1057 Stack 树状数组】的更多相关文章

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)(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).…
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…
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…
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…
目录 题目大意 题目分析 题目大意 要求维护一个栈,提供压栈.弹栈以及求栈内中位数的操作(当栈内元素\(n\)为偶数时,只是求第\(n/2\)个元素而非中间两数的平均值).最多操作100000次,压栈的数字\(key\)范围是[1,100000]. 题目分析 前两个操作用\(stack\)就好. 求中位数.暴力做法即使用上优先队列也是稳稳的超时.考虑树状数组. 压栈时,将\(key\)值对应的位置加1.弹栈减1. 求中位数,可以二分求出\(sum[1:p]==(n+1)/2\)最小的\(p\),…
题意:给你n盘歌碟按照(1....n)从上到下放,接着m个询问,每一次拿出x碟,输出x上方有多少碟并将此碟放到开头 直接想其实就是一线段的区间更新,单点求值,但是根据题意我们可以这样想 首先我们倒着存  n--1,接着每次询问时把放碟子放到最后,这样我们要开一个映射数组映射每个碟子在哪个位置 其中我们需要使用树状数组维护每个绝对位置是否有碟子(有些碟子已经放到了后面了),再使用区间求和就好了 #include<set> #include<map> #include<queue…