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PRIME1 - Prime Generator |
Peter wants to generate some prime numbers for his cryptosystem. Help him! Your task is to generate all prime numbers between two given numbers!
Input
The input begins with the number t of test cases in a single line (t<=10). In each of the next t lines there are two numbers m and n (1 <= m <= n <= 1000000000, n-m<=100000) separated by a space.
Output
For every test case print all prime numbers p such that m <= p <= n, one number per line, test cases separated by an empty line.
Example
Input: 2 1 10 3 5 Output: 2 3 5 7 3 5Warning: large Input/Output data, be careful with certain languages (though most should be OK if the algorithm is well designed)
Information
After cluster change, please consider PRINT as a more challenging problem.| Added by: | Adam Dzedzej |
| Date: | 2004-05-01 |
| Time limit: | 6s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All except: NODEJS PERL6 |
hide comments
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2020-05-18 12:54:35
AC in one go! Good question Par sab farzi ka bhaukaal dikhate h "AC in one go" likh kr to maine bhi dikha diya Par maza nhi aaya |
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2020-05-15 16:41:59
should i get all the inputs before processing? |
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2020-05-14 08:39:51
No need of sieve....run from 2 to square root and check if n is 1 |
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2020-05-12 11:22:06
The Osqrt(N) approach took 3.63 sec |
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2020-05-10 21:50:56
Hey!no need for any complex logic. Use simple logic to traverse from m to n and then check from 2 to root(m) with two loops and it will be accepted. runtime o(nroot(n)). |
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2020-05-09 20:20:49
Since the range difference is 10^5,we need not create a historical array to save result from 1 - N (MAX - 10^9) But creating a historical array of size 10^ 5 for every test case can be expensive.Any idea how to go ahead? Last edit: 2020-05-09 20:21:39 |
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2020-05-08 09:18:46
Use segmented Seive guyz ... its meant to be done that way |
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2020-05-05 21:32:42
I got tle error when i use sieveof eratosthenes |
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2020-04-30 09:14:13
you can use segmented sieve to solve this problem complexity of this is around max (root(n) , n-m)) * loglogn which is always less than o(n) for this problem Idea is first generate all prime from 1 to root(n) and store that in a list and then check prime for number between range m to n using same like sieve |
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2020-04-28 04:03:06
Even with somewhat optimized algorithm (restrict scan to 6k+-1 and primes up to sqrt(n)) it's not possible to avoid timeout in python! That's OK for a challenge but not for a introductory problem. |
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