Dividing coins UVA - 562 ————01背包（平分硬币）

浅浅的花香味﹌ 2022-05-17 05:50 192阅读 0赞

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

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**分析**

第一行一个 t ，测试数据的个数  
第二行一个 n ，物品的个数  
第三行 n 个数，代表的是第 i 个物品的价值（也是重量）  
问：怎么分能使两个人拿到的的物品的价值之差最少

这一大题，可以用01背包来解决的，我们想，这么多物品让两个人来分，记物品的总价值为 sum  s u m  ，则其中一个人拿到的物品总价值肯定会是 value≤sum  v a l u e ≤ s u m   
因为物品的价值和重量是相等的

所以我们开一个最大能装重量为sum2  s u m 2 的背包，然后求这个背包能装物品的最大价值，背包的最大价值就是其中一个人拿到的物品的价值，然后，然后就得到结果了

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//#include<bits/stdc++.h>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    typedef long long ll;
    const ll MAXN=100005;
    ll dp[MAXN];
    ll v[MAXN];
    ll t,n,W;
    int main()
    {
        scanf("%lld",&t);
        while(t--)
        {
            scanf("%lld",&n);
            ll sum=0;
            for(ll i=0;i<n;i++)
            {
                scanf("%lld",&v[i]);
                sum+=v[i];
            }
            memset(dp,0,sizeof(dp));
            ll x=sum/2;
            for(ll i=0;i<n;i++)
                for(ll j=x;j>=v[i];j--)
                    dp[j]=max(dp[j],dp[j-v[i]]+v[i]);
            printf("%lld\n",sum-2*dp[x]);
        }
        return 0;
    }