There is n integer elements. Choose k elements then calculate their GCD, call result X.
What is maximal value of X?
k<=n<=3000, k<=100, elements in range 1..10^10.
Example:
3 2
120
36
100
Sample output
20
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3382 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 155 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 145 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 134 |
| 9 | maroonrk | 132 |
| 10 | cry | 130 |
There is n integer elements. Choose k elements then calculate their GCD, call result X.
What is maximal value of X?
k<=n<=3000, k<=100, elements in range 1..10^10.
Example:
3 2
120
36
100
Sample output
20
| Name |
|---|



These constraints are in a very gray area, it really depends on the time limit. So depending on the limit you may or may not be able to:
Factor each number: Sieve of Eratosthenes + 3000 * (primes below 10 ^ 5) = 3000 * 10000; Generate divisors of each number in O(divisor_count). Divisor_count is at most 2300. Keep a frequency table for the divisors and just select the biggest one that's over k.
Just a first thought. It's a fragile solution, but it may very well work. Again, the constraints are too fuzzy. Will look for something cleaner and smarter though :).
If I correctly understood your solution it wouldn't work because the gcd can be a composite number instead of a prime.
I take in account all divisors, composite or not. The factoring into primes part is only so you can generate the divisors of a number quickly, instead of trying each number up to 10 ^ 5.
Thank you, I understood.
ez
gg easy FTFY