题意：
Takahashi is going to buy N items one by one.
The price of the i-th item he buys is Ai yen (the currency of Japan).
He has M discount tickets, and he can use any number of them when buying an item.
If Y tickets are used when buying an item priced X yen, he can get the item for X 2Y (rounded down to the nearest integer) yen.
What is the minimum amount of money required to buy all the items?
Constraints
All values in input are integers.
1≤N,M≤105
1≤Ai≤109
Input
Input is given from Standard Input in the following format:
N M
A1 A2 … AN
Output
Print the minimum amount of money required to buy all the items.
Sample Input 1
Copy
3 3
2 13 8
Sample Output 1
Copy
9
We can buy all the items for 9 yen, as follows:
Buy the 1-st item for 2 yen without tickets.
Buy the 2-nd item for 3 yen with 2 tickets.
Buy the 3-rd item for 4 yen with 1 ticket.
Sample Input 2
Copy
4 4
1 9 3 5
Sample Output 2
Copy
6
Sample Input 3
Copy
1 100000
1000000000
Sample Output 3
Copy
0
We can buy the item priced 1000000000 yen for 0 yen with 100000 tickets.
Sample Input 4
Copy
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Sample Output 4
Copy
9500000000
题解：
贪心即可，每次都对价格最高的票使用一次优惠券，因此可以把价格放入优先队列，每次取队首进行处理，直到优惠券全部用完为止。
代码：
#include<iostream>
#include<queue>
using namespace std;
long long a[100005];
int main()
{
long long n,m,k,price,sum=0;
priority_queue<long long>que;
scanf("%lld%lld",&n,&m);
for(int i=0;i<n;i++)scanf("%lld",&a[i]);
for(int i=0;i<n;i++)que.push(a[i]);
while(m>0)
{
price=que.top();
price/=2;
m--;
que.pop();
que.push(price);
}
while(!que.empty())
{
sum+=que.top();
que.pop();
}
printf("%lld\n",sum);
}