SPOJ.com - Problem COMDIV

COMDIV - Number of common divisors

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You will be given T (T ≤ 10⁶) pairs of numbers. All you have to tell is the number of common divisors between the 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 ≤ 10⁶)

One integer describing number of common divisors between two numbers.

Input:
3
100000 100000
12 24
747794 238336

Output:
36
6
2

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	importme: 2019-09-28 21:42:38 My 50th on SPOJ!!
	sanket17: 2019-07-13 06:42:47 if u r using cin and cout then copy this lines ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	towhid1zaman: 2019-06-19 06:10:26 find gcd, and dont use printf
	frochbg: 2019-05-30 20:07:31 Nice problem :)
	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 happy coding
	Muhammad Annaqeeb: 2019-04-09 17:54:13 Here is a note about comparing the same solution speed across different programming languages: C++ with non-optimized cin and cout ---Time Limit Exceeded C++ with optimized cin and cout -- 0.32 second Java with non-optimized IO -- Time Limit Exceeded Java with optimized IO -- 0.52 second Free Pascal with non-optimized IO -- 0.34 second
	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, O(sqrt(gcd(a,b)))