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 - 1liters 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?
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 a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.
4 12 20 30 70 90
150
4 3 10000 1000 100 10
10
4 3 10 100 1000 10000
30
5 787787787 123456789 234567890 345678901 456789012 987654321
44981600785557577
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.
n 瓶柠檬水,第 i 瓶的体积为 2^(i - 1),价格为 v_i
你需要买 m 体积的柠檬水
求最小花费
很显然的一个贪心是算出每份柠檬水单位体积的价格然后排序
但是因为有除法 会有精度问题
所以注意到每瓶柠檬水的体积都是2的倍数
所以我们统一到体积最高进行排序
除就变成乘了(齐齐真厉害)
再根据排序后的结果从小到大贪心
因为允许你所买的柠檬水的体积超过原本所需的柠檬水的体积
通过递归求解
对于当前柠檬水
比较一下 剩下的直接买光 与 买下一份 的代价
就好了
注意一下细节 爆 int
#include <bits/stdc++.h> #define LL long long using namespace std; const int N = 110; struct Node { int x; ///体积 int y; ///cost LL z; ///统一体积后的花费 }; bool cmp(Node a, Node b) { return a.z < b.z; } Node node[N]; int tem[N]; int n, m; LL solve(int x, LL ans) { LL q; LL t; if(m <= 0){ return ans; } q = m / node[x].x; m -= q * node[x].x; ans += q * node[x].y; t = solve(x + 1, ans); if(ans + node[x].y < t){ return ans + node[x].y; } else{ return t; } } int main(int argc, char const *argv[]) { tem[0] = 1; for(int i = 1; i <= 30; i ++){ tem[i] = tem[i - 1] * 2; } while(scanf("%d%d", &n, &m) == 2){ for(int i = 0; i < n; i ++){ node[i].x = tem[i]; scanf("%d", &node[i].y); node[i].z = 1ll * node[i].y * tem[n - i - 1]; /// } sort(node, node + n, cmp); printf("%lld\n", solve(0, 0)); } return 0; }