Codeforces 994B. Knights of a Polygonal Table C++

题目地址：http://codeforces.com/problemset/problem/994/B

题目：

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 kk 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 nn and kk (1≤n≤105,0≤k≤min(n−1,10))(1≤n≤105,0≤k≤min(n−1,10)) — the number of knights and the number kk from the statement.

The second line contains nn integers p1,p2,…,pnp1,p2,…,pn (1≤pi≤109)(1≤pi≤109) — powers of the knights. All pipi are distinct.

The third line contains nn integers c1,c2,…,cnc1,c2,…,cn (0≤ci≤109)(0≤ci≤109) — the number of coins each knight has.

Output

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

Examples

input

4 2
4 5 9 7
1 2 11 33

output

1 3 46 36 

input

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

output

1 3 5 7 9 

input

1 0
2
3

output

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=2k=2 other knights. It is optimal to kill the second and the fourth knights: 2+11+33=462+11+33=46.
• The fourth knight should kill the first and the second knights: 33+1+2=3633+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.

思路：

STL大法好。

既然要找最大的值，原本想用优先队列做的，可是优先队列不能遍历，于是找到了这个东西——multiset.

关于它的用法可以参考https://blog.csdn.net/longshengguoji/article/details/8546286

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
using namespace std;
struct st
{
    long long int f;
    long long int m;
    long long int mm;
    int id;
}qs[100001];
bool cmp(st A,st B)
{
    return A.f<B.f;
}
int main()
{
    int n,a;
    while(scanf("%lld%lld",&n,&a)!=EOF)
    {
        long long int m[100005];
        multiset<long long int> s;
        memset(qs,0,sizeof(qs));
        for(int i=0;i<n;i++)
        {
            cin>>qs[i].f;
            qs[i].id=i;
        }
        for(int i=0;i<n;i++)
        {
            cin>>qs[i].m;
            qs[i].mm=qs[i].m;
        }
        sort(qs,qs+n,cmp);
        for(int i=0;i<n;i++)
        {
            multiset<long long int>::iterator it;
            for(it=s.begin();it!=s.end();it++)
            {
                qs[i].mm=qs[i].mm+*it;
            }
            s.insert(qs[i].m);
            while(s.size() > a){
                s.erase(s.begin());
            }
            m[qs[i].id]=qs[i].mm;
        }
        for(int i=0;i<n;i++)
        {
            cout<<m[i]<<" ";
        }
        cout<<endl;
    }
    return 0;
}