Educational Codeforces Round 32 E. Maximum Subsequence[折半搜索+二分+状压枚举] 好题!

E. Maximum Subsequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 .

Examples

input

Copy

4 4
5 2 4 1

output

Copy

3

input

Copy

3 20
199 41 299

output

Copy

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}.

题意:n个数中取若干个数,求sigma%m最大值为多少

思路:折半搜索+二分搜索[套路]

1.折半搜索,这题一共有35个数,很显然暴力不行.

2.先二进制暴力枚举一半,求出所有可能O(1e6),sum.

3.再二进制暴力枚举另一半,求出所有可能 sum1 

4. sum1 <m && sum <m 求 max   ((sum+sum1)%m)

5.

sum1+sum一共两种情况

一:2*m>=sum1+sum >=m 那么一定有sum越大越好

二:sum1+sum<=m 那么最大值存在于m-sum1-1

#include<bits/stdc++.h>
#define PI acos(-1.0)
using namespace std;
typedef long long ll;

const int N=1e5+50;
const int MOD=1e9+7;
const int INF=0x3f3f3f3f;

ll a[50];
vector <ll> vec;

int main(void){
    ll n,m;
    cin >>n>>m;

    for(int i=1;i<=n;i++)   scanf("%lld",&a[i]),a[i]%=m;
    ll nn=n/2;
    for(int i=0;i<=(1<<nn)-1;++i){  //二进制枚举状态
        ll sum=0;
        for(int j=0;j<=20;j++){
            int t=1&(i>>j);  //位运算判断当前数加或者不加
            if(t==1)    sum+=a[j+1];
        }
        sum%=m;
        vec.push_back(sum);
    }
    sort(vec.begin(),vec.end());
    ll mx=-1;
    for(int i=0;i<=(1<<(n-nn))-1;++i){
        ll sum=0;
        for(int j=0;j<=20;j++){
            int t=1&(i>>j);
            if(t==1)    sum+=a[j+nn+1];
        }
        sum%=m;
        auto it=prev(lower_bound(vec.begin(),vec.end(),m-sum));//小技巧,寻找<=m-sum-1中最大的,即找到 第一个>=m-sum的数 的前一个数
        mx=max(mx,*it+sum);
        mx=max(mx,(*vec.rbegin()+sum)%m);

    }
    cout << mx << endl;
    return 0;
}
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