Codeforces 913C：Party Lemonade（贪心）

C. Party Lemonade

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 ≤ 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.

题意

有n种不同规格的饮料，容量分别为2^i-1L，并给出每种饮料的价格。

现在题目要求至少购买 l升的饮料，最少需要多少钱。

思路

仔细一看，以为是背包，又读了一遍题，貌似是个sb贪心题

但是这个贪心又有点dp的感觉，因为不能只贪最便宜的，每次贪心要对两种状态进行比较，不断地在两种状态间转换

首先按照每个饮料的单价（每升多少钱）升序排序，然后会有两种状态：

1. 全部买最便宜的那种饮料，直到买的数量大于等于L
2. 最便宜的饮料买的尽可能的多，但是不要超过L，剩下的部分作为L返回第一步

代码

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
struct wzy
{
    ll bigness;
    ll money;
    double price;
}p[maxn];
bool cmp(wzy u,wzy v)
{
    return u.price<v.price;
}
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("/home/wzy/in.txt", "r", stdin);
        freopen("/home/wzy/out.txt", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    ll l;
    cin>>n>>l;
    for(int i=1;i<=n;i++)
    {
        cin>>p[i].money;
        p[i].bigness=(1LL<<(i-1));
        p[i].price=1.0*p[i].money/p[i].bigness;
    }
    sort(p+1,p+1+n,cmp);
    ll L=l;
    ll ans=INF;
    ll res1=0,res2=0;
    while(L>0)
    {
        for(int i=1;i<=n;i++)
        {
            // 全买这个饮料
            res1=res2+ceil(1.0*L/p[i].bigness)*p[i].money;
            ans=min(ans,res1);
            // 多余的部分买别的
            res2+=L/p[i].bigness*p[i].money;
            L-=(p[i].bigness*(L/p[i].bigness));
        }
    }
    ans=min(ans,res2);
    cout<<ans<<endl;
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}