Hello CodeForces How to find minimal prime divisor for n <= 10^12 if there are number of queries <= 10^5
# | User | Rating |
---|---|---|
1 | tourist | 3993 |
2 | jiangly | 3743 |
3 | orzdevinwang | 3707 |
4 | Radewoosh | 3627 |
5 | jqdai0815 | 3620 |
6 | Benq | 3564 |
7 | Kevin114514 | 3443 |
8 | ksun48 | 3434 |
9 | Rewinding | 3397 |
10 | Um_nik | 3396 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 155 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
10 | djm03178 | 152 |
Hello CodeForces How to find minimal prime divisor for n <= 10^12 if there are number of queries <= 10^5
Hi CodeForces, help me please with this -> problem
I tryed to use BFS, but i got TLE
Hi CodeForces.
I won't give link of problen and explain statement of problem, because this problem has more easier solution. But i want implement my solution idea, but i have problems with realization, and i need your help.
In problem i must calculate dp
for (int i = 1; i <= n; i++){
int mn = inf, mx = -inf;
for (int j = i; j >= 1; j--){
mn = min(mn, a[j]), mx = max(mx, a[j]);
dp[i] = max(dp[i], dp[j - 1] + (mx - mn));
}
}
Problem can not be solved in O(N^2), because N <= 10^6
I think, exist way to calculate dp mush faster -> dp[i] = max(dp[j - 1] + (mx[j] - mn[j])) 1 <= j < i
mx[j] -> maximum from j to i
mn[j] -> minimum from j to i
Thanks in advance!
Where is editorial from author???
Can you explain soluton idea with treap?
Hello CodeForces. Offensively that round #383 was rescheduled.
Please help me with this problem.
Ladies and Gentlemen, please help me -> problem.
Name |
---|