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.

## 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

dp[i] = min(dp[i - 1] * 2, arr[i]);

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 = arr;
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) dp2 = dp;
else{
dp2 = 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 << endl;
return 0;
}

posted @ 2019-09-29 20:47  moxin0509  阅读(229)  评论(0编辑  收藏  举报