COMDIV - Number of common divisors

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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

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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

sanyam19: 2018-01-02 14:48:23

AC in 1 go :)

karthik1997: 2017-12-27 16:39:58

Man , Never thought i would end up with fastest submission till date ( 0.10s )
Fast I/O + sieve_modified + binary_gcd + O( log(N) ) +factorization method for number of divisors got me this :)

Last edit: 2017-12-27 20:04:49
chetan4060: 2017-12-19 18:25:20

use sieve first then gcd
AC in one go:-)

sirjan13: 2017-11-02 21:22:19

50th =)
Be Careful with I/O.Cheers!

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