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 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 ≤ 1e9) — 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 ≤ 1e9) — 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.

SampleInput 1
4 12
20 30 70 90
SampleOutput 1
150
SampleInput 2
4 3
10000 1000 100 10
SampleOutput 2
10
SampleInput 3
4 3
10 100 1000 10000
SampleOutput 3
30
SampleInput 4
5 787787787
123456789 234567890 345678901 456789012 987654321
SampleOutput 4
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.

题目链接

思路

先观察发现仿佛跟二进制有关系，可以先得到第一个状态转移方程，计算出选择当前这个位置的这个值的最小花费是多少
假设dp[i] 为选择第i个 的最小花费
容易得到第一个状态转移方程
dp[i] = min(dp[i - 1] * 2, arr[i]);
这样可以得到当前这个位置的最小花费
但是题目要求的是大于等于L，所以吧dp[i]状态变换成大于等于i的最小花费，其实很简单只要从大到小遍历一下，保存最小值即min(dp[i], dp[i + 1])
但是考虑L所对应的二进制去dp
可以得到第二个状态转移方程
设dp2[i]为选择了当前这个位置对应的价值的最小值
vids[i]代表i这个位置是否是L对应有值的二进制

if(vids[i] == 1){
        dp2[i] = min(dp2[i - 1] + dp[i - 1] * 2, dp2[i - 1] + dp[i]);
}
else{
        dp2[i] = min(dp2[i - 1], dp[i]);
}

附上代码：

    #include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 35;
long long dp[MAXN], arr[MAXN], dp2[MAXN];
int vids[MAXN];
int main(){
    ios::sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    for(int i = 1; i <= 32; i ++){
        arr[i] = __LONG_LONG_MAX__;
    }
    for(int i = 1; i <= n; i ++){
        cin >> arr[i];
    }
    dp[1] = arr[1];
    for(int i = 2; i <= 32; i ++){
        dp[i] = min(arr[i], dp[i - 1] * 2);
    }
    for(int i = 32 - 1; i >= 1; i --){
        dp[i] = min(dp[i], dp[i + 1]);
    }
    int num = 1;
    while(k){
        if(k & 1) vids[num] = 1;
        num ++;
        k >>= 1;
    }
    if(vids[1] == 1) dp2[1] = dp[1];
    else{
        dp2[1] = 0;
    }
    for(int i = 2; i <= 32; i ++){
        if(vids[i] == 1){
            dp2[i] = min(dp2[i - 1] + dp[i - 1] * 2, dp2[i - 1] + dp[i]);
        }
        else{
            dp2[i] = min(dp2[i - 1], dp[i]);
        }
    }
    cout << dp2[32] << endl;
    return 0;
}