]),依次插入一棵初始为空的二叉排序树.输出最底层和最底层上一层的结点个数之和,例如x+y=x+y. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; typedef struct Node{ int value; int vis; int level; int visl,…

题意:输入一个正整数N(<=1e5),两个小数P和R,分别表示树的结点个数和商品原价以及每下探一层会涨幅的百分比.输出叶子结点深度最小的商品价格和深度最小的叶子结点个数. trick: 测试点1只有根节点一个点,输出P和1. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; vector<]; ]; void dfs(int x,int y){ )…

题意: 输入一个正整数N(<=100),接着输入N行每行包括0~N-1结点的左右子结点,接着输入一行N个数表示数的结点值.输出这颗二叉排序树的层次遍历. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; pair<]; ]; ; ]; void dfs(int x){ ) return ; dfs(a[x].first); ans[x]=b[++c…

题意: 输入一个地址和一个正整数N(<=100000),接着输入N行每行包括一个五位数的地址和一个结点的值以及下一个结点的地址.输出除去具有相同绝对值的结点的链表以及被除去的链表(由被除去的结点组成的链表). AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ],val[]; vector<pair<int,int> >ans,a…

题意: 输入一个正整数N(<=1e5),和两个小数r和f,表示树的结点总数和商品的原价以及每向下一层价格升高的幅度.下一行输入N个结点的父结点,-1表示为根节点.输出最深的叶子结点处购买商品的价格以及有几个深度最深的结点. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; vector<]; int mx; ]; void dfs(int x…

题意: 输入两个正整数N和K(2<=N<=200),代表城市的数量和道路的数量.接着输入起点城市的名称(所有城市的名字均用三个大写字母表示),接着输入N-1行每行包括一个城市的名字和到达该城市所能获得的快乐,接着输入M行每行包括一条道路的两端城市名称和道路的长度.输出从起点城市到目标城市"ROM"获得最大快乐且经历道路长度最短的道路条数和经历的道路长度和获得的快乐总数以及其中经过城市最少的那条路的到达每个城市所获得的平均快乐.下一行输出这条路的路径,城市名之间用"…

题意: 输入一个正整数N(<=1e5),表示共有N个结点,接着输入两个浮点数分别表示商品的进货价和每经过一层会增加的价格百分比.接着输入N行每行包括一个非负整数X,如果X为0则表明该结点为叶子结点接着输入一个整数表示该零售商进货的数量,X不为零则接着输入X个正整数表示它的下级经销商是哪些结点.输出所有零售商进货的总价.(结点从0~N-1,0为根节点即供应商) AAAAAccepted code: #include<bits/stdc++.h> using namespace std; ]…

题意: 输入一个四位的正整数N,输出每位数字降序排序后的四位数字减去升序排序后的四位数字等于的四位数字,如果数字全部相同或者结果为6174(黑洞循环数字)则停止. trick: 这道题一反常态的输入的数字是一个int类型而不是包含前导零往常采用字符串的形式输入,所以在测试点2,3,4如果用字符串输入会超时..... AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace st…

题意: 输入一个正整数N(<=100000),接着输入N个正整数(0~N-1的排列).每次操作可以将0和另一个数的位置进行交换,输出最少操作次数使得排列为升序. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ],b[]; int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(…

题意: 输入三个整数A,B,C(long long范围内),输出是否A+B>C. trick: 测试点2包括溢出的数据,判断一下是否溢出即可. AAAAAccepted code: #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 t; ci…

题目分析: 这题我在写的时候在PTA提交能过但是在牛客网就WA了一个点,先写一下思路留个坑 这题的简单来说就是需要找一条最短路->最开心->点最少(平均幸福指数自然就高了),由于本题给出的所有的城市的名称都是字符串(三个大写字母),所以我先将每个字符串计算一个hash值去代表每一个城市,接着就是用dijkstra算法去逐步判断各种情况,是直接用了一个dijkstra去计算,而没有说是先dijkstra算出最短距离再dfs所有的路径去找出最优解,因为我觉得这题局部最优就能反映到全局最优,也不知是…

最短路径 Emergency (25)-PAT甲级真题(Dijkstra算法) Public Bike Management (30)-PAT甲级真题(Dijkstra + DFS) Travel Plan (30)-PAT甲级真题(Dijkstra + DFS,输出路径,边权) All Roads Lead to Rome (30)-PAT甲级真题-Dijkstra + DFS Online Map (30)-PAT甲级真题(Dijkstra + DFS) 最短路径扩展问题 要求数最短路径有多…

At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to find the ones who have unlocked a…

PAT (Advanced Level) Practice 1006 Sign In and Sign Out (25 分) 凌宸1642 题目描述: At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of s…

1006. Sign In and Sign Out (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door…

1006 Sign In and Sign Out(25 分) At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to f…

At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door.  Given the records of signing in's and out's, you are supposed to find the ones who have unlocked…

At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in’s and out’s, you are supposed to find the ones who have unlocked a…

At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to find the ones who have unlocked a…

At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to find the ones who have unlocked a…

At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to find the ones who have unlocked a…