C 纸牌游戏

C 纸牌游戏

• 时，这时的的还没被淘汰，所以还得加上
• 时，这时的已经被淘汰了，所以降序排序后，就一直取前的数即可，快速幂写一下就行了。

#include<bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")

using namespace std;

#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define ll long long
#define ull unsigned long long
#define rint register int
#define ld long double
#define db double
#define rep(i, l, r) for (rint i = l; i <= r; i++)
#define rep1(i, a, n) for (rint i = a; i < n; i++)
#define per(i, l, r) for (rint i = l; i >= r; i--)
#define per1(i ,a, n) for (rint i = a; i > n; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define sd(x) scanf("%d",&(x))
#define slld(x) scanf("%lld",&(x))
#define sdd(x,y) scanf("%d%d",&(x),&(y))
#define sc(s) scanf("%s",(s))
#define pd(x) printf("%d\n",(x))
#define plld(x) printf("%lld\n",(x))
#define pdk(x) printf("%d ",(x))
const int inf=0x3f3f3f3f;

namespace IO{
    char ibuf[1<<21],*ip=ibuf,*ip_=ibuf;
    char obuf[1<<21],*op=obuf,*op_=obuf+(1<<21);
    inline char gc(){
        if(ip!=ip_)return *ip++;
        ip=ibuf;ip_=ip+fread(ibuf,1,1<<21,stdin);
        return ip==ip_?EOF:*ip++;
    }
    inline void pc(char c){
        if(op==op_)fwrite(obuf,1,1<<21,stdout),op=obuf;
        *op++=c;
    }
    inline ll read(){
        register ll x=0,ch=gc(),w=1;
        for(;ch<'0'||ch>'9';ch=gc())if(ch=='-')w=-1;
        for(;ch>='0'&&ch<='9';ch=gc())x=x*10+ch-48;
        return w*x;
    }
    template<class I>
    inline void write(I x){
        if(x<0)pc('-'),x=-x;
        if(x>9)write(x/10);pc(x%10+'0');
    }
    class flusher_{
    public:
        ~flusher_(){if(op!=obuf)fwrite(obuf,1,op-obuf,stdout);}
    }IO_flusher;
}
using namespace IO;
ll qm (ll a, ll b ,ll c){
    ll ret=1%c;
    while(b){
        if(b&1)
            ret=ret*a%c;
        a=a*a%c;
        b=b>>1;
    }
    return ret;
}
const int N=1e6+10;
const int mod=1e9+7;
ll n,m,x,y,a[N],ans1,ans2,ans;
int main()
{
    n=read();m=read();x=read();y=read();
    rep(i, 1, n) {
        a[i]=read();
    }
    sort(a+1,a+1+n,greater<ll>());
    rep(i, 1, n){
        if(i<=x) ans1=(ans1+a[i])%mod;
        if(i>x && i<=x+y) ans2=(ans2+a[i])%mod;
    }
    if(m < 2){
        rep(i, x+y+1, n){
            ans=(ans+a[i])%mod;
        }
    }
    write((ans%mod+ans1*qm(3,m,mod)%mod+ans2*qm(2,m,mod)%mod)%mod);
    return 0;
}