10831 - Gerg's Cake

[Eduard]

Can somebody give hint how to solve this poblem?
Thanks

[mf]

Quadratic residues and Legendre's/Jacobi's symbols.

You can assume that integer p is always either a prime (including 2), or unity -- this follows from the first paragraph.

[Larry]

Here are my inputs and outputs, is there anything wrong?

Code: Select all

1 3
1024 17
2 101
0 1
3000000000 5
15 11
721761 7
115 2
6137 2
4 2
0 2
-1 -1

Code: Select all

Yes
Yes
No
Yes
No
Yes
No
Yes
Yes
Yes
Yes

[mf]

Your output is correct except for the case "3000000000 5". 3000000000 is congruent to 0 (mod 5), and zero is a quadratic residue modulo 5, so the answer for that case should be "Yes".

Before evaluating Legendre symbol in this problem, you should first reduce the value of a modulo p.

[Larry]

Thanks, that was what I was looking for.

[Emilio]

Hello!

Only one question?

Could anyone say me what is wrong in my code?

Code: Select all

#include <iostream>
using namespace std;

main ()
{
    long long a, p, x, sq;
    
    while (cin >> a >> p && (a!=-1 || p!=-1))
    {
        x=a%p;
        sq=0;
        while (sq*sq<x) sq++;
        if (sq*sq==x) cout << "Yes\n";
        else cout << "No\n";
    }
}

Any test cases will be appreciated.

Thnaks!

[mf]

Here are some test cases. My program prints "Yes" to all of them.

Code: Select all

2 7
9 7
16 7
3 11
5 11
31337 2147483629
2147483628 2147483629
-1 -1

[Emilio]

Hello and thanks mf!

I have used Legendre's symbol, and get correct output for all the input of this thread.
Maybe I haven't computed right Legendre's symbol.
If someone can explain me a bit how have computed this to solve the problem, I will be happy

By other hand I have here some test cases, if someone can say if the output is correct I will be happy too

input:

Code: Select all

2 7 
9 7 
16 7 
3 11 
5 11 
31337 2147483629 
2147483628 2147483629 
1 3 
1024 17 
2 101 
0 1
300000000 5
15 11 
721761 7 
115 2 
6137 2 
4 2 
0 2 
3 2
5 2
2 3
2 5
2 2
234 23
234 31
485940 101
2384 101
3843 101
3848 41
3848 31
48885557 31
31 41
347583495 7
23477767 23
347823 101
3478234 2147483629 
115 3
115 2
3478293 17
2348884 19
1234123 17
1234123 19
1234123 23
-1 -1

output:

Code: Select all

Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
No
Yes
Yes
Yes
No
Yes
Yes
Yes
Yes
Yes
Yes
No
No
Yes
Yes
No
No
No
Yes
No
Yes
Yes
No
Yes
No
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes

[Martin Macko]

Hello and thanks mf!

I have used Legendre's symbol, and get correct output for all the input of this thread.
Maybe I haven't computed right Legendre's symbol.
If someone can explain me a bit how have computed this to solve the problem, I will be happy

By other hand I have here some test cases, if someone can say if the output is correct I will be happy too

The answer for a=31, b=41 should be Yes. (32nd case)

[Emilio]

Thanks Martin Macko!

I got AC.

I had a small bug at my algorithm. But one question, I have obtained a time of 0:00.535, and most of solvers have obtained a time about 0:00.030 or smaller, I would like know the method used. I used a method that is about Legendre's symbol and law of quadratic reciprocity, in this method I must break down a number in his prime numbers, and make transformations to obtain trivial cases that I can solve directly.

Any other hint or method will be highly-regarded.

[mf]

Legendre symbol can be computed as a^((p-1)/2) (mod p). It's just modular exponentiation.
My 0.002s solution is based on an algorithm from http://www.cacr.math.uwaterloo.ca/hac/about/chap2.pdf page 27, algorithm #2.149.

[Emilio]

Thanks mf another time!

A good link

Now, I have solved it at time of 0:00.004

[Masum]

Posted an ac code. So deleted.

[brianfry713]

That is AC code and input 31 41 should output Yes