问题
I'm trying to solve the problem PRIME1 using segmented sieve of Eratosthenes. My program works correctly with the normal sieve that is up to NEW_MAX. But there is some problem with cases n > NEW_MAX, where segmented sieving comes into play. In such cases it merely prints all the numbers. Here is the link to the code with relevant test cases: http://ideone.com/8H5lK#view_edit_box
/* segmented sieve */
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#define MAX_LIMIT 1000000000 //10^9
#define NEW_MAX 31623 /*SQUARE ROOT OF 1000000000*/
#define MAX_WIDTH 100000 //10^5
char flags[NEW_MAX+100]; /*TO PREVENT SEGMENTATION FAULT*/
void initialise(char flagarr[], long int n) //initialise all elements to true from 1 to n
{
long int i;
for (i = 1; i <= n; i++)
flagarr[i] = 't';
}
void sieve(unsigned long long m, unsigned long long n, char seg_flags[])
{
unsigned long long p, i, limit;
if (m == 1)
seg_flags[1] = 'f';
/*Seperate inner loop for p=2 so that evens are not iterated*/
for (i = 4; i >= m && i <= n; i += 2)
{
seg_flags[i-m+1] = 'f';
}
if (seg_flags == flags)
limit = NEW_MAX;
else
limit = sqrt(n);
for (p = 3; p <= limit+1; p += 2) //initial p+=2 bcoz we need not check even
{
if (flags[p] == 't')
{
for (i = p*p; i >= m && i <= n; i += p) //start from p square since under it would have been cut
seg_flags[i-m+1] = 'f'; /*p*p, p*p+p, p*p + 2p are not primes*/
}
}
}
void print_sieve(unsigned long long m,unsigned long long n,char flagarr[])
{
unsigned long long i;
if (flags == flagarr) //print non-segented sieve
{
for (i = m; i <= n; i++)
if (flagarr[i] == 't')
printf("%llu\n", i);
}
else
{
//print segmented
for (i = m; i <= n; i++)
if (flagarr[i-m+1] == 't')
printf("%llu\n", i);
}
}
int main()
{
unsigned long long m, n;
int t;
char seg_flags[MAX_WIDTH+100];
/*setting of flags for prime nos. by sieve of erasthromas upto NEW_MAX*/
initialise(flags, NEW_MAX);
flags[1] = 'f'; /*1 is not prime*/
sieve(1, NEW_MAX, flags);
/*end of initial sieving*/
scanf("%d", &t);
while (t--)
{
scanf("%llu %llu", &m, &n);
if (n <= NEW_MAX)
print_sieve(m, n, flags); //NO SEGMENTED SIEVING NECESSARY
else if (m > NEW_MAX)
{
initialise(seg_flags, n-m+1); //segmented sieving necessary
sieve(m, n, seg_flags);
print_sieve(m, n, seg_flags);
}
else if (m <= NEW_MAX && n > NEW_MAX) //PARTIAL SEGMENTED SIEVING NECESSARY
{
print_sieve(m, NEW_MAX, flags);
/*now lower bound for seg sieving is new_max+1*/
initialise(seg_flags, n-NEW_MAX);
sieve(NEW_MAX+1, n, seg_flags);
print_sieve(NEW_MAX+1, n, seg_flags);
}
putchar('\n');
}
system("pause");
return 0;
}
Update: Thanks fr your response Daniel. I implemented some of ur suggestions, my code now produces correct output :-
/*segmented sieve*/
#include<math.h>
#include<stdio.h>
#include<stdlib.h>
#define MAX_LIMIT 1000000000 /*10^9*/
#define NEW_MAX 31623 /*SQUARE ROOT OF 1000000000*/
#define MAX_WIDTH 100000 /*10^5*/
int flags[NEW_MAX+1]; /*TO PREVENT SEGMENTATION FAULT goblal so initialised to 0,true*/
void initialise(int flagarr[],long int n)
/*initialise all elements to true from 1 to n*/
{
long int i;
for(i=3;i<=n;i+=2)
flagarr[i]=0;
}
void sieve(unsigned long m,unsigned long n,int seg_flags[])
{
unsigned long p,i,limit;
/*Seperate inner loop for p=2 so that evens are not iterated*/
if(m%2==0)
i=m;
else
i=m+1;
/*i is now next even > m*/
for(;i<=n;i+=2)
{
seg_flags[i-m+1]=1;
}
if(seg_flags==flags)
limit=NEW_MAX;
else
limit=sqrt(n);
for(p=3;p<=limit+1;p+=2) /*initial p+=2 bcoz we need not check even*/
{
if(flags[p]==0)
{
for(i=p*p; i<=n ;i+=p)
/*start from p square since under it would have been cut*/
{
if(i<m)
continue;
seg_flags[i-m+1]=1;
/*p*p, p*p+p, p*p + 2p are not primes*/
}
}
}
}
void print_sieve(unsigned long m,unsigned long n,int flagarr[])
{
unsigned long i;
if(m<3)
{printf("2\n");m=3;}
if(m%2==0)
i=m+1;
else
i=m;
if(flags==flagarr) /*print non-segented sieve*/
{
for(;i<=n;i+=2)
if(flagarr[i]==0)
printf("%lu\n",i);
}
else {
//print segmented
for(;i<=n;i+=2)
if(flagarr[i-m+1]==0)
printf("%lu\n",i);
}}
int main()
{
unsigned long m,n;
int t;
int seg_flags[MAX_WIDTH+100];
/*setting of flags for prime nos. by sieve of erasthromas upto NEW_MAX*/
sieve(1,NEW_MAX,flags);
/*end of initial sieving*/
scanf("%d",&t);
while(t--)
{
scanf("%lu %lu",&m,&n);
if(n<=NEW_MAX)
print_sieve(m,n,flags);
/*NO SEGMENTED SIEVING NECESSARY*/
else if(m>NEW_MAX)
{
initialise(seg_flags,n-m+1);
/*segmented sieving necessary*/
sieve(m,n,seg_flags);
print_sieve(m,n,seg_flags);
}
else if(m<=NEW_MAX && n>NEW_MAX)
/*PARTIAL SEGMENTED SIEVING NECESSARY*/
{
print_sieve(m,NEW_MAX,flags);
/*now lower bound for seg sieving is new_max+1*/
initialise(seg_flags,n-NEW_MAX);
sieve(NEW_MAX+1,n,seg_flags);
print_sieve(NEW_MAX+1,n,seg_flags);
}
putchar('\n');
}
system("pause");
return 0;
}
but my sieve function below further taking into account ur other suggestions produces incorrect output:-
void sieve(unsigned long m,unsigned long n,int seg_flags[])
{
unsigned long p,i,limit;
p=sqrt(m);
while(flags[p]!=0)
p++;
/*we thus get the starting prime, the first prime>sqrt(m)*/
if(seg_flags==flags)
limit=NEW_MAX;
else
limit=sqrt(n);
for(;p<=limit+1;p++) /*initial p+=2 bcoz we need not check even*/
{
if(flags[p]==0)
{
if(m%p==0) /*to find first multiple of p>=m*/
i=m;
else
i=(m/p+1)*p;
for(; i<=n ;i+=p)
/*start from p square since under it would have been cut*/
{
seg_flags[i-m+1]=1;
/*p*p, p*p+p, p*p + 2p are not primes*/
}
}
}
}
回答1:
Your problem is
for (i = 4; i >= m && i <= n; i += 2)
for (i = p*p; i >= m && i <= n; i += p)
You only ever eliminate the multiples of 2 as such if the lower end of the range is 4 or less, and you only eliminate multiples of primes larger than sqrt(m). Remove the i >= m part from the loop condition and replace it with an if (i < m) { continue; } in the loop body (better, calculate the first multiple of p not less than m directly and avoid that condition completely).
And instead of using 't' and 'f' as flags, use 1 and 0 as DMR intended, that will be understood better.
Re update: This
/*Seperate inner loop for p=2 so that evens are not iterated*/
if(m%2==0)
i=m;
else
i=m+1;
/*i is now next even > m*/
for(;i<=n;i+=2)
will hurt you if m == 2. If m == 2, you must start with i = 4.
Concerning
unsigned long p,i,limit;
p=sqrt(m);
while(flags[p]!=0)
p++;
/* snip */
for(;p<=limit+1;p++)
it seems you misunderstood me, "and you only eliminate multiples of primes larger than sqrt(m)" doesn't mean you needn't eliminate multiples of smaller primes, it means that your initial version didn't, which resulted in almost all numbers in the range not eliminated. You should start the outer loop with p = 2 -or have a separate pass for the multiples of 2 and start that loop with 3, incrementing p by 2, and start the inner loop at the larger of p*p and the smallest multiple of p not less than m. Your code for the latter works, so wrapping it in an
if ((i = p*p) < m) {
/* set i to smallest multiple of p which is >= m */
}
would work (you can make it a bit faster avoiding a branch and using only one division, but the difference will be minuscule).
Finally, your choice of what you represent by 0 and 1 is uncanonical, this is C, so 0 evaluates to false in conditions and everything else to true, so the canonical replacement would have been 't' -> 1 and 'f' -> 0 and in contexts like this, where the array entries are flags, one would check
if (flags[p]) // instead of: if (flags[p] != 0)
if (!flags[p]) // instead of: if (flags[p] == 0)
also there's no need to change the array types from char[] to int[], char is an integer type too, so 0 and 1 are perfectly fine char values. The choice of type for the arrays has performance implications. On the one hand, int-sized loads and stores are typically faster than byte-sized, so that would favour int flags[] or even long int flags[] for word-sized reads and writes. On the other hand, with the smaller char flags[] you get better cache locality. You would get even better cache locality with using a single bit per flag, but that would make reading/setting/clearing flags still slower. What yields the best performance depends on the architecture and size of the sieves.
回答2:
Daniel Fischer seems to have clarified some of the main trouble points.
If you're interested in seeing some more concise code/explanation for this primes problem, check out:
http://www.swageroo.com/wordpress/spoj-problem-2-prime-generator-prime1/
来源：https://stackoverflow.com/questions/9362177/spoj-prime1-using-sieve-of-eratosthenes-in-c