Problem:
Score : 500 points

Problem Statement

Given are N integers A_1,\ldots,A_N.

We will choose exactly K of these elements. Find the maximum possible product of the chosen elements.

Then, print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Constraints

1 \leq K \leq N \leq 2\times 10^5

|A_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N K
A_1 \ldots A_N

Output

Print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

Sample Input 1

4 2
1 2 -3 -4

Sample Output 1

12

The possible products of the two chosen elements are 2, -3, -4, -6, -8, and 12, so the maximum product is 12.

Sample Input 2

4 3
-1 -2 -3 -4

Sample Output 2

1000000001

The possible products of the three chosen elements are -24, -12, -8, and -6, so the maximum product is -6.

We print this value modulo (10^9+7), that is, 1000000001.

Sample Input 3

2 1
-1 1000000000

Sample Output 3

1000000000

The possible products of the one chosen element are -1 and 1000000000, so the maximum product is 1000000000.

Sample Input 4

10 10
1000000000 100000000 10000000 1000000 100000 10000 1000 100 10 1

Sample Output 4

999983200

Be sure to print the product modulo (10^9+7).

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define int long long
#define mod 1000000007
using namespace std;
int ar[200002] ;
vector<int>ps,ne,tup ;
bool cmp(int a , int b) {return a>b ; }
int32_t main()
{
    int n,k ; cin>>n>>k ;
    int an=1 ;
    for(int i = 0 ; i < n ; i++) {cin>>ar[i] ; if(ar[i]>=0) an=0 ;}
    if(k==n)
    {
        int ans=1 ;
        for(int i = 0 ; i < n ; i++)
        {
            int x=ar[i] ;
            x=(x+mod)%mod ;
            ans=(ans*x)%mod ;
        }
        cout<<ans<<endl ;
    }
    else
    {
        if(an)
        {
           // cout<<"jkl "<<k<<endl ;
            if(k&1ll) sort(ar,ar+n,cmp) ;
            else sort(ar,ar+n) ;
            int ans=1 ;
            for(int i = 0 ; i < k ; i++)
            {
                int x=ar[i] ;
                x=(x+mod)%mod ;
                ans=(ans*x)%mod ;
            }
            cout<<ans<<endl ;
        }
        else
        {
           // cout<<"oo"<<endl ;
            int ans=1 ;
            for(int i = 0 ; i < n ; i++)
            {
                if(ar[i]>=0) ps.push_back(ar[i]) ;
                else ne.push_back(ar[i]) ;
            }
            sort(ps.begin(),ps.end()) ;
            sort(ne.rbegin(),ne.rend()) ;
          //  for(int i = 0 ; i < ps.size() ; i++) cout<<ps[i]<<" " ; cout<<endl ;
          //  for(int i = 0 ; i < ne.size() ; i++) cout<<ne[i]<<" " ; cout<<endl ;
            if(k&1)
            {
                int x=ps.back() ; ps.pop_back() ;
                ans=(ans*x*1ll)%mod ;
            }
            while((int)ps.size()>1)
            {
                int x=ps.back() ; ps.pop_back() ;
                int y=ps.back() ; ps.pop_back() ;
                int h=(x*y) ;
                tup.push_back(h) ;
            }
            while((int)ne.size()>1)
            {
                int x=ne.back() ; ne.pop_back() ;
                int y=ne.back() ; ne.pop_back() ;
                int h=(x*y) ;
                tup.push_back(h) ;
            }
            sort(tup.rbegin(),tup.rend()) ;
            //for(int i = 0 ; i < (int)tup.size() ; i++) cout<<tup[i]<<" " ; cout<<endl ;
            for(int i = 0 ; i < k/2 ; i++)
            {
                ans%=mod ; tup[i]%=mod ;
                ans=(ans*tup[i])%mod ;
            }
            cout<<ans%mod<<endl ;

        }
    }

}
// -1000000000 -1 -1000000000 -1000000000 -1000000000