Codeforces 981 D. Bookshelves [dp + 贪心]

最新推荐文章于 2020-08-09 12:52:53 发布

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最新推荐文章于 2020-08-09 12:52:53 发布

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分类专栏：动态规划动态规划---------递推DP codeforces

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本文介绍了一个关于书架分配的问题及解决方法。问题旨在通过合理的书本分配，使得多个书架的总价值（定义为所有书架价值的按位与运算结果）最大化。文章提供了完整的代码实现，采用动态规划的方法来寻找最优解。

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Mr Keks is a typical white-collar in Byteland.

He has a bookshelf in his office with some books on it, each book has an integer positive price.

Mr Keks defines the value of a shelf as the sum of books prices on it.

Miraculously, Mr Keks was promoted and now he is moving into a new office.

He learned that in the new office he will have not a single bookshelf, but exactly $k$ bookshelves. He decided that the beauty of the $k$ shelves is the bitwise AND of the values of all the shelves.

He also decided that he won't spend time on reordering the books, so he will place several first books on the first shelf, several next books on the next shelf and so on. Of course, he will place at least one book on each shelf. This way he will put all his books on $k$ shelves in such a way that the beauty of the shelves is as large as possible. Compute this maximum possible beauty.

Input

The first line contains two integers $n$ and $k$ ( $1 \leq k \leq n \leq 50$ ) — the number of books and the number of shelves in the new office.

The second line contains $n$ integers $a_{1}, a_{2}, \dots a_{n}$ , ( $0 < a_{i} < 2^{50}$ ) — the prices of the books in the order they stand on the old shelf.

Output

Print the maximum possible beauty of $k$ shelves in the new office.

  Examples
 
    input
    
     Copy
    
10 4
9 14 28 1 7 13 15 29 2 31

    output
    
     Copy
    
24

    input
    
     Copy
    
7 3
3 14 15 92 65 35 89

    output
    
     Copy
    
64

Note

In the first example you can split the books as follows:

$(9 + 14 + 28 + 1 + 7) & (13 + 15) & (29 + 2) & (31) = 24.$

In the second example you can split the books as follows:

$(3 + 14 + 15 + 92) & (65) & (35 + 89) = 64.$

代码：

#include<bits/stdc++.h>
#define ll long long

using namespace std;

int n, k;
ll a[57], sum[57];
bool dp[57][57];

bool check(ll x)
{
    memset(dp, 0, sizeof(dp));
    dp[0][0] = true;
    for(int i = 1; i <= n; ++i)
    {
        for(int j = 1; j <= k; ++j)
        {
            for(int l = 1; l <= i; ++l)
            {
                if((sum[i] - sum[l -1]&x) == x&&dp[l - 1][j -1])
                {
                    dp[i][j] = true;
                    break;
                }
            }
        }
    }
    return dp[n][k];
}


int main()
{
    //freopen("in.txt", "r", stdin);
    while(~scanf("%d%d", &n, &k))
    {
        for(int i = 1; i <= n; ++i) {
            scanf("%lld", a+i);
            sum[i] = sum[i - 1] + a[i];
        }
        ll ans = 0;
        for(ll x = (1ll<<60); x; x >>= 1)
        {
            if(check(ans|x))
            {
                ans |= x;
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}