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 (m ≤ 100) 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

Sample Output

0

1

【分析】:它所给出的n个钱币加起来sum,将sum/2当作体积,求出在sum/2下的背包最大值,sum-2*dp[sum/2]

【代码】:

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,n,x) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxn = 120;
const int maxm = 1e5 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
/*
2
3
2 3 5
4
1 2 4 6 */
int dp[maxm],a[maxm];
int n,m;
int main()
{
int t; cin>>t;
while(t--)
{
cin>>n;
int sum=0;
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++) cin>>a[i],sum+=a[i];
for(int i=1;i<=n;i++)
{
for(int j=sum/2;j>=a[i];j--)
{
dp[j] = max(dp[j],dp[j-a[i]]+a[i]);
}
}
cout<< sum - 2*dp[sum/2] <<endl;
}
}

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】的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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 figh ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. uva562 Dividing coins 01背包

    link:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...

  9. UVA 562 Dividing coins

    题目描述:给出一些不同面值的硬币,每个硬币只有一个.将这些硬币分成两堆,并且两堆硬币的面值和尽可能接近. 分析:将所有能够取到的面值数标记出来,然后选择最接近sum/2的两个面值 状态表示:d[j]表 ...

随机推荐

  1. 解决NSTimer循环引用

    NSTimer常见用法 @interface XXClass : NSObject - (void)start; - (void)stop; @end @implementation XXClass ...

  2. Python写的计算器程序(主要目的在于熟悉下正则表达式)

    import res = '1-2*((60-30-(-40/5)*(9-2*5/3-7/3*99/4*2998-10*568/14.3))+(-4*3)/16-3)'s2 = 1-2*((60-30 ...

  3. 数据库——pymysql模块的使用(13)

    1.基本用法——建立链接,获取游标,执行sql语句,关闭 建立远程链接账号和权限 mysql> grant all on *.* to '; Query OK, rows affected, w ...

  4. HDU 4189 Cybercrime Donut Investigation 线段树+思路

    参考:http://www.cnblogs.com/slon/archive/2012/03/30/2426104.html 题意:给一个有n个点的点集,有q个询问,每个询问询问一个点p,求与p曼哈顿 ...

  5. windows系统查找文件-通配符的使用

    在windows中可以使用通配符“* ”.“? ”查找文件.对于相同字符开头的单词和相同字符结尾的单词可以用“<”和“ >”通配符查找单词.1.如果要查找: 任意单个字符 :键入 ? 例如 ...

  6. PHP可变变量的简单使用

    知识点: 可变变量:简单说就是将一个变量的值用作另外一个变量的命名上,例如$a = 'b';$$a就是$b HTML代码: <!doctype html> <html> < ...

  7. ccpc 网络赛 hdu 6155

    # ccpc 网络赛 hdu 6155(矩阵乘法 + 线段树) 题意: 给出 01 串,要么询问某个区间内不同的 01 子序列数量,要么把区间翻转. 叉姐的题解: 先考虑怎么算 \(s_1, s_2, ...

  8. 获取地址栏参数 - queryString(正则表达式版本)

    获取所有query string function queryStringAll() { var reg = /(?:^|&)([^&]+)=([^&]+)(?=&|$ ...

  9. Class-dump

    What is class-dump? This is a command-line utility for examining the Objective-C runtime information ...

  10. 使用java mail的网易smtp协议 发送邮件

    package com.enation.newtest; import java.security.GeneralSecurityException; import java.util.Propert ...