codeforces Hello 2018 C. Party Lemonade(贪心)

最新推荐文章于 2020-03-05 14:38:09 发布

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最新推荐文章于 2020-03-05 14:38:09 发布

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探讨了在新年派对中如何以最低成本购买所需柠檬水量的问题，通过不同容量及价格的瓶装柠檬水组合，利用算法寻找最优解。

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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.

这个贼有意思。用二进制

#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
using ll = long long;
ll cost[50];
int main() {
   //freopen("in.txt", "r", stdin);
    int n, l;
    cin >> n >> l;
    for (int i = 1; i <= n; ++i) {
        cin >> cost[i];
    }
    
    for (int i = 1; i <= 30; ++i) {
        if (i + 1 > n || cost[i] * 2 < cost[i + 1])
            cost[i + 1] = cost[i] * 2;
    }
    
    for (int i = 30; i >= 1; --i) {
        if (cost[i + 1] < cost[i])
            cost[i] = cost[i + 1];
    }
    ll ans = 0;
    ll res = 1e18;
    for (int i = 30; i >= 0; --i) {
        if (l & (1ll << i))
            ans += cost[i + 1];
        
        res = min(ans + cost[i + 1], res);
    }
    cout << min(res, ans) << endl;
}