Party Lemonade
Party Lemonade
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
假设dp[i] 为选择第i个 的最小花费
容易得到第一个状态转移方程
dp[i] = min(dp[i - 1] * 2, arr[i]);
这样可以得到当前这个位置的最小花费
但是题目要求的是大于等于L,所以吧dp[i]状态变换成大于等于i的最小花费,其实很简单只要从大到小遍历一下,保存最小值即min(dp[i], dp[i + 1])
但是考虑L所对应的二进制去dp
可以得到第二个状态转移方程
设dp2[i]为选择了当前这个位置对应的价值的最小值
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[1] = arr[1];
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] == 1) dp2[1] = dp[1];
else{
dp2[1] = 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[32] << endl;
return 0;
}
原文地址:https://www.cnblogs.com/qq136155330/p/11609963.html
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