Filtering of prime numbers in a given range (poj 2689)

Problem: Evaluate all prime numbers from L to R (L <r <= 2147483647, R-l <= 10 ^ 6)

Solution: To determine whether X is a prime number, check whether X % (all prime numbers smaller than X) has 0...

Of course, you can also optimize X % (all prime numbers smaller than SQRT (x)

Known SQRT (2147483647) = 49361;

The number of prime numbers in the range of [2, 49361] is less than 5000.

Then we use a prime number smaller than SQRT (r) to multiply the upper coefficient for enumeration. In this way, the multiplied number is not a prime number. Small the number of these screens, and the rest is the prime number.

How can we determine this coefficient? We can know that the lower limit of this coefficient is k = L/prime [I]. Of course, if Prime [I]> = L, K is initially 2. In fact, K * prime [I] must be greater than l and closest to L.

 

Table hitting:

CL (VIS,True);For(Ll I =2; I <n; ++I ){For(Ll J = LL (I) * LL (I); j <n; j + = I) vis [J] =False;} N=0;For(Ll I =2; I <n; ++I ){If(Vis [I]) {Prime [n++] =I ;}}
 

 

 

Screen out non-prime numbers:

 Void Get_p2 (ll l, ll R) {ll I; ll K;  If (R <n ){ //  If r <= 49361, you can directly use the original prime number table. I = 0  ;  While (Prime [I] <L) ++ I;  For (N2 = 0 ; Prime [I] <= r; ++ I) P2 [n2 ++] = Prime [I];  Return ;} Cl (VIS,  True  );  For (I = 0 ; I <n & prime [I] * prime [I] <= r; ++ I ){  If (Prime [I]> = L) k = 2  ;  Else  {K = L/ Prime [I];  If (K * prime [I] <L) ++K ;}  While (K * prime [I] <= R) {vis [K * Prime [I]-l] = False  ;  //  Printf ("% LLD % d \ n", K * prime [I], K * prime [I]-l, prime [I]); ++ K ;}} N2 = 0  ;  For (I = 0 ; I <= R-l; ++ I ){ If (Vis [I]) P2 [n2 ++] = I + L ;}} 

 

Given that the original prime number table has n prime numbers, the average number of K increases is ave (K), and the time complexity is O (n * ave (k ));