QOJ.ac
QOJ
| ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
|---|---|---|---|---|---|---|---|---|---|
| #785801 | #9568. Left Shifting 3 | renew | TL | 7ms | 10604kb | Python3 | 461b | 2024-11-26 19:15:29 | 2024-11-26 19:15:47 |
Judging History
answer
def count_nanjing_substrings(s, n, k):
max_count = 0
for d in range(min(k + 1, n)):
shifted_s = s[d:] + s[:d]
count = 0
for i in range(n - 6):
if shifted_s[i:i + 7] == "nanjing":
count += 1
max_count = max(max_count, count)
return max_count
for _ in range(int(input())):
n, k = map(int, input().split())
s = input()
print(count_nanjing_substrings(s, n, k))
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 7ms
memory: 10604kb
input:
4 21 10 jingicpcnanjingsuanan 21 0 jingicpcnanjingsuanan 21 3 nanjingnanjingnanjing 4 100 icpc
output:
2 1 3 0
result:
ok 4 number(s): "2 1 3 0"
Test #2:
score: -100
Time Limit Exceeded
input:
2130 39 7 nnananjingannanjingngnanjinganjinggjina 1 479084228 g 33 2 gqnanjinggrjdtktnanjingcvsenanjin 24 196055605 ginganjingnanjingnanjing 23 3 ngnanjinganjingjinnanji 40 3 njingaaznannanjingnananjingyonwpnanjinga 40 207842908 nanjinggphconanjingkonanjinannanjinglxna 46 3 ingjingnnanjingnanjinging...
output:
3 0 3 2 2 3 3 4 3 4 0 2 4 3 2 1 1 1 4 2 0 3 3 0 0 1 0 0 0 5 4 0 1 2 1 2 2 1 1 1 3 3 1 3 2 0 1 2 4 1 2 1 2 1 2 3 0 1 0 0 1 1 3 2 2 1 0 3 1 2 1 1 4 4 1 1 1 1 0 1 1 1 1 2 0 4 4 3 1 1 2 1 1 1 1 5 1 4 0 1 2 1 3 4 3 3 3 3 1 3 2 1 3 1 2 0 0 1 0 5 0 2 0 3 1 0 2 2 3 2 1 2 0 1 1 1 2 4 1 3 2 0 1 1 2 2 2 1 0 3 ...