C. Party Lemonade
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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

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.

還是icefox巨佬強啊orz %%我好菜啊 細節太多我還是fail system test了

題意：給定2^0~2^n-1L的汽水 的花費 最後至少L的汽水的最小花費是多少

那麼首先我們把他給的數都讀進來 然後倒著dp一下 設dp[i]表示買2^i的最小花費是多少

然後得到每個正確的花費 然後針對每一位貪心考慮即可 為什麼這題可以貪心考慮 因為這個都是2的次冪 所以我 確信我如果小的次冪多選了 一定不會影響我大的次冪

所以可以這麼來搞 我把L轉換成二進位制 單獨考慮每一個二進位制位上的數怎麼去搞

假設我這個二進位制是1010110001 那麼 首先有1的位置我可以認為一定買2^i最優 比如處理其中一個子問題110001的時候 在處理最高位1的時候 我有可能選兩個這個一位置上的數 有可能就是按部就班的選一個 然後比我最高位小的地方我再遞迴下去去搞 這裡得注意一些細節 就是比如我選了最後一個1010110001的最後一個1 但是我發現 它到上一個1的位置中的某個0的位置我選了反而會更好 那麼我就需要去選一下這個位置的0 然後後面原本算出來的代價直接被替換就好

還有個小細節 我這樣算出來 比如8的費用是1000 但是我發現我買1000的費用只需要10即可 我最後還需要掃一遍 把這種情況做掉即可

#include<cmath>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
int n,L;long long cost[33];
inline ll dfs(int v){
    double len=log2(v);int len1=len;
    if(len==len1) return cost[len1];
    ll left=v-(1<<len1);ll tmp=dfs(left);int len2=log2(left);
    for (int i=len2+1;i<=len1;++i) tmp=min(tmp,cost[i]);
    return cost[len1]*2<cost[len1]+tmp?cost[len1]*2:cost[len1]+tmp;
}
int main(){
    freopen("cf.in","r",stdin);
    scanf("%d%d",&n,&L);
    for (int i=1;i<=32;++i) cost[i]=1LL<<62; 
    for (int i=0;i<n;++i) scanf("%lld",&cost[i]);
    for (int i=1;i<=32;++i) cost[i]=min(cost[i],cost[i-1]*2);
    //for (int i=0;i<30;++i) printf("%lld ",cost[i]);puts("");
    int len=log2(L);ll ans=dfs(L);
    for (int i=len+1;i<=32;++i) ans=min(ans,cost[i]);
    printf("%I64d",ans);
    return 0;
}