problem 914. X of a Kind in a Deck of Cards 题意:每个数字对应的数目可以均分为多组含有K个相同数目该数字的数组. 思路:使用 map 结构记录数组中每个元素出现的次数,该题转化为求次数的最大公约数,若所有次数的最大公约数大于或者等于 2,返回 true,否则返回 false. solution: class Solution { public: bool hasGroupsSizeX(vector<int>& deck) { //Greate…

In a deck of cards, each card has an integer written on it. Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where: Each group has exactly X cards. All the cards in eac…

In a deck of cards, each card has an integer written on it. Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where: Each group has exactly X cards. All the cards in eac…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 遍历 最大公约数 日期 题目地址: https://leetcode.com/problems/x-of-a-kind-in-a-deck-of-cards/description/ 题目描述 In a deck of cards, each card has an integer written on it. Return true if and…

原题 题目原意可转换为 两组有大于等于2的公因数 /** * @param {number[]} deck * @return {boolean} */ var hasGroupsSizeX = function(deck) { var map = {}; for (let i = 0; i < deck.length; i++) { if (map[deck[i]]) map[deck[i]] += 1; else map[deck[i]] = 1; } var min = map[deck[…

C. Hongcow Buys a Deck of Cards time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output One day, Hongcow goes to the store and sees a brand new deck of n special cards. Each individual card is eit…

In a deck of cards, each card has an integer written on it. Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where: Each group has exactly X cards. All the cards in eac…

In a deck of cards, each card has an integer written on it. Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where: Each group has exactly X cards. All the cards in eac…

In a deck of cards, each card has an integer written on it. Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where: Each group has exactly X cards. All the cards in eac…

Hongcow Buys a Deck of Cards 啊啊啊, 为什么我连这种垃圾dp都写不出来.. 不是应该10分钟就该秒掉的题吗.. 从dp想到暴力然后gg, 没有想到把省下的红色开成一维. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair&…

In a deck of cards, each card has an integer written on it. Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where: Each group has exactly X cards. All the cards in eac…

地址:http://codeforces.com/problemset/problem/744/C 题目: C. Hongcow Buys a Deck of Cards time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output One day, Hongcow goes to the store and sees a brand ne…

这是悦乐书的第352次更新,第377篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第214题(顺位题号是914).在一副牌中,每张牌上都写有一个整数. 当且仅当您可以选择X >= 2时才返回true,以便可以将整个牌组分成一组或多组牌,其中: 每组都有X张牌. 每组中的所有牌都具有相同的整数. 例如: 输入:[1,2,3,4,4,3,2,1] 输出:true 说明:可能的分区[1,1],[2,2],[3,3],[4,4] 输入:[1,1,1,2,2,2,3,3] 输…

Define “Straight” as 5 cards with consecutive numbers. Determine if the deck can be fully divided into sets of “Straight”. Example: 1, 2, 3, 4, 4, 5, 5, 6, 7, 8 -> TrueYou may assume the cards are sorted 这个是用一个hashtable,key是数字,value是出现次数 然后遍历原数组,每一个数…

给定一副牌,每张牌上都写着一个整数. 此时,你需要选定一个数字 X,使我们可以将整副牌按下述规则分成 1 组或更多组: 每组都有 X 张牌. 组内所有的牌上都写着相同的整数. 仅当你可选的 X >= 2 时返回 true. 示例 1: 输入:[1,2,3,4,4,3,2,1] 输出:true 解释:可行的分组是 [1,1],[2,2],[3,3],[4,4] 示例 2: 输入:[1,1,1,2,2,2,3,3] 输出:false 解释:没有满足要求的分组. 示例 3: 输入:[1] 输出:fal…

题意: 一一个21点游戏. 1. 有三个牌堆,分别为1X,2X,3X. 2. 纸牌A的值为1,纸牌2-9的值与牌面面相同,10(T).J.Q.K的值为10,而而joke(F)的值为 任意大大. 3. 一一列牌要按顺序放入入三个牌堆中.当某个牌堆的值超过21点时,不能在放牌;如果某个牌堆的 总值为21点时,这个排队讲会被清空;joke放上后这个牌堆的值立立即变为21点. 4. 成功放上一一张牌得50美元;成功清空一一个牌堆讲得到100*牌堆号美元,即1X得100美元,2X得 200美元,3X得30…

题意:一个特殊21点游戏 具体http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2852 题解:建一个三维dp,表示三个卡槽分别为i,j,l分时最大的收益情况. 对所有当前状态dp,将下一个可能的状态存入f, 坑:~-1==0 #define _CRT_SECURE_NO_WARNINGS #include<cstring> #include<cctype> #include<cstdlib> #i…

大意: n个红黑卡, 每天可以选择领取一块红币一块黑币, 或者买一张卡, 第$i$张卡的花费红币数$max(r_i-A,0)$, 花费黑币数$max(b_i-B,0)$, A为当前红卡数, B为当前黑卡数, 求买完所有卡最少天数. 这题挺巧妙的, 刚开始看花费的范围太大一直在想怎么贪心... 实际上注意到减费最多只有120, 可以按照减费进行dp即可 这题CF大神的最优解写了个模拟退火ORZ #include <iostream> #include <algorithm> #inc…

这题的难点在于状态的设计 首先显然是个状压,需要一维表示卡的状态,另一维如果设计成天数,难以知道当前的钱数,没法确定是否能够购买新的卡,如果设计成钱数,会发现状态数过多,空间与时间都无法承受.但是可以发现,如果没有买卡的钱会因当前卡数变化而变化这个条件的话,买卡的钱是一定的,而我们因拥有卡而省的钱不会超过120(1+2+3+...+15).所以可以将状态设计成f[i][j]表示卡的状态为i,省了j个红币,能省多少个蓝币. 然后就结束了... 注意所有下标为状态的数组的大小T T... #incl…

题意 你有\(n\)个物品,物品和硬币有\(A\),\(B\)两种类型,假设你有\(M\)个\(A\)物品和\(N\)个\(B\)物品 每一轮你可以选择获得\(A, B\)硬币各\(1\)个,或者(硬币足够)花\(\max(a_i - M, 0)\)个\(A\),\(\max(b_i - N, 0)\)个\(B\)买\(i\)这个物品 问买到所有物品最少要多少轮 题解 巧妙的\(dp\),考虑间接设计状态 \(f[S][A] = B\)表示\(S\)这个集合买过了,\(A\)类花\(\sum a…

题意:现在有n张卡片(n <= 16), 每一轮你可以执行两种操作中的一种.1:获得一张红色令牌和一张蓝色令牌.2:购买一张卡片(如果可以买的话),购买的时候蓝色卡片可以充当蓝色令牌,红色同理,但是购买后只消耗令牌,不消耗卡片.问最少多少轮可以购买全部卡片. 思路1:状压DP.我们发现卡片可以减少令牌的使用,如果不考虑卡片的话,总花费其实是固定的.所以,只要我们算出了通过令牌最多可以减免多少花费,就可以得到答案了. 设dp[i][j]为卡片的够买状态为i时,其中红色卡片的花费减免了j,蓝色卡片花…

请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如:[Swift]LeetCode156.二叉树的上下颠倒 $ Binary Tree Upside Down 请下拉滚动条查看最新 Weekly Contest!!! Swift LeetCode 目录 | Catalog 序        号 题名Title 难度     Difficulty  两数之…

Double Patience Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 694   Accepted: 368 Case Time Limit: 1000MS   Special Judge Description Double Patience is a single player game played with a standard 36-card deck. The cards are shuffled a…

敲完三题挂机一小时.....  也没懂DE什么意思  rank600上了一波分... A. Hongcow Learns the Cyclic Shift 给一个字符串,每次可以把最后一个字符拿到开头  问能形成多少种.. 暴力模拟  set去重... B. Hongcow Solves A Puzzle 判断矩形即可... C. Hongcow Builds A Nation 并查集求最大块  然后把未标记的块放进最大块里  最后的连边数-最初的   为我们添加的最多可能 D. Hongcow…

下面用枚举类型来实现一副扑克牌的类: //定义一个玩牌的类 function Card(suit,rank){ function inherit(p){ if(p==null) throw TypeError(); if(Object.create) return Object.create(p); var t = typeof p; if(t!=="object" && t!=="function") throw TypeError(); func…

//实现枚举类型,扑克牌应用 function creatEnum(p){     //构造函数     var Enumeration = function(){throw 'can not Instantiate Enumerations';};     //重写原型并将原型赋值给变量proto     var proto = Enumeration.prototype = {         constructor:Enumeration,         toString:functio…

Stat2.2x Probability(概率)课程由加州大学伯克利分校(University of California, Berkeley)于2014年在edX平台讲授. PDF笔记下载(Academia.edu) PRACTICE PROBLEMS FOR THE MIDTERM PROBLEM 1 In a group of 5 high school students, 2 are in 9th grade, 2 are in 10th grade, and 1 is in 12th…

题目描述 Most of you have played card games (and if you haven’t, why not???) in which the deck of cards is randomized by shuffling it one or more times.A perfect shuffle is a type of shuffle where the initial deck is divided exactly in half, and the two hal…

18.2 Write a method to shuffle a deck of cards. It must be a perfect shuffle—in other words, each of the 52! permutations of the deck has to be equally likely. Assume that you are given a random number generator which is perfect. 这道题让我们实现一个洗牌的算法,实际上洗…

Integrating the FlyCapture SDK for use with OpenCV CStereoGrabber_Bumblebee.h OpenCV with PGR Flycapture cameras http://www.cis.fordham.edu/twiki/pub/Main/FRCVCodeStereoServer/stereoCamera.cpp http://robocup.wtb.tue.nl/svn/techunited/trunk/dev/Rene/B…