D2. Equalizing by Division (hard version)

D2. Equalizing by Division (hard version)

涉及下标运算一定要注意下标是否越界！！！

思路，暴力判断以每个数字为到达态最小花费

#include<bits/stdc++.h>
using namespace std;
#define sc(x) scanf("%I64d",&x);
#define read(A) for(int i=1;i<=n;i++)scanf("%I64d",&A[i]);
#define int long long
#define P pair<int,int>
#define fi first
#define se second
#define endl '
'
#define ll long long
#define maxn 200000+10
int n,m,T;
int A[maxn];
int B[200005];
int ch(int x,int y)
{
    for(int i=1;i<=y;i++){
        x/=2;
    }
    return x;
}
int Ans=1e18;
int check(int x,int t)
{
    int _x=x;
    if(x==0)
    {
        int ans=0;
        for(int i=B[0]; i<n; i++)
        {
            int c=A[i];

            while(c)
            {
                ans++;
                c/=2;
                if(ans>Ans){
                    return 1e18;
                }
            }
            t++;
            //cout<<t<<m<<endl;
            if(t==m)
            {
                return ans;
            }
        }
    }
    int ans=0;
    int k=2;
    int y=1;
    x*=2;
    while(x<=200000&&t<m)
    {
        for(int i=0; i<k; i++)
        {
            if(ch(x+i,y)!=_x)break;
            if(x+i>200000)break;
            if(x+i<=200000&&B[x+i]>=m-t)
            {
                ans+=(m-t)*y;
                if(ans>Ans)return 1e18;
                t=m;
                return ans;
            }
            else
            {
                ans+=(B[x+i])*y;
                 if(ans>Ans)return 1e18;
                t+=B[x+i];
            }

        }
        x*=2;
        y++;
        k*=2;
    }
    if(m<=t)return ans;
    else
        return 1e18;
}
signed main()
{
    sc(n);
    sc(m);
    for(int i=0; i<n; i++)
    {
        sc(A[i]);
        //cout<<A[i]<<endl;
        B[A[i]]++;
        if(B[A[i]]>=m)
        {
            puts("0");
            return 0;
        }
    }
    sort(A,A+n);
    int t=0;

    for(int i=0; i<=100000; i++)
    {
        t=check(i,B[i]);
       // if(t<ans)cout<<i<<endl;
        Ans=min(t,Ans);
    }
    cout<<Ans<<'
';


}