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最新推荐文章于 2025-04-12 22:04:03 发布

Vmorish

最新推荐文章于 2025-04-12 22:04:03 发布

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分类专栏：动态规划基础

本文链接：https://blog.youkuaiyun.com/cherish0222/article/details/71926153

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本文介绍了一道经典的0-1背包问题，并提供了一个实际案例分析。通过寻找能被等分为两部分的最大金额，该问题旨在帮助理解如何利用动态规划解决这类问题。

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DESCRIPTION

Kile and Pogi have found N banknotes on the street. After making sure that the original owner is nowhere to be found, they decided to split the banknotes amongst themselves. In the end they want to “win” the same amount of money sothey split the banknotes in such a way. Of course, the sum of the banknotes nobody ends up having is the least possible one. Since they couldn’t just leave the remaining banknotes on the street, they decided to go to a nearby casino and put everything on red, hoping that they would end up getting twice the money they bet. The roulette decided on the (lucky,for this time) number 13 and our heroes decided to split the money they won. The payout is such that Kile and Pogi will always be able to split the money they won into two equal parts. Because of the immense adrenaline rush, the boys have lost their mathematical abilities. Help them figure out how much money each of them is taking home.

INPUT

The first line of input contains the integer N

(1 ≤ N ≤ 500) that denotes the number of banknotes on the street. Each of the following N lines contains a single positive integer ci that denotes the value of the ith banknote in kunas (kn). The total sum of money will not exceed 100 000 kn.

OUTPUT

You must output the amount of money each of them took home.

SAMPLE INPUT

4 2 3 1 6

5 2 3 5 8 13

SAMPLE OUTPUT

分析：描述的这么一大堆的东西和题意无关！？原谅我六级还没过……看了题解之后，才知道这是0-1背包的问题，求组成的最大n且n为偶数，且也可以组成n/2，最终结果为（sum-n）+n/2

参考代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;
const int maxn = 500+10;
const int maxm = 100000+10;
int n;
int c[maxn];
int dp[maxm];

int main()
{
    while( ~scanf("%d",&n))
    {
        int sum = 0;
        for( int i = 1; i <= n; i++)
        {
            scanf("%d",&c[i]);
            sum += c[i];
        }
        memset(dp,0,sizeof(dp));
        dp[0] = 1;
        for( int i = 1; i <= n; i++)
        {
            for( int j = sum; j >= c[i]; j--)
                if( dp[j-c[i]])
                    dp[j] = 1;
        }

        //for( int i = 1; i <= sum; i++)
          //  printf("%d_%d\n",i,dp[i]);

        int tmp = 0;
        for( int i = sum; i >= 0; i--)
        {
            if( !(i&1) && dp[i] && dp[i>>1])
            {
                tmp = (i>>1);
                break;
            }
        }
        //printf("%d %d ",sum,tmp);
        printf("%d\n",sum-tmp);

    }

    return 0;
}