POJ-2689-Prime Distance（筛法）

Language: Default Prime Distance Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15777   Accepted: 4194 Description The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers.  Your program is given 2 numbers: L and U (1<=L< U<=2,147,483,647), and you are to find the two adjacent primes C1 and C2 (L<=C1< C2<=U) that are closest (i.e. C2-C1 is the minimum). If there are other pairs that are the same distance apart, use the first pair. You are also to find the two adjacent primes D1 and D2 (L<=D1< D2<=U) where D1 and D2 are as distant from each other as possible (again choosing the first pair if there is a tie). Input Each line of input will contain two positive integers, L and U, with L < U. The difference between L and U will not exceed 1,000,000. Output For each L and U, the output will either be the statement that there are no adjacent primes (because there are less than two primes between the two given numbers) or a line giving the two pairs of adjacent primes. Sample Input 2 17 14 17 Sample Output 2,3 are closest, 7,11 are most distant. There are no adjacent primes. Source

题意：求给定区间内的质数距离最小的一对和质数距离最大的一对。

由于区间过大，不可能直接求出1e9之间的所有质数，但是我们知道 2,147,483,647之间的质数可以用sqrt（2,147,483,647）内的质数筛出来，所以可以先求出sqrt（2,147,483,647） 约5W内的质数， 因为给定区间长度少于1e6，所以直接用5w内质数筛出来区间内质数就可以。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define INF 0x3f3f3f3f
#define LL long long
#define bug cout<<"bug\n"
using namespace std;
#define MAXN 50005
const int M = 1e6+7;
int num_prime;
bool not_prime[MAXN*30];
int prime[MAXN];
long long a,b;
long long prime_2[M];
void get_prime()
{
    long long i,j;
    memset(not_prime,0,sizeof(not_prime));
    num_prime=0;
    not_prime[1]=1;
    for(i=2; i=a)
                not_prime[j-a]=1;
    }
    if(a==1)
        not_prime[0]=1;
}
int main()
{
    get_prime();
    while(scanf("%I64d%I64d",&a,&b)!=EOF)
    {
        long long min=INF,max=-INF;
        long long minl,minr,maxl,maxr;
        get_primein_ab();
        int num_prime2=0;
        for(int i=0; i<=b-a; ++i)
            if(!not_prime[i])
                prime_2[++num_prime2]=i+a;
        if(num_prime2<=1)
            printf("There are no adjacent primes.\n");
        else
        {
            for(int i=1; imax)
                {
                    max=prime_2[i+1]-prime_2[i];
                    maxl=prime_2[i];
                    maxr=prime_2[i+1];
                }
                if(prime_2[i+1]-prime_2[i]