【CodeForces】---Party Lemonade（贪心&&思维&&递归）

题目链接:Click Here

                C. Party Lemonade
                time limit per test1 second
                memory limit per test256 megabytes
                inputstandard input
                outputstandard 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.

题意：n种杯子，每一种无穷多个，每一种容积不同，呈2的指数次幂上升，且价格不同。问买体积为L的柠檬水，需要最少多少钱。
分析：贪心 ，递归
第i种的容积是第i-1种的2倍，所以可根据此更新价格。若买两瓶第i-1种比买一瓶第i种便宜，那就用花费少的那个。价格数组c[i],c[i]=min(c[i],2*c[i-1]),从n到30的c[i]=c[i-1]*2.
柠檬水的容积都是2的指数次幂。c[i]每一步都是更优，容积指数越大越好，所以先用循环找到容积指数尽可能大的情况，即小于L的最大2^i，确定一个i，再依次递归往下找指数小的。
取完一个i的剩余的容积x,比x大的最小的容积2^j,在j到i之间，选出最小的一个就好了。
最后注意，可能一开始就取容量大于L的是更优的情况，所以循环从flag到30或许可以找出最小的结果哦，min更新cnt。

这里写代码片
#include<iostream>
#include<cstring>
using namespace std;
typedef long long LL;
LL c[35];
LL dfs(LL x)        //递归 
{
    int pos=31;
    for(int i=0;i<=30;i++)
    {
        if(x<(1<<i))
        {
            pos=i-1;//pos记录比L小的最大的容量指数 
            break;
        }
    }
    if(x==(1<<pos))         //当前所需容量等于pos所指容量，返回其价格
        return c[pos];
    //不相等时，说明还有剩余哦，
    LL res=x-(1<<pos);      //所需容量比pos所指容量多的部分
    LL sum=dfs(res);        //对这一部分进行递归，
    int pos1=31;
    for(int i=0;i<=30;i++)
    {
        if(res<(1<<i))
        {
            pos1=i-1;//pos1记录比res小的最大容量指数
            break;
        }
    }
    for(int i=pos1+1;i<=pos;i++)        //在pos1+1~pos之间c[i]和sum，找较小者，更新c[i] 
        sum=min(sum,c[i]);
    return c[pos]+sum;
}                   //递归最后返回容量x在0~k中所需的最小的价格 
int main()
{
    int n;
    LL L;
    while(cin>>n>>L)
    {
        for(int i=0;i<n;i++)
            cin>>c[i];
        for(int i=1;i<n;i++)
            c[i]=min(c[i],c[i-1]*2);        //更新为更优 
        for(int i=n;i<=30;i++)
            c[i]=c[i-1]*2;//更新，体积是2倍，价格也是呢 
        int flag=31;
        for(int i=0;i<=30;i++)
        {
            if(L<(1<<i))
            {
                flag=i;         //循环找到第一个大于L的容积对应的二次指数，用flag记录
                break;
            }
        }
        LL cnt=dfs(L);              //递归找到0~flag最小的花费 
        for(int i=flag;i<=30;i++)       //杯子容量比L大时，是否存在更小的花费呢 
            cnt=min(cnt,c[i]);
        cout<<cnt<<endl;
    }   
    return 0;
}