题目描述: 给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict,判定 s 是否可以被空格拆分为一个或多个在字典中出现的单词. 说明: 拆分时可以重复使用字典中的单词. 你可以假设字典中没有重复的单词. 示例 1: 输入: s = "leetcode", wordDict = ["leet", "code"] 输出: true 解释: 返回 true 因为 "leetcode" 可以被拆分成 "le

Cat VS Dog Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 2046    Accepted Submission(s): 719 Problem DescriptionThe zoo have N cats and M dogs, today there are P children visiting the zoo, ea

称号: Cat vs. Dog Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 219 Accepted Submission(s): 86   Problem Description The latest reality show has hit the TV: ``Cat vs. Dog''. In this show, a bunch

HDU 3829 Cat VS Dog / NBUT 1305 Cat VS Dog(二分图最大匹配) Description The zoo have N cats and M dogs, today there are P children visiting the zoo, each child has a like-animal and a dislike-animal, if the child's like-animal is a cat, then his/hers dislike

Cat VS Dog Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others) Problem Description The zoo have N cats and M dogs, today there are P children visiting the zoo, each child has a like-animal and a dislike-animal, if the

Cat VS Dog Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 4383    Accepted Submission(s): 1602 Problem Description The zoo have N cats and M dogs, today there are P children visiting the zoo,

Cat vs. Dog Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1422    Accepted Submission(s): 534 Problem Description The latest reality show has hit the TV: ``Cat vs. Dog''. In this show, a bunch

Cat VS Dog Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 4751    Accepted Submission(s): 1757 Description: The zoo have N cats and M dogs, today there are P children visiting the zoo, each ch

Cat VS Dog Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 3829 Description The zoo have N cats and M dogs, today there are P children visiting the zoo, each child has a like-animal and a dislik

Cat vs. Dog Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2279    Accepted Submission(s): 886 Problem Description  The latest reality show has hit the TV: ``Cat vs. Dog''. In this show, a bun

题目链接:https://vjudge.net/problem/HDU-3829 Cat VS Dog Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 4118    Accepted Submission(s): 1493 Problem Description The zoo have N cats and M dogs, toda

题意:每个人有喜欢的猫和不喜欢的狗.留下他喜欢的猫他就高心,否则不高心.问最后最多有几个人高心. 思路:二分图求最大匹配 #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<cstdlib> #include<string> #include<cmath> #include<vector> #def

这篇博客光看标题或许就会产生疑问,某个对象,只要它存在某个属性,且值不是空的,不就能直接用点方法获取吗,为什么要拆分成key和value多此一举呢?下面,我用一个例子告诉大家,既然这方法是存在的,那就有它存在的价值. 有一个对象,比如说是仓库清单:model.苹果:100斤,香蕉:50斤,梨子:80斤........(共50种货物) 现在我要建立一个tableView表格,一个分区,50个单元格,每个cell的内容是:货物种类    存有多少 cell肯定是根据IndexPatch.row来取值

Cat VS Dog Problem Description The zoo have N cats and M dogs, today there are P children visiting the zoo, each child has a like-animal and a dislike-animal, if the child's like-animal is a cat, then his/hers dislike-animal must be a dog, and vice v

题目是:给出一个数字(10,000-100,000,000),把这个数字拆分成4段,怎样使得4段的乘积最小.比如12345拆分成1*2*3*45=270, 10000=1*00*0*0=0. 解题分析稍后给出... My Code: #include <iostream> #include <string> using namespace std; int dp[5][20]; int num(const string &str,int b,int e) { b--; e-

JAVA装箱和拆箱 从Java SE5开始就提供了自动装箱的特性,如果要生成一个数值为10的Integer对象,只需要这样就可以了.原文链接: http://www.cnblogs.com/dolphin0520/p/3780005.htmlInteger i = 10; //装箱int n = i; //拆箱 含义:装箱就是自动将基本数据类型转换为包装器类型:拆箱就是  自动将包装器类型转换为基本数据类型. 原理:装箱的时候自动调用的是Integer的valueOf(int)方法.而在拆箱的时

Cat VS Dog 题意:一群小朋友去动物园,如果每个小朋友喜欢的动物是猫,那么不喜欢的动物一定是狗,反之也是.现在动物园的管理者要拿走一些动物,如果拿走的是某个小朋友不喜欢的动物,那这个小朋友就非常开心,反之,如果是某个小朋友喜欢的动物,这个小朋友就非常的不开心,问那完后最多有几个小朋友会非常开心. 暑假最后一场个人赛,可还行,有点凉凉~~ 讲真看出是二分图最大匹配的题目,但是没学最大独立集,死活建不出图来,就到底还是自己的能力不行啊! 最小覆盖: 定义:假如选了一个点就相当于覆盖了以它为端

public function tt_add(){ $res = $,); echo array_sum($res); echo '----' . count($res); dump($res); exit; } /*** * @param $amount '总数' * @param $count '需要拆分数据个数' * @return array|mixed '把总数amount拆分已标准差最小的标准,平均拆分成count个整数' */ public function arr_avg($am

Java装箱和拆箱的基本概念及使用 要理解装箱和拆箱的概念,就要理解Java数据类型 装箱:把基本类型用它们相应的引用类型包装起来,使其具有对象的性质.int包装成Integer.float包装成Float 拆箱:和装箱相反,将引用类型的对象简化成值类型的数据 Integer a = 100;                  这是自动装箱  (编译器调用的是static Integer valueOf(int i))int     b = new Integer(100); 这是自动拆箱 看下

一.装箱.拆箱定义 如果一个int型量被传递到需要一个Integer对象的地方,那么,编译器将在幕后插入一个对Integer构造方法的调用,这就叫做自动装箱.而如果一个Integer对象被放到需要int型量的地方,则编译器将幕后插入一个队intValue方法的调用,这就叫做自动拆箱. public static void main(String[] args) { // 装箱 Integer i1 = Integer.valueOf(1); // 自动装箱 Integer i2 = 1;// 默

坑在哪里?   我们都知道Java的八种基本数据类型:int, short, long, double, byte, char, float, boolean   分别有各自对应的包装类型:Integer, Short, Long, Double, Byte, Character, Float, Boolean   并且二者之间可以相互直接赋值,例如: // 基本数据类型赋值给封装类 inta = 1; Integer b = a; // 封装类型赋值给基本数据类型 Character c =