Dividing+POJ+01背包问题
来源:互联网 发布:网红喵大仙的淘宝店 编辑:程序博客网 时间:2024/04/30 03:43
Dividing
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 59202 Accepted: 15291
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 just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value. Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.
Input
Each line in the input file describes one collection of marbles to be divided. The lines contain six non-negative integers n1 , . . . , n6 , where ni is the number of marbles of value i. So, the example from above would be described by the input-line "1 0 1 2 0 0". The maximum total number of marbles will be 20000.
The last line of the input file will be "0 0 0 0 0 0"; do not process this line.
The last line of the input file will be "0 0 0 0 0 0"; do not process this line.
Output
For each collection, output "Collection #k:", where k is the number of the test case, and then either "Can be divided." or "Can't be divided.".
Output a blank line after each test case.
Output a blank line after each test case.
Sample Input
1 0 1 2 0 0 1 0 0 0 1 1 0 0 0 0 0 0
Sample Output
Collection #1:Can't be divided.Collection #2:Can be divided.
解决方案:此题可用01多重背包的模型来解决,最终结果若sum-dp[sum/2]==dp[sum/2],即为“Can be divided.”,否则
"Can't be divided."
0 0
- Dividing+POJ+01背包问题
- POJ 1014 Dividing (多重背包问题)
- poj 1014 Dividing 多重背包问题
- POJ 1014 Dividing 多重背包,二进制压缩成一般01背包问题
- POJ 1014 Dividing 多重背包,二进制压缩成一般01背包问题
- POJ 1014 Dividing 背包
- POJ 1014 Dividing(多重背包 + 二进制优化 + 01背包)
- POJ 1014 Dividing(多重背包转换成01背包)
- uva562 Dividing coins 01背包 平衡问题
- UVA 562Dividing coins(01 背包问题)
- Dividing 多重背包问题
- POJ 1014 && HDU 1059 Dividing(完全背包问题)
- POJ 1014 Dividing (多重背包问题+递归)【模板】
- poj 1014 Dividing 多重背包
- Poj 1014 Dividing//多重背包
- Poj 1014 Dividing(多重背包)
- POJ 1014 Dividing 多重背包
- POJ 1014 Dividing 多重背包
- 变长数据结构及其应用
- Oracle层次查询的基本用法
- linux内核的生成过程, vmlinux调试分析
- 【140817】类似红警画面的VC游戏源码,一个模块
- ISO给UIImageView增加点击事件
- Dividing+POJ+01背包问题
- 我的太鼓
- Oracle 中PLSQL的ftp应用
- srvbuildres: error while loading shared libraries: libXm.so.3
- IOS单例模式(Singleton)介绍
- 【索引】Maths - Simple Geometry
- 我的太鼓达人2
- « AFNetworking 图片的本地缓存问题 Get application bundle seed ID in iOS » AFNetworking 使用总结 (用法+JSON解析)
- poj 1681 极角排序(只能向左拐的虫子)