Avito Code Challenge 2018 D. Bookshelves

D. Bookshelves

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 $k$ ($1\le k\le n\le 50$) — 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:

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

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

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

 

题意：

            给 n 个数  ，要求将 n 个数分为 k 个区域， 使每个区域元素和的and和最大，求最大值。

思路：既然是and和最大，那么我们从最高位开始枚举 答案该位为 1 的时候，然后通过dp转移一哈。

 

过题代码

#include <bits/stdc++.h>
using namespace std;
const int maxn=70;
typedef long long ll;
ll sum[maxn],n,k,x,ans=0;
bool dp[maxn][maxn];
int main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie();
    cin>>n>>k;
    sum[0]=0;
    for (int i=1; i<=n; i++) cin>>x,sum[i]=sum[i-1]+x;
    for (int p=60; p>=0; p--){
        memset(dp,false,sizeof(dp));
        ll a=(ans|(1<<p));
        for (int i=1; i<=n; i++)
            for (int j=1; j<i; j++){
                if(j==1) dp[i][j]=((sum[i]&a)==a);
                else
                    for (int s=1; s<i; s++)
                        if(dp[s][j-1])
                            dp[i][j]|=(((sum[i]-sum[s])&a)==a);
            }
        if(dp[n][k]) ans=a;
    }
    cout<<ans<<endl;
    return 0;
}