Codeforces Hello 2018 C. Party Lemonade（思维）

Description

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

题目大意

现在有m种饮品，第i种饮品的价格为p[i]，容量为$2^{i-1}$，现在输出准备至少n容量的饮品至少需要多少钱。

解题思路

由于每相邻两种饮品的容量为2倍关系，所以一种情况下如果两份 i 饮品的价格小于一份 i+1 饮品的价格，然后更新 i+1 种饮品的价格；另一种情况下如果 i+1 种饮品的价格小于 i 种饮品的价格，那么变更新 i 种饮品的价格。以此遍历两次将每种饮品对应的价格更新至最小值。然后对总容量n的二进制表示下的每一位进行判断，如果当前位为1 ，那么便购买当前位对应容量下的饮料，计算最终总价钱。

代码实现

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define maxn 37
ll p[maxn];
int main()
{
    ios::sync_with_stdio(false);
    int m,n;
    cin>>m>>n;
    for(int i=0;i<m;i++)
        cin>>p[i];
    for(int i=1;i<m;i++)
        p[i]=min(p[i],p[i-1]*2);
    for(int i=m-2;i>=0;i--)
        p[i]=min(p[i],p[i+1]);
    int mark=0;
    for(int i=0;i<31;i++)
        if((1<<i)>n)
        {
            mark=i;
            break;
        }
    for(int i=m;i<=mark;i++)
        p[i]=p[i-1]*2;
    ll ans=0;
    for(int i=0;i<=mark;i++)
    {
        if(ans>p[i]) ans=p[i];
        if(n&(1<<i)) ans+=p[i];
    }
    cout<<ans<<endl;
    return 0;
}