COMDIV - Number of common divisors
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.
Input
First line of input: T (Number of test cases)
In next T lines, each have one pair A B (0 < A, B ≤ 106)
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
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Abdulsalam Abdo Helal:
2015-09-15 14:30:30
scanf and prinf is recommended to pass time
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Abishek:
2015-09-10 10:00:03
fast i/o and basic math ! |
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Medo:
2015-08-20 16:03:11
If you are sure of your complexity. Change to Printf scanf, and make sure all your variables are declared as int not long long. |
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prakash_reddy:
2015-08-18 15:56:38
Nice problem.... :) |
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sy_117:
2015-08-15 16:04:40
AC in first go.....Basic Math nothing else! |
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Indian Cyber Army (IndiShell):
2015-08-08 11:16:29
My code is running perfectly on Dev C++ but here its showing runtime error with floating point exception. I have used long long int. can you please point out the error to me.
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Sue:
2015-06-25 11:55:55
AC in the first go ~
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Varun Gambhir:
2015-06-15 12:58:44
All those getting TLE with sqrt(n) approach and using C/C++, use scanf printf instead of cin and cout. |
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excursionist:
2015-05-15 22:31:40
Loved it !!
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[Lakshman]:
2015-03-10 06:15:02
@Muhammad Annaqeeb: When pyramid cluster was functional the time limit was 6s now the problem has been moved to CUBE cluster, time limit is reduced to 0.6s. As CUBE is 20x times faster than Pyramid. So O(sqrt(gcd(a,b))) can easily pass, it is also recommended to use scanf & printf. |
| 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 |
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