CF888E：Maximum Subsequence（中途相遇）

最新推荐文章于 2020-12-13 19:51:07 发布

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本文介绍了一种通过折半搜索和二分法求解给定数列中若干个数的和取模最大值的问题，提供了完整的C++代码实现。

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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}.

题意：给一个数列和m，在数列任选若干个数的和取模m最大。

思路：2的35次方会超时，考虑折半搜索，前后分别枚举，最后二分取最大值即可。

# include <bits/stdc++.h>
# define pb push_back
# define mp make_pair
using namespace std;
typedef long long LL;
LL a[40], n, m;
set<LL>s, t;
void dfs(int cur, LL sum)
{
    if(cur == n/2+1)
    {

        sum = sum%m;
        s.insert(sum);
        return;
    }
    dfs(cur+1, sum);
    dfs(cur+1, sum+a[cur]);
}
void dfs2(int cur, LL sum)
{
    if(cur == n+1)
    {
        sum = sum%m;
        t.insert(sum);
        return;
    }
    dfs2(cur+1, sum);
    dfs2(cur+1,sum+a[cur]);
}
int main()
{
    scanf("%lld%lld",&n,&m);
    for(int i=1; i<=n; ++i) scanf("%lld",&a[i]);
    dfs(1, 0);
    dfs2(n/2+1, 0);
    LL ans = 0;
    for(auto it=s.begin(); it!=s.end(); ++it)
    {
        LL tmp = *it;
        auto pos = t.lower_bound(m-tmp);
        if(pos != t.begin())
        {
            --pos;
            if(*pos + tmp <= m)
            ans = max(ans, *pos + tmp);
        }
    }
    printf("%lld\n",ans);
    return 0;
}