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### 10392 - Factoring Large Numbers

Posted: Wed Mar 31, 2004 10:35 pm
Can anybody give me a hint on how to solve this problem?

The worst case is 32bitprime * 32bitprime, which means I have to know all primes up to 32bit long...

How can I compute them (fast)????

It I had 5Gig of memory, I would be able to do it in under 5min..  Any suggestions?

Posted: Thu Apr 01, 2004 2:49 am
Try naive.. it's fast enough for this problem..

Posted: Thu Apr 01, 2004 4:25 am
Larry wrote:Try naive.. it's fast enough for this problem..
You mean divide the number by two and then all odds?

Posted: Thu Apr 01, 2004 8:42 am
You don't have to know all primes in 32bit range. It's enough to know primes in 16bit range In this range it's very easy to compute all primes ...
Thik about it Best regards
DM

Posted: Thu Apr 01, 2004 5:05 pm
Dominik Michniewski wrote:You don't have to know all primes in 32bit range. It's enough to know primes in 16bit range In this range it's very easy to compute all primes ...
Thik about it Best regards
DM
As far as I understand, if input is a 64bit prime, I have to try to divide it by all primes up to 32bit before I can be sure I was given a prime.(unless I use probabalistic prime testing)
Am I wrong?

Posted: Thu Apr 01, 2004 7:32 pm
Use the method of sqrt to factorize the given number(The bangla method ).It's really easy.Try to make this code efficient by some small changes.It's really easy and simple.

Posted: Thu Apr 01, 2004 8:24 pm
You have right minskcity, I misread problem description ...
But as I see in my code, it's really not necessary to compute all primes in 32bit range ... Naive algorithm is still nice for this problem I observe thatyou can use iterative algorithm : when number is prime, you should check only numbers in range 2..sqrt(number). and number is shrinking after each step, so sqrt(number) go down Maybe this help you ...

Best regards
DM

Posted: Fri Apr 02, 2004 12:49 am
alu_mathics wrote:Use the method of sqrt to factorize the given number(The bangla method ).It's really easy.Try to make this code efficient by some small changes.It's really easy and simple.
Factorizing an arbitrary 64bit number reasonably fast is not "easy and simple": my programs was accepted by judge in 0.3sec, but if I give it 1000 big primes as input it will run for 10 hours or so...
This problem reminded me another "easy" problem that asked to find a square root of 100-digits number, but the actual input did not contain numbers over 35 digits long.

### Re: 10392 - Factoring Large Numbers

Posted: Mon Aug 11, 2008 10:59 pm
Hello there.

I'm getting W.A in this problem, can someone give me some test cases?

below is my code.

Thanks for any help.

Code: Select all

``````#include <iostream>
#include <math.h>

using namespace std;

int main(void) {

long long int n;
long long int sq;
long long int i=0, c=0;

while (cin >> n && n >= 0) {
if (i) cout << endl;

c = n;
while((c % 2) == 0) {
cout << "    2" << endl;
c = c / 2;
}

i = 3;
while (i <= (sqrt(c) + 1)) {
if ((c % i) == 0) {
cout << "    " << i << endl;
c = c / i;
}
else {
i = i + 2;
}
}

if (c > 1) cout << "    " << c << endl;
}

return 0;
}
``````

### Re: 10392 - Factoring Large Numbers

Posted: Wed Jan 07, 2009 8:25 am
Instead of outputting a blank line between test cases, you should output a blank line after each test case (including the last one). On the other hand, thanks for your program. I was implementing the same "naive" idea but got TLE, which I thought was reasonable at first. Then I saw your program got WA instead of TLE and realized checking "i<=sqrt(c)+1" was actually faster than checking say "i*i<=c".

### Re: 10392 - Factoring Large Numbers

Posted: Thu Nov 12, 2009 4:33 pm
I am having a formatting output issue and I can't seem to find where it is.

Anyone know?

Code: Select all

``````import java.io.*;

public class Main {
public static void main(String[] args) throws IOException {

System.in));

while (true) {

if (s.equals("-1")) break;

long m = Long.parseLong(s);
long n = 0;
long i = 0;

n = m;

while(n%2 == 0)
{
System.out.println("    2");
n = n/2;
}
i = 3;
while (i <= (Math.sqrt(n))+1)
{
if ((n % i) == 0) {
System.out.println("    " +i);
n = n / i;
}
else {
i = i + 2;
}
}
if (n > 1)
{
System.out.println("    "+n);

}
}

}

}
``````

### Re: 10392 - Factoring Large Numbers

Posted: Sun May 08, 2011 8:25 pm
I have used ordinary prime factoring algo,i generated primes upto 10^7 (though it wasnt necessary)
Try this:

Code: Select all

``````9999999999999999
3242345235676578
5345234788786864
4325243658766786
-1

``````
out:

Code: Select all

``````    3
3
11
17
73
101
137
5882353

2
3
3
3
60043430290307

2
2
2
2
17
19
31
33364343783

2
7
308945975626199

``````

### Re: 10392 - Factoring Large Numbers

Posted: Thu Sep 08, 2011 4:52 pm
Hi, I'm getting WA. I tried previous cases mentioned in this forum and produced same result, but I still got WA.
My algorithm is to calculate all primes less than 10^6 and try dividing the input.
Anyone can help me? Probably with some I/O cases
Thanks

EDIT : got AC already.. I missed printing "\n" after the last testcase. It was very strange, should be PE instead of WA.

### Re: 10392 - Factoring Large Numbers

Posted: Thu Nov 10, 2011 12:50 pm
Each positive number from the input must be factored and all factors (other than 1) printed out. The factors must be printed in ascending order with 4 leading spaces preceding a left justified number, and followed by a single blank line.

### Re: 10392 - Factoring Large Numbers

Posted: Sun Apr 07, 2013 10:58 pm
Sample Input- Output:

Input:

Code: Select all

``````1000007
10000009
2387468238648623864
761253475173152355522
77777777777812222222222
1000000000000223657
237424528
23482397493264962346826
3245986234596236523695665
2734823482584528458248
235429469249269462946926496
3593057032750375730570
``````
Output:

Code: Select all

``````    29
34483

23
434783

2
2
2
641
55331
8414359573

4936968151060739266

6304763052752609166

31
32258064516136247

2
2
2
2
11
1349003

``````