OJ Board

Alright:

output for 1:

1 is not prime.

output for 2:

2 is prime.

output for 13:

13 is emirp.

thanks a lot for your help

Same as mine except the one for input 1 (mine is prime, but it doesn't matter, as it's not an input, which is stated 1 < N < 1000000).

If you don't mind, could you send your codes through Private Message so that I can have a look at it? I am now very curious and it's interesting to me~

I have received your codes~

I have not taken a in-depth look at it, but I just have done an interesting thing: I use a program to generate a file consisting of 2,3,4,5,...,999999, then use this file as the input file of my program and your program. Finally, I diff the output file generated by the two program and found that..... NO DIFFERENCE (at least found by the diff program in UNIX)

I can be sure that your program is of no problem~
Or something not releted to the problem introduces the WA~ e.g. submitted with wrong problem number? Haha.....

I just checked and I had submitted with the right problem number .....
just resubmitted it and it works now
Thanks for your help

I'm a bit confused about emirp ,can anyone define emirp more
clearly???

why 2 is not emirp ?????

thx in advance.

I quote from the problem statement:

An Emirp (Prime spelt backwards) is a Prime that gives you a different Prime when its digits are reversed

2 is not different than 2 when reversed...

thx kmhasan, that clears up the whole thing now

if anyone can tell me why
this program is wrong
.... I would be very grateful.

#include<stdio.h>
#include<math.h>
int chkprime(unsigned long);

int main()
{
int k=0,cnt=0,j,in;
unsigned long n,rf=0,xf,fnum[50],num;
while(scanf("%lu",&n)!=EOF)
{
num=n;
if(!(n%2) || !(n%3)) {k=0;if(n==2 || n==3) k=1;}
else k=chkprime(n);
if(k==0) {printf("%lu is not prime.\n",num);continue;}
while(n!=0)
{ xf=n%10;
fnum[cnt]=xf;
n=n/10;
cnt++;
}
for(j=cnt-1,in=0;j>=0;j--,in++)
rf=rf+fnum[j]*pow(10,in);
if(!(rf%2) || !(rf%3)) {k=0;if(rf==2 || rf==3) k=1;}
else k=chkprime(rf);
if(k==0 ||(num==rf)) printf("%lu is prime.\n",num);
else printf("%lu is emirp.\n",num);
cnt=0;
k=0;
rf=0;
}
return 0;
}
int chkprime(unsigned long n)
{
unsigned long d;

unsigned long i,in,p=n-1;
d=1;
if(p==1) i=1;
for(in=4;in<=2147483648;in=in*2)
if(p<in) {i=in/2;break;}
for(;i>=1;i=i/2)

{
d=((d*d)%n);
if(p&i)
d=((d*2)%n);
}
if(d!=1) return 0;
if(n==3) return 1;
else if(n==5) return 1;
for(p=3;p<=sqrt(n);p+=2)
if(!(n%p)) return 0;
return 1;

}

Try This Test Case :
Note : that 2 is not emirp, but prime!

Sample input:
2
1
10
11
71
9001
10009901
1321
1099933

Sample Output
2 is prime.
1 is prime.
10 is not prime.
11 is prime.
71 is emirp.
9001 is emirp.
10009901 is not prime.
1321 is emirp
1099933 is emirp.

Hope this Helps.
GOOD LUCK
RED SCORPION

An Emirp (Prime spelt backwards) is a Prime that gives you a different Prime when its digits are reversed.

You should check if the mirror is different before the second prime test.

logical error i guess. modify the following code:
code to modify:
[cpp]if(isPrime(b))
if(a!=b)
printf("%lld is emirp.\n",a);
else
printf("%lld is prime.\n",a);
[/cpp]
code to replace with:
[cpp]if(isPrime(b)) {
if(a!=b)
printf("%lld is emirp.\n",a);
else
printf("%lld is prime.\n",a);
}
else
printf("%lld is prime.\n",a);
[/cpp]
and i think you should use long instead of long long.

good luck
-sohel

anyone can help me
[cpp]//10235
#include<stdio.h>
#include<string.h>
main()
{
unsigned long n=0,tr1=0,tr2=0,in1=0,in2=0,i=0,j=0,num=2;
char in[100000];
while(scanf("%s",in)==1)
{
n=strlen(in);
j=n-1;
while(i<n)
{
in1=(in-'0')+in1*10;
in2=(in[j]-'0')+in2*10;
i++;
j--;
}
while(tr1==0&&num<in1)
{
if(in1%num==0)
tr1=1;
else
num++;
}
num=2;
if(in1==in2)
tr2=1;
while(tr2==0&&num<in2)
{
if(in2%num==0)
tr2=1;
else
num++;
}
if(tr1==1)
printf("%ld is not prime.\n",in1);
else if(tr1==0&&tr2==1)
printf("%ld is prime.\n",in1);
else if(tr1==0&&tr2==0)
printf("%ld is emirp.\n",in1);
tr1=0;
tr2=0;
in1=0;
in2=0;
num=2;
i=0;
}
}[/cpp]

Hi,

Change the limits of your while loops to sqrt(in1)+1 and sqrt(in2)+1.

If the TLE still, try precalc the prime numbers and put them on a big array
(sieve of eratostenes).

Bye.

Red Scorpion wrote:Try This Test Case :
Note : that 2 is not emirp, but prime!

Sample input:
2
1
10
11
71
9001
10009901
1321
1099933

Sample Output
2 is prime.
1 is prime.
10 is not prime.
11 is prime.
71 is emirp.
9001 is emirp.
10009901 is not prime.
1321 is emirp
1099933 is emirp.

Hope this Helps.
GOOD LUCK
RED SCORPION

1 is not prime!!

1 is not prime!!

it doesn't matter actually. problem statement says:

Assume that 1< N< 1000000.

so, 1 is not in the judge input

good luck
-sohel