题目来源:http://poj.org/problem?id=1014 题目大意: Marsha和Bill拥有一些弹珠.但是这些弹珠的价值不一样.每个弹珠的价值都是1到6之间的自然数.他们希望把这些弹珠分为两份,每份的总价值相等.然而,有时候是不存在这样的划分的(即使总的价值为偶数).比如弹珠的价值分别为1,3,4,4.写一个程序判断一些弹珠是否可以被分为价值相等的两份. 输入:每行代表一个测试用例,含6个非负整数.n1,...n6.ni表示价值为i的弹珠有多少个.测试用例最多20000个.输入…

Dividing Time Limit: 1000MS Memory Limit: 10000K Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had…

转载请注明出处:http://blog.csdn.net/u012860063 题目链接:id=1014">http://poj.org/problem?id=1014 Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. Th…

http://poj.org/problem?id=1014 Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, b…

#include <cstdio> #include <algorithm> #include <cstring> using namespace std; int T,A[10]; bool f[100010]; inline int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&…

Problem Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could…

图论 图论解题报告索引 DFS poj1321 - 棋盘问题 poj1416 - Shredding Company poj2676 - Sudoku poj2488 - A Knight's Journey poj1724 - ROADS(邻接表+DFS) BFS poj3278 - Catch That Cow(空间BFS) poj2251 - Dungeon Master(空间BFS) poj3414 - Pots poj1915 - Knight Moves poj3126 - Prim…

一.需求:计算网页访问量前三名 import org.apache.spark.rdd.RDD import org.apache.spark.{SparkConf, SparkContext} /** * 需求:计算网页访问量前三名 * 用户:喜欢视频 直播 * 帮助企业做经营和决策 * * 看数据 */ object UrlCount { def main(args: Array[String]): Unit = { //1.加载数据 val conf:SparkConf = new Spa…

http://poj.org/problem?id=1014 (题目链接) 题意 给出有分别价值为1,2,3,4,5,6的6种物品,输入6个数字,表示相应价值的物品的数量,问一下能不能将物品分成两份,是两份的总价值相等. solution 多年以前写的程序了,现在才写博客= =.这道题一看就是多重背包,所以我们用二进制把它拆分成01背包就很好做了,不知道的话就看<背包九讲>吧.. 首先我们把6种物品的总价值记为S,如果S%2==1,那么显然是无解的.考虑S%2==0的情况.我们把每个种物品用二…

这是悦乐书的第305次更新,第324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173题(顺位题号是728).自分割数是一个可被其包含的每个数字整除的数字.例如,128是自分割数,因为128%1 == 0,128%2 == 0,128%8 == 0.此外,不允许自分割数包含数字零.给定数字的下限和上限,输出每个可能的自分割数的数组,如果可能,包括边界.例如: 输入:left = 1,right = 22 输出:[1,2,3,4,5,6,7,8,9,11,12,1…

Dividing Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 63013   Accepted: 16315 Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbl…

有价值为1~6的宝物各num[i]个,求能否分成价值相等的两部分. #include <iostream> #include <cstring> #include <string> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <queue> #include <map> #def…

题目意思 有六种不同的石子,权值为\(1\)~\(6\),给出六种石子的数量,求能否将石子分成权值相等的两份. 解析 这题可以直接用多重背包写, 因为仔细想想, 能够平均分成两份, 也就是能将一部分石子拼成总权值的二分之一. 那么,直接用可行性DP写就行了. (对于可行性DP请自行上百度吧qwq(因为这又是另一个故事了) 上代码吧: #include <iostream> #include <cstdio> #include <cstring> using namesp…

POJ 排序的思想就是根据选取范围的题目的totalSubmittedNumber和totalAcceptedNumber计算一个avgAcceptRate. 每一道题都有一个value,value = acceptedNumber / avgAcceptRate + submittedNumber. 这里用到avgAcceptedRate的原因是考虑到通过的数量站的权重可能比提交的数量占更大的权重,所以给acceptedNumber乘上了一个因子. 当然计算value还有别的方法,比如POJ上…

http://www.onjava.com/pub/a/onjava/2001/05/30/optimization.htmlComparing the performance of LinkedLists and ArrayLists (and Vectors) (Page last updated May 2001, Added 2001-06-18, Author Jack Shirazi, Publisher OnJava). Tips: ArrayList is faster than…

Dividing Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14861    Accepted Submission(s): 4140 Problem Description Marsha and Bill own a collection of marbles. They want to split the collection…

3.3. Data TypesJava is a strongly typed language(强类型语音). This means that every variable must have a declared type(每个变量都必须声明类型). There are eight primitive types in Java(Java有8种原始类型). Four of them are integer types; two are floatingpoint number types;…

Part 1 reference:http://jaxenter.com/lambdas-in-java-8-part-1-49700.html Get to know lambda expressions in Java 8. Few things excite a community of software developers more than a new release of their chosen programming language or platform. Java dev…

reference from:http://www.programcreek.com/2013/10/top-10-questions-about-java-exceptions/ This article summarizes the top 10 frequently asked questions about Java exceptions. 1. Checked vs. Unchecked In brief, checked exceptions must be explicitly c…

Dividing Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 18190    Accepted Submission(s): 5080 Problem Description Marsha and Bill own a collection of marbles. They want to split the collection…

1.1 NIO Buffers - Class java.nio.Buffer NIO data transfer is through the so-called buffers implemented in java.nio.Buffer class. A Buffer is similar to an array, except that it's implemented much more efficiently by closely coupled with the underlyin…

排序需要掌握的有冒泡排序,插入排序和选择排序.时间为O(N*N). 冒泡排序: 外层循环从后往前,内存循环从前往后到外层循环,相邻数组项两两比较,将较大的值后移. 插入排序: 从排序过程的中间开始(程序从第二个数组项开始a[1]),此时队列已经拍好了一部分.此时,将后边的数组项一次插入到已经排好序的部分队列中. 选择排序: 从第一个数组项开始,找到包括该数组项在内的所有往后数组项中的最小项与当前项进行交换,其实相当于依次将最小值往前冒泡. 示例代码: package chap03.BubbleS…

点击打开链接链接 Dividing Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17544    Accepted Submission(s): 4912 Problem Description Marsha and Bill own a collection of marbles. They want to split the c…

总体阅读了Long的源码,基本跟Integer类类似,所以特别全部贴出源码,直接注释进行理解. // final修饰符 public final class Long extends Number implements Comparable<Long> { /** * A constant holding the minimum value a {@code long} can * have, -2⁶³. */ // 最小值-负值 @Native…

Integer定义,final不可修改的类 public final class Integer extends Number implements Comparable<Integer> 常量定义 /** * A constant holding the minimum value an {@code int} can * have, -2³¹. */ @Native public static final int MIN_VALUE = 0x8…

有两种类型的异常:一种是checked异常一种是unchecked异常,在这篇文章中我们将利用实例来学习这两种异常,checked的异常和unchecked异常最大的区别就是checked去唱是在编译时检查的而unchecked异常是在运行时检查的. 什么是checked异常呢? checked异常在编译时检查,这意味着如果一个方法抛出checked异常,那么它应该使用try-catch块或者使用throws关键字来处理这个异常,否则的话程序会报编译错误,命名为checked异常是因为是在编译时…

要说清楚Java浮点数的取值范围与其精度,必须先了解浮点数的表示方法与浮点数的结构组成.因为机器只认识01,你想表示小数,你要机器认识小数点这个东西,必须采用某种方法.比如,简单点的,float四个字节,前两个字节表示整数位,后两个字节表示小数位(这就是一种规则标准),这样就组成一个浮点数.而Java中浮点数采用的是IEEE 754标准. IEEE 754 标准 更多详见:https://baike.baidu.com/item/IEEE%20754 IEEE 754 标准是IEEE二进位浮点数…

沉淀再出发:java中的equals()辨析 一.前言 关于java中的equals,我们可能非常奇怪,在Object中定义了这个函数,其他的很多类中都重载了它,导致了我们对于辨析其中的内涵有了混淆,再加上和“==”的比较,就显得更加的复杂了. 二.java中的equals() 2.1.Object.java中的equals()     让我们来看一下Object.java中的equals().     首先是Object的定义: /* * Copyright (c) 1994, 2012, O…

RXD and dividing Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1893    Accepted Submission(s): 809 Problem Description RXD has a tree T, with the size of n. Each edge has a cost.Define f(S) …

地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=6060 题目: RXD and dividing Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 522    Accepted Submission(s): 219 Problem Description RXD has a tr…