Div2--B. Knights of a Polygonal Table（坑题）

B. Knights of a Polygonal Table

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Unlike Knights of a Round Table, Knights of a Polygonal Table deprived of nobility and happy to kill each other. But each knight has some power and a knight can kill another knight if and only if his power is greater than the power of victim. However, even such a knight will torment his conscience, so he can kill no more than $k$ other knights. Also, each knight has some number of coins. After a kill, a knight can pick up all victim's coins.

Now each knight ponders: how many coins he can have if only he kills other knights?

You should answer this question for each knight.

Input

The first line contains two integers $n$ and $k$ $(1\le n\le {10}^5,0\le k\le min(n-1,10\left)\right)$ — the number of knights and the number $k$ from the statement.

The second line contains $n$ integers $p_1,p_2,\dots ,p_n$ $(1\le p_i\le {10}^9)$ — powers of the knights. All $p_i$ are distinct.

The third line contains $n$ integers $c_1,c_2,\dots ,c_n$ $(0\le c_i\le {10}^9)$ — the number of coins each knight has.

Output

Print $n$ integers — the maximum number of coins each knight can have it only he kills other knights.

Examples

input

Copy

4 2
4 5 9 7
1 2 11 33

output

Copy

1 3 46 36 

input

Copy

5 1
1 2 3 4 5
1 2 3 4 5

output

Copy

1 3 5 7 9 

input

Copy

1 0
2
3

output

Copy

3 

Note

Consider the first example.

• The first knight is the weakest, so he can't kill anyone. That leaves him with the only coin he initially has.
• The second knight can kill the first knight and add his coin to his own two.
• The third knight is the strongest, but he can't kill more than $k=2$ other knights. It is optimal to kill the second and the fourth knights: $2+11+33=46$.
• The fourth knight should kill the first and the second knights: $33+1+2=36$.

In the second example the first knight can't kill anyone, while all the others should kill the one with the index less by one than their own.

In the third example there is only one knight, so he can't kill anyone.

题意：n个骑士，每个骑士可以杀死比他武力值低的骑士，并获得他杀死的骑士的金币，每个骑士最多杀死k个骑士，输出每个骑士最大金币数

思路：排序加暴力  时间 O(1e5 * n *lgn )数据大 用long long

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e5+5;

typedef long long ll;

struct node{
    ll val,id,power,v;
}ss[maxn];

int v[15],pos;

int cmp(node x,node y)
{
    return x.power<y.power;
}

int cmp1(node x,node y)
{
    return x.id<y.id;
}

int cmp2(int x,int y)
{
    return x>y;
}

int main()
{
    int x,y;
    scanf("%d %d",&x,&y);
    for(int i = 0;i<x;i++){
        scanf("%lld",&ss[i].power);
        ss[i].id = i;
    }
    for(int i = 0;i<x;i++){
        scanf("%lld",&ss[i].val);
    }
    pos = -1;
    sort(ss,ss+x,cmp);
    memset(v,0,sizeof(v));
    for(int i = 0;i<x;i++){
        if(pos != 10 ){
            pos ++;
        }
        sort(v,v+pos+1,cmp2);
        ss[i].v = 0;
        for(int j = 0;j < y && j < pos;j++){
            ss[i].v += v[j];
        }
        ss[i].v += ss[i].val;
        v[pos] = ss[i].val > v[pos] ? ss[i].val : v[pos] ;
    }
    sort(ss,ss + x,cmp1);
    for (int i = 0;i<x;i++){
        printf("%lld ",ss[i].v);
    }
    return 0;
}