COMDIV - Number of common divisors

no tags 

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.


First line of input: T (Number of test cases)
In next T lines, each have one pair A B (0 < A, B <= 10^6)


One integer describing number of common divisors between two numbers.


100000 100000
12 24
747794 238336

hide comments
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!
Find the gcd of two numbers and then, number of divisors. Optimise in both the cases along with faster i/o.

Last edit: 2018-08-28 22:04:53
puneetgargmnc: 2018-08-17 11:11:34

use Fast I/O using better mathematical logic , no need to use scanf printf..

srjsunny: 2018-02-20 20:36:06

use scanf and printf , no need to use seive

viniet_sw: 2018-01-02 15:14:00

bhosadi waalo ac in one go ka kya show off krte rehte ho gand utha ke har jagah

Added by:Mir Wasi Ahmed
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