Description

N柱磚,希望有連續K柱的高度是一樣的. 你可以選擇以下兩個動作 1:從某柱磚的頂端拿一塊磚出來,丟掉不要了. 2:從倉庫中拿出一塊磚,放到另一柱.倉庫無限大. 現在希望用最小次數的動作完成任務.

Input

第一行給出N,K. (1 ≤ k ≤ n ≤ 100000), 下面N行,每行代表這柱磚的高度.0 ≤ hi ≤ 1000000

Output

最小的動作次數

Sample Input

Sample Output

#include<cmath>
#include
 
<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;
int lowbit(int x){return x&-x;}
ll c[2100000],s[2100000];
void change(ll x,ll d)
{
    while(x<=1100000)
    {
        c[x]+=d;
        x+=lowbit(x);
    }
}
ll getshuliang(ll x)
{
    ll sum
 
=0;
    while(x!=0)
    {
        sum+=c[x];
        x-=lowbit(x);
    }
    return sum;
}
ll getpaiming(ll sum)
{
    ll x=0,now=0;
    for(ll i=(1ll<<22);i;i>>=1)
    {
        if(x+i<=1100000&&now+c[x+i]<sum)
        now+=c[x+i],x+=i;
    }
    ++x;
    return x;
}
 
void gai(ll x,ll d)
{
    while(x<=1100000)
    {
        s[x]+=d;
        x+=lowbit(x);
    }
}
ll getsum(ll x)
{
    ll sum=0ll;
    while(x)
    {
        sum+=s[x];
        x-=lowbit(x);
    }
    return sum;
}
ll dd[1100000];
int n,k;
ll a[110000];
int main()
{
    freopen("akc.in","r",stdin);
    freopen("akc.out","w",stdout);
    ll ans=1ll<<61;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)scanf("%lld",&a[i]),a[i]++;
    int pos=1;
    for(int i=1;i<k;i++)dd[a[i]]++,change(a[i],1),gai(a[i],a[i]);
    for(int i=k;i<=n;i++)
    {
        change(a[i],1),gai(a[i],a[i]);dd[a[i]]++;
        ll X;
        if(k%2==1)X=getpaiming(k/2+1);
        else X=(getpaiming(k/2)+getpaiming(k/2+1))/2;

        ll sum1=getsum(X-1),sum2=getsum(getpaiming(k))-sum1-dd[X]*X;
        ll geshu1=getshuliang(X-1),geshu2=k-geshu1-dd[X];
        
        ans=min(ans,X*geshu1-sum1+sum2-X*geshu2);
        if(ans==0ll)break;
        change(a[pos],-1),gai(a[pos],-a[pos]);dd[a[pos]]--;
        pos++;
    }
    printf("%lld\n",ans);
    return 0;
}

by_lmy

BZOJ1112: [POI2008]磚塊Klo