Kolya and Movie Theatre
Recently, Kolya found out that a new movie theatre is going to be opened in his city soon, which will show a new movie every day for $$$n$$$ days. So, on the day with the number $$$1 \le i \le n$$$, the movie theatre will show the premiere of the $$$i$$$-th movie. Also, Kolya found out the schedule of the movies and assigned the entertainment value to each movie, denoted by $$$a_i$$$.However, the longer Kolya stays without visiting a movie theatre, the larger the decrease in entertainment value of the next movie. That decrease is equivalent to $$$d \cdot cnt$$$, where $$$d$$$ is a predetermined value and $$$cnt$$$ is the number of days since the last visit to the movie theatre. It is also known that Kolya managed to visit another movie theatre a day before the new one opened — the day with the number $$$0$$$. So if we visit the movie theatre the first time on the day with the number $$$i$$$, then $$$cnt$$$ — the number of days since the last visit to the movie theatre will be equal to $$$i$$$.For example, if $$$d = 2$$$ and $$$a = [3, 2, 5, 4, 6]$$$, then by visiting movies with indices $$$1$$$ and $$$3$$$, $$$cnt$$$ value for the day $$$1$$$ will be equal to $$$1 - 0 = 1$$$ and $$$cnt$$$ value for the day $$$3$$$ will be $$$3 - 1 = 2$$$, so the total entertainment value of the movies will be $$$a_1 - d \cdot 1 + a_3 - d \cdot 2 = 3 - 2 \cdot 1 + 5 - 2 \cdot 2 = 2$$$.Unfortunately, Kolya only has time to visit at most $$$m$$$ movies. Help him create a plan to visit the cinema in such a way that the total entertainment value of all the movies he visits is maximized.
The first test case is explained in the problem statement.In the second test case, it is optimal not to visit any movies.In the third test case, it is optimal to visit movies with numbers $$$2$$$, $$$3$$$, $$$5$$$, $$$6$$$, so the total entertainment value of the visited movies will be $$$45 - 6 \cdot 2 + 1 - 6 \cdot 1 + 39 - 6 \cdot 2 + 11 - 6 \cdot 1 = 60$$$.
Each test consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The description of the test cases follows.The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$d$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le m \le n$$$, $$$1 \le d \le 10^9$$$).The second line of each set of input data contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$-10^9 \le a_i \le 10^9$$$) — the entertainment values of the movies.It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.
For each test case, output a single integer — the maximum total entertainment value that Kolya can get.
from heapq import heappush, heappop, heappushpop

def run():
    (n, m, d) = list(map(int, input().split()))
    a = list(map(int, input().split()))
    rsum = 0
    c = 0
    h = []
    ret = 0
    for (idx, x) in enumerate(a):
        if x < 0:
            continue
        rsum += x
        if c < m:
            c += 1
            heappush(h, x)
        else:
            t = heappushpop(h, x)
            rsum -= t
        tmp = rsum - (idx + 1) * d
        if tmp > ret:
            ret = tmp
    print(ret)
TT = int(input())
for _ in range(TT):
    run()
def read_int():
    return int(input())

def read_list():
    return list(map(int, input().split()))

def main():
    t = int(input())
    while t > 0:
        solve()
        t -= 1
import heapq

def solve():
    (_, m, d) = read_list()
    a = read_list()
    st = [0]
    size = 1
    ans = 0
    current_sum = 0
    for (i, val) in enumerate(a):
        if val < 0:
            continue
        current_sum += val
        heapq.heappush(st, val)
        size += 1
        if size > m:
            current_sum -= heapq.heappop(st)
            size -= 1
        if current_sum - (i + 1) * d > ans:
            ans = current_sum - (i + 1) * d
    print(ans)
if __name__ == '__main__':
    main()
from heapq import heappush, heapreplace

def main():
    t = int(input())
    for _ in range(t):
        (n, m, d) = map(int, input().split())
        a = list(map(int, input().split()))
        h = []
        (ans, s) = (0, 0)
        for (i, ai) in zip(range(1, n + 1), a):
            if ai < 0:
                continue
            if len(h) < m:
                heappush(h, ai)
                s += ai
            elif h[0] < ai:
                s += ai - h[0]
                heapreplace(h, ai)
            ans = max(ans, s - d * i)
        print(ans)
if __name__ == '__main__':
    main()