897 - Anagrammatic Primes

[abc]

What changed after rejudge? Is there any anaglemtic primes after 991? What are the traps?

[HidWalker]

I Found some I/O, may be you can try...

Input:
1
2
3
4
5
6
7
8
9

Output:
2
3
5
5
7
7
0
0
0


Hope it helps

[Larry]

I also get the same input, though I also get WA. I find that there is 22 of these primes.. is there more?

[..]

Yes, my accepted program also found 22 anaglemtic prime.

[Larry]

I corrected my error.. I had a typo in one part.. whoops.. =)

[L I M O N]

What changed after rejudge? Is there any anaglemtic primes after 991? What are the traps?

no , there is no ana. prime after 991. There are only 22 ana. primes.
i think there is no trips in this problem. Just do what's the problem say.
how do you consider the case "You must find the first anagrammatic prime which is larger than n and less than the next power of 10 greater than n." ???

L I M O N

[InOutMoTo]

We can know that from n > 10, if n is a ana.prime ,
the digits of n can only contain 1, 3, 7, 9.
(since even number & 5 obviously cant be ana.prime)

But, does anyone know why numbers bigger than 1000
with digits of 1, 3, 7, 9 can't be ana. prime?
(or just by this program result?)

I try to find it , but failed.

[L I M O N]

We can know that from n > 10, if n is a ana.prime ,
the digits of n can only contain 1, 3, 7, 9.
(since even number & 5 obviously cant be ana.prime)
But, does anyone know why numbers bigger than 1000
with digits of 1, 3, 7, 9 can't be ana. prime?
(or just by this program result?)

9 isn't a prime number.
how do you sure that there is no ana. prime after 1000 ???

L I M O N

[Observer]

Well there ARE bigger anagrammatic primes!!!

http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=004023

From this we know that 1111111111111111111 is also an anagrammatic prime @_@

[InOutMoTo]

TO LIMON : I didn't say 9 is ana.prime. I say : "If n is a ana.prime ,
the digits of n can only contain 1, 3, 7, 9. "

TO Observer: Thanks for your data

[Larry]

Observer: I've meant to say, "in the range", though I left it out.. thanks for the interesting data!

[L I M O N]

TO LIMON : I didn't say 9 is ana.prime. I say : "If n is a ana.prime ,
the digits of n can only contain 1, 3, 7, 9. "

Sorry InOutMoTo,
i really misunderstand.
L I M O N

[Kallol]

only 22 anagramatic primes IN THE RANGE ...

they are :
0->10
----------
1. 2
2. 3
3. 5
4. 7

10->100
-----------
5. 11
6. 13
7. 17
8. 31
9. 37
10. 71
11. 73
12. 79
13. 97

100->1000
---------------
14. 113
15. 131
16. 199
17. 311
18. 337
19. 373
20. 733
21. 919
22. 991

if input is greater than 991 , ans would be 0.

precalculation makes it an easy problem .

[clare]

Hello everyone!
I am getting WA with this code Could someone plz give me a hint.
From the previous post I understood that anagrammatic primes are only in the range from 1 to 1000. Is that right?

However, here is my code:

#include<iostream>
#include<string>
#include<algorithm>
#include<cmath>
#include<stdio.h>

using namespace std;

bool is_prime(long long int result)
{
bool prime = true;
for(long long int i = 2;prime and i*i<=result; i++)
{
if(result%i==0)
prime = false;
}
return prime;
}

string to_string(long long int n)
{
string result="";
while(n/10!=0)
{
result+=(n%10) + '0';
n/=10;
}
result+=(n%10)+'0';
return result;
}

string reverse(string num)
{
string result="";
for(int i=num.size()-1; i>=0; i--)
result+=num[i];
return result;
}

long long int to_int(string num)
{
long long int result = 0;
int times = 1;
for(int i = num.size()-1; i>=0; i--)
{
result+=(num[i]-'0')*times;
times*=10;
}
return result;
}

bool check(string num)
{
bool result = true ;
long long int number = to_int(num);
result = is_prime(number);
for(int i = 0; i<num.size(); i++)
{
for(int j=1; result and j<num.size(); j++)
{
swap(num[j-1], num[j]);
number = to_int(num);
result = is_prime(number);
if(result==false)
goto stop;

}
}
stop:
return result;
}

int main()
{
long long int array[50];
array[0]=2; array[1]=3; array[2]=5; array[3]=7;
int a =4;
for(long long int i = 10; i<=1000; i++)
{
string string_num = to_string(i);
bool ok = check(string_num);
if(ok)
{
array[a]=i;
a++;
}
}
long long number;
while(cin>>number and number !=0)
{
if(number>1000)
cout<<0<<endl;
string broj = to_string(number);
double len = broj.size();
long long int power = static_cast<long long int>(pow(10, len));
bool go = true;
for(int g=0; go and g<a; g++)
{
if(array[g]>number and array[g]<power)
{
cout<<array[g]<<endl;
go = false;
}
}
}
}

Thanx!

[smile2ka10]

Don't you think that 1 is a power of 10?! ;)