PAT 1109 Group Photo】的更多相关文章

1109 Group Photo(25 分) Formation is very important when taking a group photo. Given the rules of forming K rows with N people as the following: The number of people in each row must be N/K (round down to the nearest integer), with all the extra peopl…
Formation is very important when taking a group photo. Given the rules of forming K rows with N people as the following: The number of people in each row must be N/K (round down to the nearest integer), with all the extra people (if any) standing in…
1109 Group Photo (25 分) Formation is very important when taking a group photo. Given the rules of forming K rows with N people as the following: The number of people in each row must be N/K (round down to the nearest integer), with all the extra peop…
Formation is very important when taking a group photo. Given the rules of forming K rows with N people as the following: The number of people in each row must be N/K (round down to the nearest integer), with all the extra people (if any) standing in…
Formation is very important when taking a group photo. Given the rules of forming K rows with N people as the following: The number of people in each row must be N/K (round down to the nearest integer), with all the extra people (if any) standing in…
Formation is very important when taking a group photo. Given the rules of forming K rows with N people as the following: The number of people in each row must be N/K (round down to the nearest integer), with all the extra people (if any) standing in…
Formation is very important when taking a group photo. Given the rules of forming K rows with N people as the following: The number of people in each row must be / (round down to the nearest integer), with all the extra people (if any) standing in th…
简单模拟. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<queue> #include<stack> #include<string> #include<algorithm> #include<iostream> using namespace std…
题意:n个人,要拍成k行排队,每行 n/k人,多余的都在最后一排. 从第一排到最后一排个子是逐渐增高的,即后一排最低的个子要>=前一排的所有人 每排排列规则如下: 1.中间m/2+1为该排最高: 2.其他人按各自降序顺序,轮流排到中间最高的左边和右边: 举个例子 190 188 186 175 170 — — 190 — — — 188 190 — — — 188 190 186 — 175 188 190 186 — 175 188 190 186 170 3.当个子一样高时,名字按字典序顺序…
题意: 输入两个整数N和K(N<=1e4,K<=10),分别表示人数和行数,接着输入N行每行包括学生的姓名(八位无空格字母且唯一)和身高([30,300]的整数).按照身高逆序,姓名字典序升序将学生从高到矮排列,将学生均分为N行输出,最先输出的一行人数包括除不尽的余数学生,每行中间(如果这一行学生人数为偶数则靠右的为中间)的学生最高,然后依次左边一位学生最高,右边一位学生最高,例如190, 188, 186, 175, 170->175, 188, 190, 186, 170. tric…
Source: PAT A1109 Group Photo (25 分) Description: Formation is very important when taking a group photo. Given the rules of forming K rows with Npeople as the following: The number of people in each row must be / (round down to the nearest integer),…
Formation is very important when taking a group photo. Given the rules of forming K rows with Npeople as the following: The number of people in each row must be / (round down to the nearest integer), with all the extra people (if any) standing in the…
Formation is very important when taking a group photo. Given the rules of forming K rows with N people as the following: The number of people in each row must be N/K (round down to the nearest integer), with all the extra people (if any) standing in…
题目大意: 有n个矩形在地上排成一列,不可重叠,已知他们的宽度w和高度h,现在使至多[n / 2]个矩形旋转90度,问最后可以用多小的矩形恰好覆盖这n个矩形,求满足条件的最小矩形面积. n, w, h <= 1000. 分析: 数据范围比较小,可以枚举答案矩形的高度H,判断是否能只旋转至多[n / 2]个矩形使得n个矩形的高度均不超过H,再用剩下的操作次数尽量使得总宽度变少. 贪心旋转,尽量把w比h大很多的矩形旋转,所以可以对矩形按照{w - h}排序,贪心旋转即可. #include<cst…
贪心枚举最后方案中最大的h,设为maxh若某个人i的wi与hi均大于maxh,则此方案不可行若某个人恰有一个属性大于maxh,则可确定他是否换属性剩下的人按wi-hi从大到小排序后贪心选择O(nlogn+n2) #include<iostream> #include<cstdio> #include<vector> #include<algorithm> using namespace std; ; int n,w[maxn],h[maxn],hh,ans=…
AK爷GhostCai的电脑又蓝屏了Orz 贪心题,确定一个maxh,限定h不大于一个值.枚举maxh. check的时候的细节很多: 1.h>maxh但w<maxh交换的时候需要占用交换名额. 2.cnt<0了就应该return -1了... 突然忘记还有啥细节 代码如下 #include <iostream> #include <algorithm> #include <cstdio> using namespace std; const int…
传送门 解题思路 这道题要用到贪心的思路,首先要枚举一个h的最大值,之后check.如果这个东西的w[i]与h[i]都大于枚举的值就直接return false,如果w[i]比这个值小,h[i]比这个值大,就将h[i]与w[i]交换,注意并不能直接交换因为后面还要用到,要在答案中将h[i]加上.如果交换次数比n/2大也return false,如果w[i]与h[i]均比这个值小,则将w[i]-h[i]加入数组.后面将它排序,选出可以被交换的更新答案. 代码 #include<iostream>…
2019/4/3 1063 Set Similarity n个序列分别先放进集合里去重.在询问的时候,遍历A集合中每个数,判断下该数在B集合中是否存在,统计存在个数(分子),分母就是两个集合大小减去分子. // 1063 Set Similarity #include <set> #include <map> #include <cstdio> #include <iostream> #include <algorithm> using name…
博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…
树(23) 备注 1004 Counting Leaves   1020 Tree Traversals   1043 Is It a Binary Search Tree 判断BST,BST的性质 1053 Path of Equal Weight   1064 Complete Binary Search Tree 完全二叉树的顺序存储,BST的性质 1066 Root of AVL Tree 构建AVL树,模板题,需理解记忆 1079 Total Sales of Supply Chain…
Source: PAT (Advanced Level) Practice Reference: [1]胡凡,曾磊.算法笔记[M].机械工业出版社.2016.7 Outline: 基础数据结构: 线性表:栈,队列,链表,顺序表 树:二叉树的建立,二叉树的遍历,完全二叉树,二叉查找树,平衡二叉树,堆,哈夫曼树 图:图的存储和遍历 经典高级算法: 深度优先搜索,广度优点搜索,回溯剪枝 贪心,并查集,哈希映射 最短路径(只考察过单源),拓扑排序(18年9月第一次涉及相关概念,未正式考过),关键路径(未…
B. Photo to Remember Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/522/problem/B Description One day n friends met at a party, they hadn't seen each other for a long time and so they decided to make a group photo together…
- 为何需要微数据 长篇加累版牍,不好理解 微标记来标注其中内容,让其容易识辨 - RDFa Resource Description Framework http://www.w3.org/TR/microdata http://www.whatwg.org/specs/web-apps/current-work/multipage/links.html#microdata http://www.w3.org/TR/rdfa-in-html --- itemscope 标记 <articlei…
vim自带一个目录浏览器,使用命令:E就可以调出来,实际上就是浏览器的名字就是"网络读写"netrw vim也自带了 补全功能, 启动键是 "ctrl_N" 或者用 ctrl_p也可以, 不管是第一个还是后面第几个,反正你可以补全就好了 . 如果有多个补全(开头的字符相同的多个), 则会显示^N^P 还有更多的补全在 ctrl_x中(6种): ctrl_x 和 ctrl_D 是宏定义补全 ctrl_x ctrl_f 是文件名补全(file) ctrl_x ctrl_…
figure和figcaption是一对经常被一起使用的语义化标签.如果你还没有看过规范中的定义,现在有机会在你的项目中使用它们了.如果你不知道怎么用,下面是关于如何正确使用它们的一些提示. figure元素经常用于image: <figure> <img src="dog.jpg" alt="Maltese Terrier"> </figure> figure元素表示内容的自包含单元.这个含义是,如果你把元素无论是移出文档或移动…
[From] https://blog.csdn.net/hany3000/article/details/51033610 ScalaTest测试框架 2016年04月01日 02:49:35 阅读数:4718 ScalaTest 是一个开源测试框架,旨在令测试更加高效.其开发者是Bill Venners(Artima主编).Bill Venners这样描述ScalaTest: ScalaTest是比JUnit和TestNG更加高阶的测试编写工具,这个Scala应用在JVM上运行,可以测试Sc…
Reordering the Cows 时间限制: 1 Sec  内存限制: 64 MB提交: 18  解决: 7[提交][状态][讨论版] 题目描述 Farmer John's N cows (1 <= N <= 100), conveniently numbered 1..N, are standing in a row.  Their ordering is described by an array A, where A(i) is the number of the cow in p…
今天在写MySQ的SQL语句的时候遇到了一个奇怪的问题 select count(*) from subsitealbum t1, photo t2,files t3 where t1.SourceAlbumId = t2.FolderId and t2.ID = t3.ID and t2.FolderId = t3.FolderID and SiteId in (select SubsiteID from channelsubsites where ChannelID='yszw') GROU…
1062 Talent and Virtue (25 分) About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people's talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a "sage(圣…
代码地址如下:http://www.demodashi.com/demo/13339.html 1.livePhoto简介 livePhoto是iOS 9.0 之后系统相机提供的拍摄动态照片的功能,但是仅在6S+,iOS 9.0+设备可用.拍摄完livePhoto之后,只需要在相册按压livePhoto相片即可动态的播放.livePhoto还可以设置为动态壁纸.如果只能用相机拍摄的livePhoto设置为动态壁纸,这不能满足我们的需求了.如果可以将视频转换为livePhoto那就完美了.如果要实…