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 $2^{i - 1}$ 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 $2^{i-1}$ ，花费分别为 $c_i$ ，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 $2^{n-1} \times 2 = 2^n$ ，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 $2^i$ 所需要的最小花费，记为 $a_i$ 。
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;
}
