[codeforces888E]Maximum Subsequence

最新推荐文章于 2024-12-06 19:48:15 发布

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最新推荐文章于 2024-12-06 19:48:15 发布

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本文探讨了一种算法，用于解决从一组整数中选择子集的问题，目标是使子集元素之和对特定模数取模后的结果最大化。通过折半搜索策略，将问题分解为两部分进行求解，最终输出最优解。

time limit per test : 1 second
memory limit per test : 256 megabytes
You are given an array a consisting of $n$ integers, and additionally an integer $m$ . You have to choose some sequence of indices $b_1, b_2, ..., b_k (1 ≤ b_1 < b_2 < ... < b_k ≤ n)$ in such a way that the value of is maximized. Chosen sequence can be empty.
Print the maximum possible value of $(∑1kabi)(\sum^k_1 {a_{b_i}})$ $m o d$ $m$ .

Input

The first line contains two integers $n$ and $m (1 ≤ n ≤ 35, 1 ≤ m ≤ 10^9)$ .
The second line contains $n$ integers $a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9)$ .

Output
Print the maximum possible value of $(∑1kabi)(\sum^k_1 {a_{b_i}})$ $m o d$ $m$ .

Examples

Input

4 4
5 2 4 1

Output

Input

3 20
199 41 299

Output

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 $b = {3}$ .

题意：
从 $n$ 个数字里面选任意个，使得他们的和对 $m$ 取模之后最大。输出这个最大值

题解：
折半搜索。先搜出前半段的所有和的结果，存在set里面，然后去后半段搜索的时候，假设当前状态的和取模后是 $x$ ，那么我们就在set里面找小于等于m-x-1的最大数字。（记得用*--upper_bound(x)）

tips:
upper_bound找的是大于x的第一个。
lower_bound找的是大于等于x的第一个。

#include<bits/stdc++.h>
#define LiangJiaJun main
using namespace std;
int n,m,c[2];
int a[44][2];
int calc(int status,int r,int sit){
    int ret=0;
    for(int i=1;i<=r;i++){
        if(status&(1<<(i-1))){
            ret=(ret+a[i][sit])%m;
        }
    }
    return ret;
}
set<int>tr;
int w33ha(){
    tr.clear();
    int half=n/2;
    c[0]=half;
    c[1]=n-half;
    for(int i=1;i<=n;i++){
        int x;
        scanf("%d",&x);
        if(i<=half)a[i][0]=x;
        else a[i-half][1]=x;
    }
    int ans=0;
    for(int S=0;S<(1<<c[0]);S++){
        int now=calc(S,c[0],0);
        tr.insert(now);
    }
    for(int S=0;S<(1<<c[1]);S++){
        int now=calc(S,c[1],1);
        int G=(*--tr.upper_bound(m-now-1));
        ans=max(ans,G+now);
    }
    printf("%d\n",ans);
    return 0;
}
int LiangJiaJun(){
    while(scanf("%d%d",&n,&m)!=EOF)w33ha();
    return 0;
}