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…

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)…

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…

# 函数原型detectMultiScale(gray, 1.2,3,CV_HAAR_SCALE_IMAGE,Size(30, 30)) # gray需要识别的图片 # 1.03:表示每次图像尺寸减小的比例 # 5:表示每一个目标至少要被检测到4次才算是真的目标(因为周围的像素和不同的窗口大小都可以检测到人脸) # CV_HAAR_SCALE_IMAGE表示不是缩放分类器来检测,而是缩放图像,Size(30, 30)为目标的最小最大尺寸…

C#  时间格式处理,获取格式: 2014-04-12T12:30:30+08:00 一.获取格式: 2014-04-12T12:30:30+08:00 方案一:(局限性,当不是当前时间时不能使用)   string time = System.DateTime.Now.ToString("s");   //    2014-04-12T12:30:30 string timeZone = DateTime.Now.ToString("o");     //2014…

java.time.format.DateTimeParseException: Text '2019-10-11 12:30:30' could not be parsed at index 10 at java.time.format.DateTimeFormatter.parseResolved0(DateTimeFormatter.java:1949) at java.time.format.DateTimeFormatter.parse(DateTimeFormatter.java:1…

PAT乙级:1057 数零壹 (20分) 题干 给定一串长度不超过 105 的字符串,本题要求你将其中所有英文字母的序号(字母 a-z 对应序号 1-26,不分大小写)相加,得到整数 N,然后再分析一下 N 的二进制表示中有多少 0.多少 1.例如给定字符串 PAT (Basic),其字母序号之和为:16+1+20+2+1+19+9+3=71,而 71 的二进制是 1000111,即有 3 个 0.4 个 1. 输入格式: 输入在一行中给出长度不超过 105.以回车结束的字符串. 输出格式: 在…

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…

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…

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…

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…

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…

给定一串长度不超过 1 的字符串,本题要求你将其中所有英文字母的序号(字母 a-z 对应序号 1-26,不分大小写)相加,得到整数 N,然后再分析一下 N 的二进制表示中有多少 0.多少 1.例如给定字符串 PAT (Basic),其字母序号之和为:16+1+20+2+1+19+9+3=71,而 71 的二进制是 1000111,即有 3 个 0.4 个 1. 输入格式: 输入在一行中给出长度不超过 1.以回车结束的字符串. 输出格式: 在一行中先后输出 0 的个数和 1 的个数,其间以空格分隔…

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…

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…

给定一串长度不超过 10^​5的字符串,本题要求你将其中所有英文字母的序号(字母 a-z 对应序号 1-26,不分大小写)相加,得到整数 N,然后再分析一下 N 的二进制表示中有多少 0.多少 1.例如给定字符串 PAT (Basic),其字母序号之和为:16+1+20+2+1+19+9+3=71,而 71 的二进制是 1000111,即有 3 个 0.4 个 1. 输入格式: 输入在一行中给出长度不超过 10^​5.以回车结束的字符串. 输出格式: 在一行中先后输出 0 的个数和 1 的个数,…

给定一串长度不超过 1 的字符串,本题要求你将其中所有英文字母的序号(字母 a-z 对应序号 1-26,不分大小写)相加,得到整数 N,然后再分析一下 N 的二进制表示中有多少 0.多少 1.例如给定字符串 PAT (Basic),其字母序号之和为:16+1+20+2+1+19+9+3=71,而 71 的二进制是 1000111,即有 3 个 0.4 个 1. 输入格式: 输入在一行中给出长度不超过 1.以回车结束的字符串. 输出格式: 在一行中先后输出 0 的个数和 1 的个数,其间以空格分隔…

07-图5 Saving James Bond - Hard Version   (30分) This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of lan…

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…

This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the center of a lake filled with crocodile…

This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the center of a lake filled with crocodile…

This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the center of a lake filled with crocodile…

06-图4. Saving James Bond - Hard Version (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was capture…