Cards and Joy CodeForces - 999F （贪心+set）

There are $n$

There are $k \cdot n$

The players have to distribute all the cards in such a way that each of them will hold exactly $k$

Print the maximum possible total joy levels of the players after the cards are distributed. Note that the sequence $h_{1}, \dots, h_{k}$

Input

The first line of input contains two integers $n$

The second line contains $k \cdot n$

The third line contains $n$

The fourth line contains $k$

Output

Print one integer — the maximum possible total joy levels of the players among all possible card distributions.

Examples

Input

4 3
1 3 2 8 5 5 8 2 2 8 5 2
1 2 2 5
2 6 7

Output

Input

3 3
9 9 9 9 9 9 9 9 9
1 2 3
1 2 3

Output

Note

In the first example, one possible optimal card distribution is the following:

Player $1$
Player $2$
Player $3$
Player $4$

Thus, the answer is $2 + 6 + 6 + 7 = 21$

In the second example, no player can get a card with his favorite number. Thus, the answer is $0$

$0$

x=a[i]%m

然后我们就要从set中找到那个让a[i]增加值最小的那个取模后的数，

如果x比set中所有的数大，那么我们必须让它加一点数然后变成%m后是set中最小数才可以成本最低。

否则我们需要用到set中的一个函数lower_bound()

这个函数的作用想必都知道，就说在set中找到第一个大于等于x的数。

注意函数返回的是一个set的迭代器，*iterator 才是取值。

然后把a[i]进行改变，最后输出答案。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll n,m;
ll a[maxn];
ll b[maxn];
ll c[maxn];
int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    gbtb;
    cin>>n>>m;
    ll num=n/m;
    repd(i,1,n)
    {
        cin>>a[i];
    }
    set<int> st;
    repd(i,0,m-1)
    {
        st.insert(i);
    }
    ll x;
    ll y;
    ll ans=0ll;
    repd(i,1,n)
    {
        x=a[i]%m;
        if(x>(*st.rbegin()))
        {
            y=*st.begin();
        }else
        {
            y=*st.lower_bound(x);
        }
        b[y]++;
        if(b[y]==num)
        {
            st.erase(y);
        }
        a[i]+=((y-x)+m)%m;
        ans+=((y-x)+m)%m;
    }
    cout<<ans<<endl;
    repd(i,1,n)
    {
        cout<<a[i]<<" ";
    }
    cout<<endl;
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}

Cards and Joy CodeForces - 999F （贪心+set）

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