Caso # | Resultado | Tiempo | Memoria |
---|---|---|---|
#1 |
Correcto
|
0.027 s | 3 KBi |
#2 |
Correcto
|
0.041 s | 5 KBi |
#3 |
Correcto
|
0.031 s | 3 KBi |
#4 |
Correcto
|
0.029 s | 3 KBi |
#5 |
Correcto
|
0.028 s | 3 KBi |
#6 |
Correcto
|
0.025 s | 3 KBi |
#7 |
Correcto
|
0.03 s | 3 KBi |
#8 |
Correcto
|
0.029 s | 7 KBi |
#9 |
Correcto
|
0.03 s | 3 KBi |
#10 |
Correcto
|
0.027 s | 3 KBi |
#11 |
Correcto
|
0.029 s | 3 KBi |
#12 |
Correcto
|
0.026 s | 3 KBi |
#13 |
Correcto
|
0.03 s | 3 KBi |
#14 |
Correcto
|
0.024 s | 3 KBi |
#15 |
Correcto
|
0.026 s | 3 KBi |
#16 |
Correcto
|
0.039 s | 4 KBi |
#17 |
Correcto
|
0.159 s | 11 KBi |
#18 |
Correcto
|
0.131 s | 11 KBi |
#19 |
Correcto
|
0.142 s | 12 KBi |
#20 |
Correcto
|
0.119 s | 12 KBi |
#21 |
Correcto
|
0.174 s | 12 KBi |
#22 |
Correcto
|
0.126 s | 11 KBi |
#23 |
Correcto
|
0.15 s | 11 KBi |
#24 |
Correcto
|
0.117 s | 11 KBi |
"""Using bottom up dynamic programming""" def find(a): n = len(a) dp = [0] * n dp[0] = a[0] best = dp[0] for j in range(1, n): dp[j] = max(a[j], dp[j - 1] + a[j]) best = max(best, dp[j]) return best n = int(input()) a = [int(x) for x in input().split()] print(find(a))