N柱砖,希望有连续K柱的高度是一样的. 你可以选择以下两个动作 1:从某柱砖的顶端拿一块砖出来,丢掉不要了. 2:从仓库中拿出一块砖,放到另一柱.仓库无限大. 现在希望用最小次数的动作完成任务.

第一行给出N,K. (1 ≤ k ≤ n ≤ 100000), 下面N行,每行代表这柱砖的高度.0 ≤ hi ≤ 1000000

最小的动作次数

#include <bits/stdc++.h>#define ll long long
#define inf 0x7ffffffff
#define il inline
#define debug puts("233")
#define int long longnamespace io {#define in(a) a=read()#define out(a) write(a)#define outn(a) out(a),putchar('\n')#define I_int intinline I_int read() {I_int x = 0 , f = 1 ; char c = getchar() ;while( c < '0' || c > '9' ) { if( c == '-' ) f = -1 ; c = getchar() ; }while( c >= '0' && c <= '9' ) { x = x * 10 + c - '0' ; c = getchar() ; }return x * f ;}char F[ 200 ] ;inline void write( I_int x ) {if( x == 0 ) { putchar( '0' ) ; return ; }I_int tmp = x > 0 ? x : -x ;if( x < 0 ) putchar( '-' ) ;int cnt = 0 ;while( tmp > 0 ) {F[ cnt ++ ] = tmp % 10 + '0' ;tmp /= 10 ;}while( cnt > 0 ) putchar( F[ -- cnt ] ) ;}#undef I_int}
using namespace io ;using namespace std ;#define N 100010int n , k , root = 1 ,tot = 0;
struct fhq {int siz , val , rnk , lc , rc , sum ;
}t[N];void up(int rt) { t[rt].siz = t[t[rt].lc].siz + t[t[rt].rc].siz + 1;t[rt].sum = t[t[rt].lc].sum + t[t[rt].rc].sum + t[rt].val;
}void split(int &a , int &b , int val , int rt) {if(!rt) {a = b = 0;return;}if(t[rt].val <= val) a = rt , split(t[a].rc , b , val , t[rt].rc) ;else b = rt , split(a , t[b].lc , val , t[rt].lc) ;up(rt);
}void merge(int a , int b , int &rt) {if(!a || !b) {rt = a + b;return;}if(t[a].rnk < t[b].rnk) rt = a , merge(t[a].rc , b , t[rt].rc) ;else rt = b , merge(a , t[b].lc , t[rt].lc) ;up(rt);
}int new_node(int val) {t[++tot] = (fhq) {1 , val , rand() , 0 , 0 , val} ;return tot ;
}void insert(int val) {int x = 0 , y = 0 , z = new_node(val) ;split(x , y , val - 1, root) ;merge(x , z , x) ; merge(x , y , root) ;
}void erase(int val) {int x = 0 , y = 0 , z = 0;split(x , y , val , root); split(x , z , val - 1 , x);merge(t[z].lc , t[z].rc , z) ; merge(x , z , x) ; merge(x , y , root) ;
}int find_kth(int rnk , int rt) {while(1) {if(t[t[rt].lc].siz + 1 < rnk) rnk -= t[t[rt].lc].siz + 1, rt = t[rt].rc;else if(t[t[rt].lc].siz >= rnk) rt = t[rt].lc ;else return t[rt].val; }
}void print(int rt) {if(t[rt].lc) print(t[rt].lc) ;cout<<t[rt].val<<" ";if(t[rt].rc) print(t[rt].rc) ;
}int query(int val) {int x = 0 , y = 0 , z = 0 ;split(x , y , val , root) ;split(x , z , val - 1 , x) ;int s1 = 1ll * (t[x].siz - 1)* val , s2 = 1ll * t[y].siz * val ;int a1 = t[x].sum , a2 = t[y].sum ;
//    cout<<a1<<" "<<a2<<" "<<endl;
    merge(x , z , x) ; merge(x , y , root) ;
//    cout<<s1-a1+a2-s2<<endl;return s1 - a1 - s2 + a2 ;
}int a[N] ;signed main() {
/*    freopen("1112.in","r",stdin);freopen("1112.out","w",stdout);*/srand((unsigned)time(0)) ;n = read() , k = read() ;new_node(0) ; t[1].siz = 0 ;for(int i = 1 ; i <= n ; i ++) a[i] = read() ;ll ans = 2e18 , M = 0 , L = 0 , R = 0; a[0] = 0;for(int i = 0 ; i < k ; i ++) insert(a[i]) ;int l = 0 , r = k - 1 ;while(r < n) {++l , ++r ;erase(a[l - 1]) ; insert(a[r]) ;int mid = find_kth((k>>1)+2 , root) ;ll tmp = query(mid) ;if(tmp < ans) {M = mid ; L = l ; R = r;ans = tmp;}
//        print(root);puts("");
//        cout<<l<< " "<<r<<" "<<mid<<endl;
    }outn(ans);
/*    for(int i = 1 ; i < L ; i ++) outn(a[i]) ;for(int i = L ; i <= R; i ++) outn(M) ;for(int i = R + 1 ; i <= n; i ++) outn(a[i]) ;*/return 0;
}