Codeforces Hello 2018 C. Party Lemonade 贪心优先队列

Posted 2022-12-12 ProLightsfxjh

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 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.

Source

Hello 2018

My Solution

题意：有n种饮料，每种的一份 2^(i-1)升花费ci 卢布，要求总共买L升，花最少的钱，求出最小的花费。

贪心、优先队列、乱搞

首先把饮料的单价(ci / 2^(i-1))和标号(i)丢如小根堆，

然后维护ans = 0 为已购买的饮料的最小花费, resL = L 表示剩余需要购买的饮料, precost = 9e18表示剩下的直接买一份(买的那份可能比resL多)时的最小花费。

每次取出堆顶，此时的饮料是最廉价的，

如果每份的容量大于剩余的，则刷新precost，此时如果 ans被刷新过，则 precost = min（precost， ans + ci）；否则 precost = ci。

如果每份的容量小于剩余的，则直接买 resL / 2^(i-1)份， ans += resL * ci / 2^(i - 1),并且resL %= 2^(i-1)，如果还有剩余，则需要刷新precost， precost = min（precost， ans + ci）。

最后 ans = min（ans，precost）。

按照这个顺序贪心，可以确保每次都采用的最优策略。

时间复杂度 O(nlogn)

空间复杂度 O(n)

#include <iostream>
#include <cstdio>
#include <queue>
using namespace std;
typedef long long LL;
typedef pair<double, LL> di;
const int MAXN = 1e6 + 8;

priority_queue<di, vector<di>, greater<di>> pq;

LL c[36], vo[36];

int main()

    #ifdef LOCAL
    freopen("c.txt", "r", stdin);
    //freopen("c.out", "w", stdout);
    int T = 4;
    while(T--)
    #endif // LOCAL
    ios::sync_with_stdio(false); cin.tie(0);

    LL n, L, i;
    cin >> n >> L;
    vo[0] = 1;
    for(i = 1; i <= n; i++)
        cin >> c[i];
        vo[i] = vo[i-1] * 2;
        pq.push(di(c[i]*1.0 /vo[i-1], i));
    

    LL ans = 0, resL = L, precost = 9e18;
    while(!pq.empty())
        i = pq.top().second;
        //cout << i << " " << vo[i-1] << " " << pq.top().first << endl;
        if(resL >= vo[i-1])
            ans += (LL(resL / vo[i-1])) * c[i];
            resL = resL % vo[i-1];
            precost = min(precost, ans + c[i]);
        
        else if(ans == 0)
            //cout << "?" << endl;
            precost = min(precost, c[i]);
        
        else
            precost = min(precost, ans + c[i]);
        

        if(resL == 0) break;

        pq.pop();
    
    ans = min(precost, ans);
    cout << ans << endl;



    #ifdef LOCAL
    while(!pq.empty()) pq.pop();
    cout << endl;
    
    #endif // LOCAL
    return 0;

  Thank you!

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