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 (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…

超时算法,利用2的特殊性,用2个multiset来维护.单个multiset维护没法立即找到中位数. 其实也可以只用1个multiset,用一个中位指针,++,--来维护中位数. #include<iostream> #include<stdio.h> #include<math.h> #include<string.h> #include<stdlib.h> #include<algorithm> #include<queue…

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…

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…

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…

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…