Prime number Sieve nbut1454 Lux

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Test instructions: Gives the interval [l,r],r-l<=1e6, the maximum and minimum of the difference between the adjacent prime numbers in the interval, and if there is no two prime number, the output-1

Idea: Sift through all prime numbers in the [L,r] interval, and then iterate over

#include <map> #include <set> #include <cmath> #include <stack> #include <queue> #include <cstdio> #include <string> #include <vector> #include <cstring> #include <iostream># include<algorithm> #include <functional> #define FIN freopen ("Input.txt", "R", stdin) #define FOUT freopen (" Output.txt "," w+ ", stdout) using namespace Std;typedef long Long ll;typedef pair<int, int>pii;const int MX = 1e6 + 5;co NST int INF = 0x3f3f3f3f;bool sign[mx];int subprim[mx], prim[mx], TOTLEPRM = 0;void Getprim (int n) {memset (sign,0,size    of (sign));    SIGN[1] = 1;        for (int i = 2; I <= n; i++) {if (sign[i]) continue;        prim[totleprm++] = i;        if (LL) i * i > N) continue;        for (int j = i * i; j <= N; j + = i) {sign[j] = 1; }}}int Getsubprim (int a, int b) {int totl = 0, I;    LL J;    Memset (sign, true, sizeof);    if (a < 2) a = 2;    LL L = b-a + 1; for (i = 0; i < TOTLEPRM; i++) {iF ((j = prim[i] * (a/prim[i)) < a) J + = Prim[i];        if (J < prim[i]*prim[i]) j = prim[i] * Prim[i];    for (; J <= B; j + = Prim[i]) sign[j-a] = false;        } for (i = 0; i < L; i++) if (Sign[i]) {subprim[totl++] = a + i; } return totl;}    int main () {Getprim (100000);    int L, R;//fin;        while (~SCANF ("%d%d", &l, &r)) {int minl = 0, MinR = INF, MaxL = 0, MaxR = 0;        int rear = Getsubprim (L, R);            if (Rear < 2) {printf (" -1\n");        Continue } for (int i = 0; i < rear-1; i++) {if (Minr-minl > subprim[i + 1]-subprim[i]) Minl = Subprim            [i], MinR = subprim[i + 1];        if (Maxr-maxl < Subprim[i + 1]-subprim[i]) MaxL = Subprim[i], MaxR = subprim[i + 1];    } printf ("%d%d%d%d\n", Minl, MinR, MaxL, MaxR); } return 0;}

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Prime number Sieve nbut1454 Lux