$D e s c r i p t i o n$

$I n p u t$

$O u t p u t$

$S a m p l e$ $I n p u t$ #1

2 1 1 1
5 3

$S a m p l e$ $I n p u t$ #1

2 2 0 0
5 2

$S a m p l e$ $O u t p u t$ #1

$S a m p l e$ $O u t p u t$ #2

$H i n t$

$T r a i n$ $o f$ $T h o u g h t$

其实只用排个序
然后 $(\sum^{1}_{x}A[i]) * 3 ^ m + (\sum^{x + 1}_{x + y}A[i]) * 2 ^ m$
如果m == 1
再加上 $(\sum^{x+y+1}_{n}A[i])$
因为前x个肯定是贡献最大的，所以x个名额直接给前x个就行了
双倍的也是一样
所以直接x + 1到x + y的和乘上2 ^ m就行了
记得如果m等于1的时候
要把剩下的加上

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#define ll long long
//#define mod 1e9+7
using namespace std;

const int mod = 1e9 + 7;

int n, m, Sum, l, r, x, y;
int A[100005];

int power(int a,int b)
{
    
    
	int Ans = 1;
	a = a % mod;
	while(b)
	{
    
    
		if(b & 1) Ans = (ll) Ans * a % mod;
		a = (ll)a * a % mod;
		b >>= 1;
	}
	return Ans;
}

bool nm(int i, int j)
{
    
    return i > j;}

int main()
{
    
    
	scanf("%d%d%d%d", &n, &m, &l, &r);
	for(int i = 1; i <= n; ++i)
		scanf("%d", &A[i]);
	sort(A + 1, A + n + 1, nm);
	x = power(3, m), y = power(2, m);
	for(int i = 1; i <= l; ++i)
		Sum = ((ll)Sum + A[i] * x) % mod;
	for(int i = l + 1; i <= l + r; ++i)
		Sum = ((ll)Sum + A[i] * y) % mod;
	if(m == 1)
	{
    
    
		for(int i = l + r + 1; i <= n; ++i)
			Sum = ((ll)Sum + A[i]) % mod;
	}
	printf("%d", Sum);
	return 0;
}

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