Clever Use of poj 2689 prime number Filtering

Question:

Returns an interval [L, R] to obtain the shortest prime number and the longest distance in the interval. (R <2*10 ^ 9, R-l <= 10 ^ 6)

The number of factors allowed by number theory can be obtained within the root number. Therefore, we can obtain the prime number in 5*10 ^ 4. Then, we use a screening method to find the sum of numbers in [L, R], and the rest is the prime number. Discretization is used here to save a number X-l in the array. Because it cannot be saved directly, but we find that the interval has small characteristics. Therefore, we can think of discretization.

 

# Include <iostream> # include <algorithm> # include <cstdio> # include <cstring> # include <cmath> using namespace STD; typedef long ll; const int maxn = 50000; int primes [maxn]; bool VST [maxn]; int notprimes [1000010]; int POS [1000010]; int top, PCNT; void Init () {Top = 0; memset (VST, 0, sizeof (VST); VST [0] = VST [1] = 1; for (INT I = 2; I <maxn; ++ I) if (! VST [I]) {primes [top ++] = I; for (Int J = I + I; j <maxn; j + = I) VST [J] = 1;} // printf ("Top: % d \ n", top);} void solve (int l, int R) {memset (notprimes, 0, sizeof (notprimes); If (L = 1) L = 2; // prevent all prime numbers in this interval from being screened !!!!! For (INT I = 0; I <top & (LL) primes [I] * primes [I] <= r; ++ I) {// filter factor int S = L/primes [I] + (L % primes [I]> 0 ); // The minimum multiple of the current prime number reaches l s = (S = 1? 2: S); // prevents all prime numbers in this interval from being screened !!!!! For (Int J = s; (LL) J * primes [I] <= r; ++ J) {If (LL) J * primes [I]> = L) // sum notprimes [J * primes [I]-l] = 1; // equivalent to discretization} PCNT = 0; For (INT I = 0; I <= R-l; ++ I) {If (! Notprimes [I]) {pos [PCNT ++] = I + L; // printf ("I --> % d \ n", I + l );}} if (PCNT <2) {puts ("there are no adjacent primes. ");} else {int minl, minr, maxl, maxr, MINV = 999999, maxv =-1; for (INT I = 1; I <PCNT; ++ I) {If (Pos [I]-pos [I-1]> maxv) {maxv = POS [I]-pos [I-1]; maxl = POS [I-1]; maxr = POS [I];} If (Pos [I]-pos [I-1] <MINV) {MINV = POS [I]-pos [I-1]; minl = POS [I-1]; minr = POS [I];} printf (" % D, % d are closest, % d, % d are most distant. \ n ", minl, minr, maxl, maxr) ;}} int main () {Init (); int L, R; while (~ Scanf ("% d", & L, & R) {solve (L, R);} return 0 ;}

 

Clever Use of poj 2689 prime number Filtering