贪心-CodeForces 913C

<题目重现>CodeForces - 913C 

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 2^i - 1 liters and costs c_i 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, ..., c_n (1 ≤ c_i ≤ 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.

Example

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.

<解题思路>

<关键操作：输入处理-找单价最优>

对输入的元素进行筛选与覆盖，若后面的单价优与当前的元素就将其覆盖，例如：1体积的物体价值20，输入2时的价值为50，很明显第二个单价为25>20，我们既然要价格最低，2个空间选50的不如选2个1空间20的。

 

<各变量的解释>

f：当前物品所占用的空间；

k：使用当前物体将所剩余的空间填满“可能”需要的空间（可能少1，可能正好）；

t：使用k个当前物体放入背包后背包还剩余的空间；

Ans：当前状态背包填满或移除所放入的物品的总价值（多或正好）；

Sum：当前状态背包中放入的物品的总价值（只少不多）；

 

！有人会说ans和sum万一ans永远多，sum永远少，最后肯定取不到正确值，NO，不要忘记有一个物品的空间占用是1，最终必将其正好填满。

#include
#include
#include

using namespace std;

typedef long long ll;

int n;
ll m;
ll ans=LONG_LONG_MAX;
int a[35];

int main(){
    cin>>n>>m;
    cin>>a[0];
    for(int i=1;i>a[i];
        a[i]=min(2*a[i-1],a[i]);//筛选最优
    }
    ll sum=0;
    ll t=m;
    for(int i=n-1;i>=0;i--){
        ll f=pow(2,i);
        ll k=t/f;
        t-=k*f;
        ans=min(sum+(k+1)*a[i],ans);
        sum+=k*a[i];
    }
    ans=min(ans,sum);
    cout<

<测试状态>（根据题目第一组样例）

<提交结果>