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最新推荐文章于 2020-07-22 20:14:03 发布

转载最新推荐文章于 2020-07-22 20:14:03 发布 · 64 阅读

本文探讨了如何公平地将一袋最多100枚、面值各不相同的硬币分配给两个人的问题。通过动态规划的方法，实现了寻找最小差异值的目标。

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Description

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 ( $m \le 100$ ) 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

Sample Output

0
1

dp题，dp[i]的只有 0和1，0表示不能组成钱数i,1表示可以。转移方程：if(dp[i]) dp[j+coin[i]]=1; 这样把所有的可能组成结果找出来，

然后从sum/2向下找（向上也行），第一个找到的就是所求的结果，表示某一个人能过拿到的钱数，然后进行进一步计算。

#include<bits/stdc++.h>
using namespace std;
int dp[500*100+100];
int main()
{
    int n;
    cin>>n;
    int coin[105];
    while (n--)
    {
        memset(dp,0,sizeof(dp));
        long long int sum=0;
        int m;
        cin>>m;
        for(int i=0;i<m;i++)
        {
            cin>>coin[i];
            sum+=coin[i];
        }
        dp[0]=1;
        for(int i=0;i<m;i++)
        {
            for(int j=sum;j>=0;j--)
                if(dp[j])
                    dp[j+coin[i]]=1;
        }
        int j=sum/2;
        while(!dp[j]) j--;
        printf("%d\n",abs(sum-2*j));

    }
    return 0;
}