Problem:
 Score : 700 points    Problem Statement There are N panels arranged in a row in Takahashi's house, numbered 1 through N . The i -th panel has a number A_i written on it. Takahashi is playing by throwing balls at these panels.  Takahashi threw a ball K times. Let the panel hit by a boll in the i -th throw be panel p_i . He set the score for the i -th throw as i \times A_{p_i} .  He was about to calculate the total score for his throws, when he realized that he forgot the panels hit by balls, p_1,p_2,...,p_K . The only fact he remembers is that for every i  (1 ≦ i ≦ K-1) , 1 ≦ p_{i+1}-p_i ≦ M holds. Based on this fact, find the maximum possible total score for his throws.      Constraints  1 ≦ M ≦ N ≦ 100,000  1 ≦ K ≦ min(300,N)  1 ≦ A_i ≦ 10^{9}       Partial Scores  In the test set worth 100 points, M = N .  In the test set worth another 200 points, N ≦ 300 and K ≦ 30 .  In the test set worth another 300 points, K ≦ 30 .         Input The input is given from Standard Input in the following format:  N  M  K  A_1  A_2 … A_N       Output Print the maximum possible total score for Takahashi's throws.        Sample Input 1 5 2 3
10 2 8 10 2      Sample Output 1 56  The total score is maximized when panels 1,3 and 4 are hit, in this order.       Sample Input 2 5 5 2
5 2 10 5 9      Sample Output 2 28  This case satisfies the additional constraint M = N for a partial score.       Sample Input 3 10 3 5
3 7 2 6 9 4 8 5 1 1000000000      Sample Output 3 5000000078

Solution (Python):
import sys
import numpy as np
import numba
from numba import njit, b1, i4, i8
from numba.types import Omitted

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

@njit((i8[:], i8, i8), cache=True)
def main(A, M, K):
    
    INF = 1 << 60
    dp = A.copy()

    q = np.empty(len(A) + 100, np.int32)
    q[:2] = 2, 2

    def push(q, x):
        q[q[1]] = x
        q[1] += 1

    def is_empty(q):
        return q[0] == q[1]

    for k in range(2, K + 1):
        newdp = np.full_like(A, -INF)
        q[:2] = 2, 2
        for i in range(len(A)):
            if not is_empty(q):
                x = dp[q[q[0]]]
                newdp[i] = x + A[i] * k
            x = dp[i]
            while not is_empty(q) and dp[q[q[1] - 1]] < x:
                q[1] -= 1
            push(q, i)
            while q[q[0]] <= i - M:
                q[0] += 1
        dp = newdp
    return dp.max()

N, M, K = map(int, readline().split())
A = np.array(read().split(), np.int64)

print(main(A, M, K))

Solution (Python):
from collections import deque
N, M, K = map(int, input().split())
A = list(map(int, input().split()))


dp = [0] * (N + 1)
for j in range(K):
    newDP = [0] * (N + 1)
    que = deque()  
    for i in range(j, N - K + j + 1):
        while que and que[-1][0] < dp[i]:
            que.pop()
        que.append((dp[i], i))
        while que and que[0][1] <= i - M:
            que.popleft()
        newDP[i+1] = que[0][0] + (j+1)*A[i]
    dp = newDP

print(max(dp))

Solution (Python):
import sys
def solve():
    readline = sys.stdin.readline

    N, M, K = map(int, readline().split())
    *A, = map(int, readline().split())

    S = A[:]
    T = [0]*N
    Q0 = [0]*N; Q1 = [0]*N
    for k in range(2, K+1):
        s = t = 0

        for i in range(k-1, N):
            a = S[i-1]
            while s < t and Q1[t-1] <= a:
                t -= 1
            Q0[t] = i-1; Q1[t] = a
            t += 1

            T[i] = Q1[s] + A[i]*k

            if s < t and Q0[s] <= i-M:
                s += 1
        S, T = T, S
    print(max(S))
solve()

Solution (Python):
from sys import stderr

def reads():
  return [int(x) for x in input().split()]

N, M, K = reads()
A = reads()

dp = [[0] * N, list(A)]
deq = [0] * N
for j in range(2, K+1):
  l = r = 0
  b = j & 1
  prev = dp[1-b]; curr = dp[b]
  for i in range(N):
    
    while l < r and deq[l] < i-M:
      l += 1
    
    curr[i] = (j * A[i] + prev[deq[l]]) if (i >= j-1 and l < r) else 0
    
    while l < r and prev[deq[r-1]] <= prev[i]:
      r -= 1
    deq[r] = i; r += 1

print(max(dp[K & 1]))