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[];
 long long cprime[];
 long long  isprime[maxn+];
 long long  vis[maxn+];
 long long  q;
 void getprime(){
     //memset(isprime,0,sizeof(isprime));
     q =-;
     long long  i,j;
     isprime[] = isprime[] =;
     for(i=;i<maxn;i++){
         if(!isprime[i])
             prime[++q] = i;
         for(j=;j<=q&&prime[j]*i<maxn;j++){
             isprime[prime[j]*i] =;
             if(i%prime[j]==)
                 break;
         }
     }
 }

 long long  qt;
 void getprime1(long long  a, long long  b){
     qt =-;
     if(b<maxn){
         for(long long  i=a;i<=b;i++){
             if(!isprime[i])
                 cprime[++qt] = i;
         }
     }else{
         memset(vis,,sizeof(vis));
         long long  i,k;
         for(i=;i<=b-a;i++){
             vis[i] =;
         }
         for(i=;prime[i]*prime[i]<=b&&i<=q;i++){
             k = a/prime[i];
             if(k*prime[i]<a) k++;
             if(k<=) k++;
             while(k*prime[i]<=b){
                 vis[k*prime[i]-a] =;
                 k++;
             }
         }
         for(i=;i<=b-a;i++){
             if(vis[i]==)
                 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 =-;
         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=;i<qt;i++){
             long long  temp = cprime[i+]-cprime[i];
             if(temp>maxnum){
                 maxnum =temp;
                 max1 = cprime[i];
                 max2 = cprime[i+];
             }
             if(temp<minnum){
                 minnum =temp;
                 min1 = cprime[i];
                 min2 = cprime[i+];
             }
         }
         if((qt+)<=)
             cout<<"There are no adjacent primes."<<endl;
         else
             cout<<min1<<","<<min2<<" are closest, "<<max1<<","<<max2<<" are most distant."<<endl;
     }
     return ;
 }