2019年个人训练赛第一场-E - Equalizing by Division (hard version)

标签：sum   rac   namespace   asn   msu   form   iostream   数通   als   

The only difference between easy and hard versions is the number of elements in the array.

You are given an array aa consisting of nn integers. In one move you can choose any aiai and divide it by 22 rounding down (in other words, in one move you can set ai:=⌊ai2⌋ai:=⌊ai2⌋).

You can perform such an operation any (possibly, zero) number of times with any aiai.

Your task is to calculate the minimum possible number of operations required to obtain at least kk equal numbers in the array.

Don‘t forget that it is possible to have ai=0ai=0 after some operations, thus the answer always exists.

Input

The first line of the input contains two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — the number of elements in the array and the number of equal numbers required.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤2⋅1051≤ai≤2⋅105), where aiai is the ii-th element of aa.

Output

The first line of the input contains two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — the number of elements in the array and the number of equal numbers required.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤2⋅1051≤ai≤2⋅105), where aiai is the ii-th element of aa.

Exanple

5 3
1 2 2 4 5

1

5 3
1 2 3 4 5

2

5 3
1 2 3 3 3

0

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdio> 
#define INF 0x3f3f3f3f
using namespace std;

priority_queue<int,vector<int>,greater<int> > q[(int)3e5];

int main(){
    int x,y,a;
    scanf("%d%d",&x,&y);
    for(int i=0;i<x;i++){
        scanf("%d",&a);
        q[a].push(0);
        int cnt=1;
        while(a){    
            a/=2;
            q[a].push(cnt);
            cnt++;
        }
    }
    
    int asn=INF;
    for(int i=0;i < 3e5;i++){
        if(q[i].size()<y)
            continue;
        else{
            int sum=0;
            for(int j=0;j<y;j++){
                    sum=sum+q[i].top();
                    q[i].pop();
            }
            asn=min(asn, sum);
    }
}

    printf("%d\n",asn);
    return 0;
}

2019年个人训练赛第一场-E - Equalizing by Division (hard version)

标签：sum   rac   namespace   asn   msu   form   iostream   数通   als   

原文地址：https://www.cnblogs.com/gbtj/p/11487809.html