ProjectEuler problem 35

[Tiberiu Gal]

Good point there, unfortunately it's not that big a gain in this 
particular instance.

thank you

On Wednesday, 16 May 2012 at 12:46:07 UTC, Andrea Fontana wrote:
> What about some logic optimizations?
>
> F.E. if number contains one of these digit 0,2,4,6,8,5 is not 
> circular of course ..
>
>
> On Wednesday, 16 May 2012 at 09:26:45 UTC, Tiberiu Gal wrote:
>> hi
>>
>> many claim their code solves the problem in order of ms ( 
>> c/pascal/haskell code)
>>
>> my D code takes ~1.5  seconds.
>> Can it be made faster ( without pointers )?
>> it runs the same with or without caching the primes, and most 
>> of the time it spends finding primes; isCircularPrime can be 
>> optimized a bit, obviously, but it's not the bottleneck.
>>
>> thanks
>>
>> ===========================================
>> module te35;
>>
>> import std.stdio, std.math, std.conv;
>>
>> const Max = 1_000_000;
>>
>> byte[Max] primes;
>>
>> void main() {
>> 	primes[] = -1;
>> 	int cnt;
>> 	foreach(i; 2 .. Max) {
>> 		//writefln("is %s prime ? %s ", i, i.isPrime);
>> 		if( i.isPrime && i.isCircularPrime) {
>> 			cnt++;
>> 			//writefln("\t\tis %s, circular prime ? %s ", i, 
>> i.isCircularPrime);
>> 		}
>> 	}
>>
>> 	writeln(cnt);
>> 	
>> }
>>
>> bool isPrime(int n) {
>> 	byte x = 0;
>> 	if( ( x = primes[n]) != -1) return (x == 1);
>>
>> 	if( n < 2 && n > 0 ) {
>> 		primes[n] = 0;
>> 		return true;
>> 	}
>> 	
>> 	//int s = cast(int) (sqrt( cast(real) n) ) + 1;
>> 	for(int i=2; i*i < n + 1; i++) {
>> 		if( n %i == 0 ) {
>> 			primes[n] = 0;
>> 			return false;
>> 		}
>> 	}
>> 	
>> 	primes[n] = 1;
>> 	return true;
>>
>> }
>>
>> bool isCircularPrime( int n) {
>> 	
>> 	auto c = to!string(n);
>> 	for(int i ; i < c.length; i++){
>> 		c = c[1 .. $] ~ c[0];
>> 		if( !(to!int(c) ).isPrime )
>> 			return false;
>> 	}
>> 	return true;
>> }