CodeForces 364 B. Free Market

最新推荐文章于 2019-07-10 17:44:26 发布

最新推荐文章于 2019-07-10 17:44:26 发布 · 154 阅读

本文介绍了一个关于物品交换的问题——FreeMarket。在这个问题中，John需要通过一系列的交换获得总价值最大的物品集合，并且要满足他的正义感准则。文章详细解释了交换规则，并给出了一种算法实现方案。

John Doe has recently found a "Free Market" in his city — that is the place where you can exchange some of your possessions for other things for free.

John knows that his city hasnitems in total (each item is unique). You can bring any number of items to the market and exchange them for any other one. Note that each item is one of a kind and that means that you cannot exchange set{a, b}for set{v, a}. However, you can always exchange setxfor any sety, unless there is itemp, such thatpoccurs inxandpoccurs iny.

For each item, John knows its valueci. John's sense of justice doesn't let him exchange a set of itemsxfor a set of itemsy, ifs(x) + d < s(y)(s(x)is the total price of items in the setx).

During one day John can exchange only one set of items for something else. Initially, he has no items. John wants to get a set of items with the maximum total price. Find the cost of such set and the minimum number of days John can get it in.

Input

The first line contains two space-separated integersn,d(1 ≤ n ≤ 50,1 ≤ d ≤ 104) — the number of items on the market and John's sense of justice value, correspondingly. The second line containsnspace-separated integersci(1 ≤ ci ≤ 104).

Output

Print two space-separated integers: the maximum possible price in the set of items John can get and the minimum number of days needed to get such set.

Sample test(s)

input
3 2
1 3 10

output
4 3

input
3 5
1 2 3

output
6 2

input
10 10000
10000 9999 1 10000 10000 10000 1 2 3 4

output
50010 6

Note

In the first sample John can act like this:

Take the first item (1 - 0 ≤ 2).
Exchange the first item for the second one (3 - 1 ≤ 2).
Take the first item (1 - 0 ≤ 2).

有没有相交的部分不重要，预处理出所有可能的组合重量，两两都是可以通过去掉相同的部分一次交换得到

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

bool dp[51*10001];

int main()
{
    int n,d,x,cnt=0,sum=0,w=0;
    cin>>n>>d;
    dp[0]=1;
    for(int i=0;i<n;i++)
    {
        cin>>x;
        for(int j=(sum+=x);j>=x;j--)
        {
            if(dp[j-x]) dp[j]=true;
        }
    }
    while(true)
    {
        int j=w+d;
        while(!dp[j]&&j>w) j--;
        if(j==w) break;
        w=j;cnt++;
    }
    cout<<w<<" "<<cnt<<endl;
    return 0;
}