poj 2689 大范围内素数筛选

/**
给定一定范围求其内的素数

注意：
**/

#include <iostream>
#include <math.h>
#include <cstring>
using namespace std;
#define maxn 1000000
long long  prime[500000];
long long cprime[500000];
long long  isprime[maxn+100];
long long  vis[maxn+100];
long long  q;
void getprime(){
    //memset(isprime,0,sizeof(isprime));
    q =-1;
    long long  i,j;
    isprime[0] = isprime[1] =1;
    for(i=2;i<maxn;i++){
        if(!isprime[i])
            prime[++q] = i;
        for(j=0;j<=q&&prime[j]*i<maxn;j++){
            isprime[prime[j]*i] =1;
            if(i%prime[j]==0)
                break;
        }
    }
}

long long  qt;
void getprime1(long long  a, long long  b){
    qt =-1;
    if(b<maxn){
        for(long long  i=a;i<=b;i++){
            if(!isprime[i])
                cprime[++qt] = i;
        }
    }else{
        memset(vis,0,sizeof(vis));
        long long  i,k;
        for(i=0;i<=b-a;i++){
            vis[i] =1;
        }
        for(i=0;prime[i]*prime[i]<=b&&i<=q;i++){
            k = a/prime[i];
            if(k*prime[i]<a) k++;
            if(k<=1) k++;
            while(k*prime[i]<=b){
                vis[k*prime[i]-a] =0;
                k++;
            }
        }
        for(i=0;i<=b-a;i++){
            if(vis[i]==1)
                cprime[++qt] = i+a;
        }
    }

}
int main()
{
    getprime();
    //for(int i=0;i<10;i++)
    //    cout<<prime[i]<<endl;
    long long  a,b;
    while(cin>>a>>b){
        long long  maxnum =-1;
        long long  minnum =0x3f3f3f3f;
        getprime1(a,b);
        long long  max1,max2,min1,min2;
       // for(int i=0;i<=qt;i++)
        //    cout<<cprime[i]<<endl;
        for(long long  i=0;i<qt;i++){
            long long  temp = cprime[i+1]-cprime[i];
            if(temp>maxnum){
                maxnum =temp;
                max1 = cprime[i];
                max2 = cprime[i+1];
            }
            if(temp<minnum){
                minnum =temp;
                min1 = cprime[i];
                min2 = cprime[i+1];
            }
        }
        if((qt+1)<=1)
            cout<<"There are no adjacent primes."<<endl;
        else
            cout<<min1<<","<<min2<<" are closest, "<<max1<<","<<max2<<" are most distant."<<endl;
    }
    return 0;
}