Maximum Subsequence CodeForces - 888E （二分+状压）

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 is maximized. Chosen sequence can be empty.

Print the maximum possible value of .

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 .

Example
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 is equal to 7 (and that’s 3 after taking it modulo 4).

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

这题的数据范围是35，去枚举可能是不可能的，但是如果n为17或者18，也是可以的，所以这里用状压，把数据先分成两部分，然后把其中一部分用状压枚举所有可能（记得取模），然后排序去重，然后去枚举另一半的组合也是状压去枚举（v1），然后在左边去二分查找值(v2)，应该这样去找，使得v1+v2最大，当然最大只能为m-1。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int a[262148];
int p[40];
int findd(int limit,int num)//返回的是最接近num的值，但是不会大于num
{
    int l=0,r=limit-1;
    if(a[r]<=num)
        return a[r];
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(a[mid]==num)
            return a[mid];
        if(a[mid]>num)
            r=mid-1;
        else
            l=mid+1;
    }
    return a[l-1];
}
int main()
{
    int n,m;
    int k1=0;
    int k2=0;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        scanf("%d",p+i),p[i]%=m;
    int limit=n/2;
    for(int i=0;i<(1<<limit);i++)
    {
        int t=0;
        for(int j=1;j<=limit;j++)
            if(i&(1<<(j-1)))
                t=(t+p[j])%m;
        a[k1++]=t;
    }
    sort(a,a+k1);
    int num=unique(a,a+k1)-a;

    int left=n-limit;
    int ans=0;
    for(int i=0;i<(1<<left);i++)//枚举
    {
        int temp=0;
        for(int j=1;j<=left;j++)
            if(i&(1<<(j-1)))
            temp=(temp+p[limit+j])%m;
        int now=m-1-temp;
        int get=findd(num,now);
        //cout<<temp<<" "<<get<<endl;
        ans=max(ans,get+temp);
        if(ans==m-1)//如果最大就没必要举继续找了
            break;
    }
    printf("%d\n",ans);
    return 0;
}