Codeforces Global Round 25
C
Ticket Hoarding
Codeforces
1400
原题
← BD →
翻译中…

As the CEO of a startup company, you want to reward each of your $k$ employees with a ticket to the upcoming concert. The tickets will be on sale for $n$ days, and by some time travelling, you have predicted that the price per ticket at day $i$ will be $a_i$. However, to prevent ticket hoarding, the concert organizers have implemented the following measures:

翻译中…
  • A person may purchase no more than $m$ tickets per day.
  • If a person purchases $x$ tickets on day $i$, all subsequent days (i.e. from day $i+1$ onwards) will have their prices per ticket increased by $x$.
翻译中…

For example, if $a = [1, 3, 8, 4, 5]$ and you purchase $2$ tickets on day $1$, they will cost $2$ in total, and the prices from day $2$ onwards will become $[5, 10, 6, 7]$. If you then purchase $3$ more tickets on day $2$, they will cost in total an additional $15$, and the prices from day $3$ onwards will become $[13, 9, 10]$.

翻译中…

Find the minimum spending to purchase $k$ tickets.

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Input

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Each test contains multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases. The description of the test cases follows.

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The first line of each test case contains three integers $n$, $m$, and $k$ ($1 \le n \le 3 \cdot 10^5, 1 \le m \le 10^9, 1 \le k \le \min(nm, 10^9)$) — the number of sale days, the maximum amount of ticket purchasable each day, and the number of tickets to be bought at the end.

翻译中…

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$) — the price per ticket for each of the upcoming $n$ days.

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It is guaranteed that the sum of $n$ over all test cases does not exceed $3 \cdot 10^5$.

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Output

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For each test case, print one integer: the minimum amount of money needed to purchase exactly $k$ tickets.

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Note

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In the first test case, one optimal way to buy $3$ tickets is as follows:

翻译中…
  • Buy $0$ tickets on the first day. The prices per ticket for the remaining days are $[6, 4, 2]$.
  • Buy $0$ tickets on the second day. The prices per ticket for the remaining days are $[4, 2]$.
  • Buy $1$ ticket on the third day with cost $4$. The price per ticket for the remaining day is $[3]$.
  • Buy $2$ tickets on the fourth day with cost $6$.
翻译中…

In the second test case, there is only one way to buy $8$ tickets:

翻译中…
  • Buy $2$ tickets on the first day with cost $16$. The prices per ticket for the remaining days are $[8, 6, 4]$.
  • Buy $2$ tickets on the second day with cost $16$. The prices per ticket for the remaining days are $[8, 6]$.
  • Buy $2$ tickets on the third day with cost $16$. The price per ticket for the remaining day is $[8]$.
  • Buy $2$ tickets on the fourth day with cost $16$.
翻译中…

样例

Input #1
4
4 2 3
8 6 4 2
4 2 8
8 6 4 2
5 100 1
10000 1 100 10 1000
6 3 9
5 5 5 5 5 5
Output #1
10
64
1
72
InputCopy OutputCopy
题目信息
题号
1951C
时间限制
2 s
内存限制
256 MB
标签
贪心
数学
排序

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。