codeforces Hello 2018 C. Party Lemonade(贪心)

C. Party Lemonade

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 r
e4c2
equired 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;
}