codeforces 888 E. Maximum Subsequence（好题）

最大子序列和问题

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本文介绍了一个关于寻找整数数组中特定子序列的问题，该子序列使得其元素和除以给定整数后的余数最大化。文章提供了一个解决此问题的C++实现代码示例，该算法采用两阶段选择子序列的方法并利用排序和二分查找来提高效率。

You are given an array a consisting of n integers, and additionally an integer m. You have to choose some sequence of indices b1, b2, ..., bk (1 ≤ b1 < b2 < ... < bk ≤ n) in such a way that the value of $\text{[math]}$ is maximized. Chosen sequence can be empty.

Print the maximum possible value of $\text{[math]}$ .

Input

The first line contains two integers n and m (1 ≤ n ≤ 35, 1 ≤ m ≤ 109).

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print the maximum possible value of $\text{[math]}$ .

Examples

input
4 4
5 2 4 1

output
3

input
3 20
199 41 299

output
19

Note

In the first example you can choose a sequence b = {1, 2}, so the sum $\text{[math]}$ is equal to 7 (and that's 3 after taking it modulo 4).

In the second example you can choose a sequence b = {3}.

AC code

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

int n,m;
vector<int> A,X,Y;

void gen(int l, int r, vector<int> &X)
{
    X.push_back(0);
    for (int i=l; i<=r; ++i)
    {
        int s = X.size();
        for (int x=0; x<s; ++x)
            X.push_back((A[i]+X[x])%m);
    }
}

int main()
{
    cin >> n >> m;
    A.resize(n);
    for (int i=0; i<n; ++i)
        cin >> A[i];

    int mid = n/2;
    gen(0,mid,X);
    gen(mid+1,n-1,Y);
    sort(Y.begin(),Y.end());

    int res = 0;
    for (auto ix=X.begin(); ix!=X.end(); ++ix)
        res = max(res,*ix + *(upper_bound(Y.begin(),Y.end(),m-1-*ix)-1));
    cout << res << endl;
    return 0;
}