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…

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…

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…

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…

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…

1147 Heaps (30 分)   In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal to (in a max heap) or less than or…

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(…

题目描述 Description Input Output 输出 q 行,第 i 行表示数据 Di 的答案. Sample Input 4 3 2 1 1 2 4 2 1 2 1 1 3 5 2 2 3 Sample Output 0 4 14 22 Data Constraint 题解 显然cdq+线段树,计算加上每个操作的贡献 修改取决于之后的询问,询问取决于之前的修改 然而线段树会T,所以用树状数组差分一下 开两个树状数组,询问1~t表示成s1+s2*t,修改t就在t处的s1减(t-1)*…