1057 Stack (30)(30 分) 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)…

PAT甲级1057. Stack 题意: 堆栈是最基础的数据结构之一,它基于"先进先出"(LIFO)的原理.基本操作包括Push(将元素插入顶部位置)和Pop(删除顶部元素).现在你应该实现一个额外的操作堆栈:PeekMedian - 返回堆栈中所有元素的中间值.对于N个元素,如果N是偶数,则将中值定义为(N / 2)个最小元素,或者如果N是奇数则将其定义为((N + 1)/ 2). 输入规格: 每个输入文件包含一个测试用例.对于每种情况,第一行包含正整数N(<= 105).然后…

1057 Stack (30 分)   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).…

1057 Stack (30 分) 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). No…

题目地址:http://pat.zju.edu.cn/contests/pat-a-practise/1057 用树状数组和二分搜索解决,对于这种对时间复杂度要求高的题目,用C的输入输出显然更好 #include <cstdio> #include <string> #include <vector> #include <stack> using namespace std; ; struct TreeArray { vector<int> cS…

分析: 考察树状数组 + 二分, 注意以下几点: 1.题目除了正常的进栈和出栈操作外增加了获取中位数的操作, 获取中位数,我们有以下方法: (1):每次全部退栈,进行排序,太浪费时间,不可取. (2):题目告诉我们key不会超过10^5,我们可以想到用数组来标记,但不支持快速的统计操作. (3):然后将数组转为树状数组,可以快速的统计,再配上二分就OK了. 2.二分中我们需要查找的是一点pos,sum(pos)正好是当前个数的一半,而sum(pos - 1)就不满足. #include <ios…

树状数组+二分. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<string> #include<stack> #include<vector> using namespace…

题目如下: 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 su…

题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805417945710592 题意:对一个栈进行push, pop和找中位数三种操作. 思路: 好久没写题.感觉傻逼题写多了稍微有点数据结构的都不会写了. pop和push操作就不说了. 找中位数的话就二分去找某一个数前面一共有多少小于他的数,找到那个小于他的数刚好等于一半的. 找的过程中要用到前缀和,所以自然而然就应该上树状数组. 要注意树状数组的界应该是1e5而…

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 supposed…

不懂树状数组的童鞋,正好可以通过这道题学习一下树状数组~~百度有很多教程的,我就不赘述了 题意:有三种操作,分别是1.Push key:将key压入stack2.Pop:将栈顶元素取出栈3.PeekMedian:返回stack中第(n+1)/2个小的数 建立一个栈来模拟push和pop,另外还需要树状数组,来统计栈中<=某个数的总个数不了解树状数组的建议学习一下,很有用的.树状数组为c,有个虚拟的a数组,a[i]表示i出现的次数sum(i)就是统计a[1]~a[i]的和,即1~i出现的次数当我要…

https://pintia.cn/problem-sets/994805342720868352/problems/994805417945710592 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 o…

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 supposed…

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 supposed…

题意: 输入一个正整数N(<=1e5),接着输入N行字符串,模拟栈的操作,非入栈操作时输出中位数.(总数为偶数时输入偏小的) trick: 分块操作节约时间 AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; string s; stack<int>sk; ],block[]; int main(){ ios::sync_with_stdio(…

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 supposed…

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 supposed…

1057. Stack (30) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 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 el…

1057 Stack (30分)   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). N…

1057. Stack 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…

1001. A+B Format (20) 注意负数,没别的了. 用scanf来补 前导0 和 前导的空格 很方便. #include <iostream> #include <cstdio> using namespace std; ]; int main() { int A,B; cin>>A>>B; A+=B; ) { A=-A; cout<<"-"; } ; while(A) { a[n++]=A%; A/=; } ;…

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; ) {…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…

树(23) 备注 1004 Counting Leaves   1020 Tree Traversals   1043 Is It a Binary Search Tree 判断BST,BST的性质 1053 Path of Equal Weight   1064 Complete Binary Search Tree 完全二叉树的顺序存储,BST的性质 1066 Root of AVL Tree 构建AVL树,模板题,需理解记忆 1079 Total Sales of Supply Chain…

浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1051. Pop Sequence (25) 题意 给定 stack 的容量,给定数据的入栈顺序:从 1 开始的正整数序列,在允许随机的出栈操作的情况下,要求判断某出栈序列是否可能. 比如,告知 stack 容量为 5,入栈序列的最大值为 7.有两个序列需要判断合理性: {1 2 3 4 5 6 7}: 这个序列是可能的,只需每次入栈时都做出栈操作. {3 2 1 7 5 6 4}: 这个序列是不可能的,其中前半部分 3 2 1 是合法的,…

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…

本文为PAT甲级分类汇编系列文章. 理论这一类,是让我觉得特别尴尬的题,纯粹是为了考数据结构而考数据结构.看那Author一栏清一色的某老师,就知道教数据结构的老师的思路就是和别人不一样. 题号 标题 分数 大意 Author 1051 Pop Sequence 25 判断一个序列是否是pop序列 CHEN, Yue 1052 Linked List Sorting 25 链表排序 CHEN, Yue 1057 Stack 30 一个有中位数功能的stack CHEN, Yue 1074 Rev…

今天开个坑,分类整理PAT甲级题目(https://pintia.cn/problem-sets/994805342720868352/problems/type/7)中1051~1100部分.语言是modern C++. 为什么要整理呢,因为我2019年9月要考PAT甲级,虽然是第一次考,虽然只学了数据结构(https://mooc.study.163.com/course/1000033001?tid=2402970002#/info),但我要冲着高分(2019年9月8日更新:满分)去. 下…

http://www.bnuoj.com/bnuoj/problem_show.php?pid=1057 [题意]:给定x的值,带入f(x)求函数值 [题解]:注意第一个数的符号可能是'+',这里把我坑死了... [code]: #include <iostream> #include <stdio.h> #include <math.h> #include <string.h> #include <algorithm> using namesp…

题目链接 1057: [ZJOI2007]棋盘制作 Time Limit: 20 Sec  Memory Limit: 162 MBSubmit: 2027  Solved: 1019[Submit][Status][Discuss] Description 国际象棋是世界上最古老的博弈游戏之一,和中国的围棋.象棋以及日本的将棋同享盛名.据说国际象棋起源于易经的思想,棋盘是一个8*8大小的黑白相间的方阵,对应八八六十四卦,黑白对应阴阳.而我们的主人公小Q,正是国际象棋的狂热爱好者.作为一个顶尖高…