Codeforces 913 C. Party Lemonade (thinking) _ Common skills

Description

A New Year's party was not a new year's party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already a become pleasant.

Your Favorite Store sells lemonade in bottles to n different volumes at different costs. A single bottle of Type I has volume 2i−1 2^{i-1} liters and costs CI roubles. The number of bottles of each type in the store can be considered infinite.

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

Input

The 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 for you have to pay in order to buy at least L liters of lemonade.

examples Input

4
20 30 70 90

examples Output

150

the

There are n n kinds of items, their size is 2i−1 2^{i-1}, the cost is CI c_i, the number of items is unlimited, now to make up the size of at least l goods, ask the minimum cost.

train of Thought

Because 2n−1x2=2n 2^{n-1} \times 2 = 2^n, we can think of whether small items made up of large items can bring a smaller cost.

So from small to large sweep to calculate the composition of the current size of 2i 2^i required the minimum cost, recorded as Ai a_i.

And then for size L l, we can convert it to binary, starting at the high level to the low,

Use now to record the cost of the high level of access, if the current bit is 1 1, now+=a[i] now+=a[i], because we can not combine the size of the goods with a larger than l l by this one,

If the current bit is 0 0, record Now+a[i] now+a[i], because at this point we only need to fill the bit to 1 1 to group the goods larger than L L.

Then find the minimum value.

AC Code

 #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; }