UVA 562 Dividing coins(dp + 01背包)
Dividing coins |
It's commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nickel, which was made of copper. They were both so eager to get it and the fighting was so fierce, they stretched the coin to great length and thus created copper-wire.
Not commonly known is that the fighting started, after the two Dutch tried to divide a bag with coins between the two of them. The contents of the bag appeared not to be equally divisible. The Dutch of the past couldn't stand the fact that a division should favour one of them and they always wanted a fair share to the very last cent. Nowadays fighting over a single cent will not be seen anymore, but being capable of making an equal division as fair as possible is something that will remain important forever...
That's what this whole problem is about. Not everyone is capable of seeing instantly what's the most fair division of a bag of coins between two persons. Your help is asked to solve this problem.
Given a bag with a maximum of 100 coins, determine the most fair division between two persons. This means that the difference between the amount each person obtains should be minimised. The value of a coin varies from 1 cent to 500 cents. It's not allowed to split a single coin.
Input
A line with the number of problems
n, followed by
n times:
- a line with a non negative integer m () indicating the number of coins in the bag
- a line with m numbers separated by one space, each number indicates the value of a coin.
Output
The output consists of
n lines. Each line contains the minimal positive difference between the amount the two persons obtain when they divide the coins from the corresponding bag.
Sample Input
2
3
2 3 5
4
1 2 4 6
题意:一堆硬币分给两个人,要求两个人得到钱差值最少,输出这个差值。
思路:01背包。硬币总价值为sum,一个人分到i,另一个人肯定分到sum - i,如果硬币尽量平分是最好的,先用01背包求出所有可能组成的值,然后从sum / 2开始找,找到一个可以组成的值作为i,他们的差值为sum - 2 * i。
代码:
#include <stdio.h>
#include <string.h> int t, n, coin[105], sum, i, j, dp[50005];
int main() {
scanf("%d", &t);
while (t --) {
scanf("%d", &n);
sum = 0;
memset(dp, 0, sizeof(dp));
dp[0] = 1;
for (i = 0; i < n; i ++) {
scanf("%d", &coin[i]);
sum += coin[i];
}
for (i = 0; i < n; i ++)
for (j = sum; j >= coin[i]; j --) {
if (dp[j - coin[i]])
dp[j] = 1;
}
for (i = sum / 2; i >= 0; i --)
if (dp[i]) {
printf("%d\n", sum - i * 2);
break;
}
}
return 0;
}
UVA 562 Dividing coins(dp + 01背包)的更多相关文章
- uva 562 Dividing coins(01背包)
Dividing coins It's commonly known that the Dutch have invented copper-wire. Two Dutch men were f ...
- UVA 562 Dividing coins (01背包)
题意:给你n个硬币,和n个硬币的面值.要求尽可能地平均分配成A,B两份,使得A,B之间的差最小,输出其绝对值.思路:将n个硬币的总价值累加得到sum, A,B其中必有一人获得的钱小于等于sum/2 ...
- UVA 562 Dividing coins【01背包 / 有一堆各种面值的硬币,将所有硬币分成两堆,使得两堆的总值之差尽可能小】
It's commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nic ...
- HDOJ(HDU).3466 Dividing coins ( DP 01背包 无后效性的理解)
HDOJ(HDU).3466 Dividing coins ( DP 01背包 无后效性的理解) 题意分析 要先排序,在做01背包,否则不满足无后效性,为什么呢? 等我理解了再补上. 代码总览 #in ...
- UVA 562 Dividing coins --01背包的变形
01背包的变形. 先算出硬币面值的总和,然后此题变成求背包容量为V=sum/2时,能装的最多的硬币,然后将剩余的面值和它相减取一个绝对值就是最小的差值. 代码: #include <iostre ...
- UVA 562 Dividing coins (01背包)
//平分硬币问题 //对sum/2进行01背包,sum-2*dp[sum/2] #include <iostream> #include <cstring> #include ...
- UVA 562 Dividing coins 分硬币(01背包,简单变形)
题意:一袋硬币两人分,要么公平分,要么不公平,如果能公平分,输出0,否则输出分成两半的最小差距. 思路:将提供的整袋钱的总价取一半来进行01背包,如果能分出出来,就是最佳分法.否则背包容量为一半总价的 ...
- UVa 562 - Dividing coins 均分钱币 【01背包】
题目链接:https://vjudge.net/contest/103424#problem/E 题目大意: 给你一堆硬币,让你分成两堆,分别给A,B两个人,求两人得到的最小差. 解题思路: 求解两人 ...
- Dividing coins (01背包)
It’s commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nic ...
随机推荐
- c++容器使用总结(转载)
目录 ==================================================== 第一章 容器 第二章 Vector和string 第三章 关联容器 第四章 迭代器 第五 ...
- Unique Paths II 解答
Question Follow up for "Unique Paths": Now consider if some obstacles are added to the gri ...
- Socket 相关的知识
1.关于PF_INET和AF_INET的区别 在写网络程序的时候,建立TCP socket: sock = socket(PF_INET, SOCK_STREAM, 0);然后在绑定本地地址或连接远程 ...
- JavaScript-打开新窗口
open()方法可以查找一个已经存在或者新建一个新的浏览器窗口. 语法:window.open([URL], [窗口名称], [参数字符串]) 参数解释: URL:可选参数,在窗口中显示网页的网址或路 ...
- Ubuntu+Eclipse+ADT+Genymotion+VirtualBox开发环境搭建
1.Eclispe安装就不说了 2.以下说说怎样安装ADT插件.有两种途径: (1)在线安装: 地址:https://dl-ssl.google.com/android/eclipse/(只是近期天朝 ...
- Struts2使用Interceptor实现权限控制的应用实例详解
Struts2使用Interceptor实现权限控制的应用实例详解 拦截器:是Struts2框架的核心,重点之重.因此,对于我们要向彻底学好Struts2.0.读源码和使用拦截器是必不可少的.少说了. ...
- Linux Tomcat7.0安装配置实践总结
一,安装JDk 先下载jdk,链接http://www.oracle.com/technetwork/java/javase/downloads/index.html,选择相对应平台的JDK.由于笔者 ...
- NET基础课--WinForm开发推荐3
用户体验 较长时间的运算:使用进度条(progress bar) 不要阻塞界面(UI)线程:使用多线程进行长时间的运算 状态栏(status bar)提示应用程序的状态 操作开始之后,用户应当能够通过 ...
- 推荐10款 好用的 Jquery 评分插件
Raty jQuery Raty这是一个能够自动生成可定制的星级评分jQuery插件.可以自定义图标,创建各种评级组合,星星数量,每一颗星星的注释,可以在当一个星星被点击时的加回调函数. 地址: Ra ...
- [AngularJS系列(4)] 那伤不起的provider们啊~ (Provider, Value, Constant, Service, Factory, Decorator)(转)
用AngularJS做项目,但凡用过什么service啊,factory啊,provider啊,开始的时候晕没晕?!晕没晕?!感觉干的事儿都差不多啊,到底用哪个啊?!别告诉我你们几个就是为了跟我炫耀兄 ...