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 2i-1- liters and costs c__i 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?

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 c_1, _c_2, …, _c__n (1 ≤ c__i ≤ 109) — the costs of bottles of different types.

Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.

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

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2i-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

1 #include
2 #define ll long long
3 #define ull unsigned long long
4 #define ms(a,b) memset(a,b,sizeof(a))
5 const int inf=0x3f3f3f3f;
6 const ll INF=0x3f3f3f3f3f3f3f3f;
7 const int maxn=1e6+10;
8 const int mod=1e9+7;
9 const int maxm=1e3+10;
10 using namespace std;
11 struct wzy
12 {
13 ll bigness;
14 ll money;
15 double price;
16 }p[maxn];
17 bool cmp(wzy u,wzy v)
18 {
19 return u.price>n>>l;
33 for(int i=1;i<=n;i++) 34 { 35 cin>>p[i].money;
36 p[i].bigness=(1LL<<(i-1)); 37 p[i].price=1.0*p[i].money/p[i].bigness; 38 } 39 sort(p+1,p+1+n,cmp); 40 ll L=l; 41 ll ans=INF; 42 ll res1=0,res2=0; 43 while(L>0)
44 {
45 for(int i=1;i<=n;i++)
46 {
47 // 全买这个饮料
48 res1=res2+ceil(1.0*L/p[i].bigness)*p[i].money;
49 ans=min(ans,res1);
50 // 多余的部分买别的
51 res2+=L/p[i].bigness*p[i].money;
52 L-=(p[i].bigness*(L/p[i].bigness));
53 }
54 }
55 ans=min(ans,res2);
56 cout<<ans<<endl;
57 #ifndef ONLINE_JUDGE
58 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
59 #endif
60 return 0;
61 }