Codeforces 913 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 ≤ 10^9) — 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 ≤ 10^9) — 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

Examples output

150

题意

有 n 种物品，其大小分别为 2i−1 ，花费分别为 ci ，物品的个数无限，现要组成大小至少为 L 的货物，问最小的花费。

思路

因为 2n−1×2=2n ，所以我们可以想到小物品组成大物品是否可以带来更小的花费。

于是从小到大扫一遍计算出组成当前大小为 2i 所需要的最小花费，记为 ai 。

然后针对大小 L ，我们可以将其转换为二进制，从高位往低位开始枚举，

用 now 记录已访问的高位中所需要的花费，若当前位为 1 ， now+=a[i] ，因为我们不能通过这一位组合出大小大于 L 的货物，

若当前位为 0 ，记录 now+a[i] ，因为此时我们只需要将该位填充为 1 即可组出大于 L 的货物。

然后找最小值即可。

AC 代码

#include <bits/stdc++.h>
#define IO ios::sync_with_stdio(false);\
cin.tie(0);\
cout.tie(0);
using namespace std;
const int maxn = 1e5+10;
const int mod = 1e9+7;
typedef long long LL;

LL n,l,a[maxn];
bitset<32> sk;
set<LL> ans;
int main()
{
IO;
cin>>n>>l;
for(int i=0; i<n; i++)
cin>>a[i];
LL now = a[0];
for(int i=1; i<32; i++)
{
now <<= 1;
a[i] = i<n?min(a[i],now):now;
now = a[i];
}
sk = l;
now = 0;
for(int i=31; i>=0; i--)
if(sk[i])
now +=a[i];
else
ans.insert(now+a[i]);
ans.insert(now);
cout<<*ans.begin()<<endl;
return 0;
}