SPOJ PRIME1 : TLE [closed]

问题

I tried implementing the segmented sieve algorithm for this [question]:http://www.spoj.pl/problems/PRIME1/ as follows :

#include <iostream>
#include <string>
#include <set>
#include<math.h>
#include<vector>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cstdio>
#define MAX 32000 // sqrt of the upper range
using namespace std;
int base[MAX];  // 0 indicates prime

vector<int> pv;   // vector of primes

int mod (int a, int b)
{
   if(b < 0)
     return mod(-a, -b);   
   int ret = a % b;
   if(ret < 0)
     ret+=b;
   return ret;
}
void sieve(){

     for(int i = 2 ; i * i < MAX ; i++ )
        if(!base[i])
           for(int j = i * i ; j <  MAX ; j += i )
                 base[j] = 1;

     for(int i = 2 ; i < MAX ; i++ )
         if(!base[i]) pv.push_back(i);

}
int fd_p(int p ,int a ,int b){  // find the first number in the range [a,b] which is divisible by prime p

/*  while(1){

        if(a % p == 0 && a !=p) break;
    a++;
    }
    return a;
*/  

    if(a != p){
        return (a + mod(-a,p)) ;

    }
    else{
     return (a + p);
    }

}
void seg_sieve(int a , int b){

    if(b < 2 ){ 
        cout << "" ;
    return;
    }
    if(a < 2){
      a = 2; 
    }
    int i,j;
    int seg_size  = b - a + 1;
    int*is_prime = new int[seg_size];
    memset(is_prime,0,seg_size*sizeof(int));

    vector<int> :: iterator p ;


    for(p = pv.begin(); p!=pv.end(); p++){
       int x = fd_p(*p,a,b);  

       for(i = x; i <= b; i += *p )
           is_prime[i - a] = 1;
      }

for(i=0; i < b - a + 1; i++)
    if(!is_prime[i])
        printf("%u\n", i + a);

 delete []is_prime ;
}


int main()
{
     sieve();
     int a,b,T;
     scanf("%d",&T);

     while(T--){
     scanf("%d%d",&a,&b);
     seg_sieve(a,b);
     printf("\n");   
     }
//     cout<<endl;
//     system("PAUSE");
     return 0;
}

I am getting TLE nevertheless .. I don't understand what other optimization would be required . Plz help ..

Edit 1 :just tried to implement fd_p() in constant time ... [failure] .. plz if u could help me with this bug..

Edit 2:Issue Resolved.

回答1:

You can get the first number in the interval [a,b] that is divisible by p in constant time. Try to do that and I think you should be good to go.

回答2:

I have solved this problem many years ago. Assume, that n-m <= 100000 All you need to calculate all Primes between 1 and sqrt(1000000000) < 40000. Than manually test each number between n and m. This will be ehough

 program prime1;
  Var
   t:longint;
   m,n:longint;
   i,j,k:longint;
   prime:array of longint;
   bool:boolean;
begin
 SetLength(prime,1);
 prime[0]:=2;
 for i:=3 to 40000
  do begin
   j:=0; bool:=true;
   while (prime[j]*prime[j]<= i ) do begin
     if (i mod prime[j] = 0) then begin
      bool:=false;
      break;
     end;
     inc(j);
   end;
   if (bool) then begin
    SetLength(prime,length(prime)+1);
    prime[length(prime)-1]:=i;
   end;
 end;
 readln(t);
 for k:=1 to t do begin
  readln(m,n);
  for i:=m to n do begin
   if (i=1) then continue;
   j:=0; bool:=true;
   while (prime[j]*prime[j]<= i ) do begin
     if (i mod prime[j] = 0) then begin
      bool:=false;
      break;
     end;
     inc(j);
   end;
   if (bool) then
     writeln(i);
  end;
  writeln;
 end;
end.

回答3:

You've left one last step of improvement to make. Work with the odds only.

We know that 2 is prime, and we know that no even (other than 2) is ever a prime. So there's no need to check them.

The sieve of Eratosthenes for odd primes is P = {3,5, ...} \ U {{p², p² + 2p, ...} | p in P}. Implementing that will be enough to get you through:

Treat 2 specially, as a separate case. Work with arrays half the normal size, where the array entry at offset i represents an odd value ao + 2*i where ao = a|1 is the least odd number not below a. That means that increment value of 2p corresponds to the increment of p in the offset in the array.
The starting odd multiple of a prime p in the offset sieve array, equal to or above p*p, is m = p*p >= ao ? p*p : ((ao+p-1)/p)*p; m = m&1 ? m : m+p;, provided that p <= sqrt_b. The corresponding offset in the sieve array is (m-ao)/2.

As a side note, your naming is confusing: is_prime is actually is_composite.

回答4:

What's wrong is that your fd_p function is far too slow, incrementing a till you find a good value to start your sieve will definitely time out since a can be in the range of 1 billion.

You have the right idea though.

See this blog post for an easier to understand explanation with working code as well:

http://www.swageroo.com/wordpress/spoj-problem-2-prime-generator-prime1/

来源：https://stackoverflow.com/questions/12003616/spoj-prime1-tle

标签

c++

algorithm

primes

sieve-of-eratosthenes

sieve