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: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…
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…
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)(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 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…
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…
题目传送门 题意:已知每个人的独一无二的身高以及排在他前面或者后面比他高的人数,问身高字典序最小的排法 分析:首先对身高从矮到高排序,那么可以知道每个人有多少人的身高比他高,那么取较小值(k[i], n - i - k[i]),若后者小于0则无解.然后可以理解为每个人前面要留出p + 1个位子给高个的人,可用线段树维护,s[rt] 表示当前线段还能空出的位子数.当然也能用树状数组+二分查找位子的方法. /**********************************************…
牛客多校第3场 J 思维+树状数组+二分 传送门:https://ac.nowcoder.com/acm/contest/883/J 题意: 给你q个询问,和一个队列容量f 询问有两种操作: 0.访问操作,如果询问的name在队列中存在的话,那么就输出队列中name对应的val值,然后将队列中name对应的元素扔到队列的尾部去,否则就直接将该元素插入到队列的尾部去 1.插入操作,得到队列中对应的name元素的v值为k,查询第k+v个元素的v值是多少 题解: 已知,对于插入操作,我们需要很快的查询…