poj 2689 区间筛素数

由于区间的右端点非常大（INT_MAX），而区间长度相对小（100W），所以考虑区间筛法，左端点为1的情况需要特判一下。

#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

typedef long long ll;
const int MAX = 9999999;
const int MIN = -1;
const int N = 50000;
const int M = 5000;
const int K = 1000001;
bool visit[N];
int prime[M];
bool big_visit[K];
int big_prime[N];
int pn;

void sieve( int n )
{
    int r = sqrt( n + 0.5 );
    memset( visit, 0, sizeof(visit) );
    visit[0] = visit[1] = 1;
    for ( int i = 2; i <= r; i++ )
    {
        if ( !visit[i] )
        {
            for ( int j = i * i; j <= n; j += i )
            {
                visit[j] = 1;
            }
        }
    }
}

void get_prime( int n )
{
    sieve(n);
    pn = 0;
    for ( int i = 0; i <= n; i++ )
    {
        if ( !visit[i] )
        {
            prime[pn++] = i;
        }
    }
}

int main ()
{
    get_prime( N - 1 );
    int l, u;
    while ( scanf("%d%d", &l, &u) != EOF )
    {
        memset( big_visit, 0, sizeof(big_visit) );
        if ( l == 1 ) big_visit[0] = 1;
        int bound = sqrt( u + 0.5 );
        for ( int i = 0; prime[i] <= bound; i++ )
        {
            int r = ( ( ll ) l + prime[i] - 1 ) / prime[i];
            if ( r < 2 ) r = 2;
            for ( ll j = ( ll ) r * prime[i]; j <= ( ll ) u; j += prime[i] )
            {
                big_visit[j - l] = 1;
            }
        }
        int cnt = 0;
        for ( int i = 0; i <= u - l; i++ )
        {
            if ( !big_visit[i] )
            {
                big_prime[cnt++] = i;
            }
        }
        if ( cnt == 1 )
        {
            printf("There are no adjacent primes.
");
            continue;
        }
        int minn = MAX, maxn = MIN, pma = -1, pmi = -1;
        for ( int i = 0; i < cnt - 1; i++ )
        {
            int d = big_prime[i + 1] - big_prime[i];
            if ( d > maxn )
            {
                maxn = d;
                pma = i;
            }
            if ( d < minn )
            {
                minn = d;
                pmi = i;
            }
        }
        int o1 = big_prime[pmi] + l, o2 = big_prime[pmi + 1] + l;
        int o3 = big_prime[pma] + l, o4 = big_prime[pma + 1] + l;
        printf("%d,%d are closest, %d,%d are most distant.
", o1, o2, o3, o4);
    }
    return 0;
}