COMDIV - Number of common divisors
You will be given T (T<=10^6) pair of numbers. All you have to tell is the number of common divisors between two numbers in each pair.
Input
First line of input: T (Number of test cases)
In next T lines, each have one pair A B (0 < A, B <= 10^6)
Output
One integer describing number of common divisors between two numbers.
Example
Input: 3
100000 100000
12 24
747794 238336
Output: 36
6
2
hide comments
|
scolar_fuad:
2019-04-13 15:42:27
Only a single cin and cout will cost you TLE don't need to seive just find gcd and then count divisor
|
|
Muhammad Annaqeeb:
2019-04-09 17:54:13
Here is a note about comparing the same solution speed across different programming languages:
|
|
suraj1611:
2019-03-18 19:48:13
using scanf and printf instead of cin and cout could help!
|
|
sahilsinghss:
2019-03-18 17:47:12
WTH How can using cout instead of printf cost a WA Last edit: 2019-03-18 17:47:35 |
|
midoriya:
2019-03-15 12:01:22
I don't what to do anymore --- should I use cin, cout or scanf ,printf
|
|
harry_shit:
2019-02-12 10:59:06
this was wayy more easy than i thought,
|
|
raghav6:
2019-01-11 13:00:54
Just use scanf and printf for I/O , avoid cin & cout in C++ costed me TLE. |
|
duet_cse16:
2018-10-30 08:55:54
What's the application of gcd in this problem?? |
|
saketag007:
2018-10-14 17:50:59
AC in 3rd go , just use fast input/output , no need to build sieve , just find factors for every test case in O(sqrt(n)) Last edit: 2018-10-14 17:51:15 |
|
jyotiradityafc:
2018-08-28 22:04:28
Spoiler Alert!
|
| Added by: | Mir Wasi Ahmed |
| Date: | 2010-10-31 |
| Time limit: | 0.600s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: ASM64 |
| Resource: | Own problem, used in UODA TST |
RSS