C. Party Lemonade
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.

Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.

You want to buy at least L liters of lemonade. How many roubles do you have to spend?

Input
The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.

The second line contains n integers c1, c2, …, cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.

Output
Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.

Examples
input
4 12
20 30 70 90
output
150
input
4 3
10000 1000 100 10
output
10
input
4 3
10 100 1000 10000
output
30
input
5 787787787
123456789 234567890 345678901 456789012 987654321
output
44981600785557577
Note
In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you’ll get 12 liters of lemonade for just 150 roubles.

In the second example, even though you need only 3 liters, it’s cheaper to buy a single 8-liter bottle for 10 roubles.

In the third example it’s best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.

一開始以為是動態規劃，結果寫到最後發現好像不是dp。。。。Orz。。。。
然後直接貪心去做吧。。
就是最便宜的使勁買，然後買到邊界的時候有兩種情況，第一種是直接再買一瓶，要麼就遞迴下去，看剩餘的在第二便宜的那裡買。
程式碼如下：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
struct node{
    ll v, cost;
    bool operator< (const node& b) const{
        return v * b.cost > b.v * cost;
    }
}arr[31];
ll dfs(ll limit, ll index)
{
    ll temp = limit / arr[index].v;
    ll cost = temp * arr[index].cost;
    if (limit - temp * arr[index].v == 0) return cost;
    else return cost + min(arr[index].cost, dfs(limit - temp * arr[index].v,index+1));
}
int main(void)
{
    arr[1].v = 1;
    int i, j;
    int n, k;
    cin >> n >> k;

    for (i = 1; i <= n; i++)
    {
        cin >> arr[i].cost;
        if (i > 1) arr[i].v = arr[i-1].v * 2;
    }
    sort(arr+1, arr+1+n);
    cout << dfs(k, 1);
}