二次筛选区间素数 POJ2689

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 const int maxn=50000+10;
 6 using namespace std;
 7 int p[maxn],a[maxn],f[1000000+10],t=0;
 8 void prepare(){
 9     int m=(int)sqrt(50000);
10     for(int i=2;i<=m;i++)
11         if(!p[i])
12             for(int j=i*i;j<=50000;j+=i)p[j]=1;
13     for(int i=2;i<=50000;i++)if(!p[i])a[++t]=i;
14 }
15 int main(){
16     prepare();
17     int L,U;
18     while(scanf("%d%d",&L,&U)!=EOF){
19         if(L<2)L=2;
20         memset(f,0,sizeof(f));
21         for(int i=1;i<=t;i++){
22             int s=(L)/a[i],t=U/a[i];
23             if(L%a[i])s++;     //这步是必须的，不然会出现j*a[i]-L<0 而出现runtime error
24             if(s<2)s=2;
25             for(int j=s;j<=t;j++)f[j*a[i]-L]=1;
26         }
27         int p=-1,max_ans=-1,min_ans=1000000000,x1,y1,x2,y2;
28         for(int i=0;i<=U-L;i++){
29             if(f[i]==0){
30                 if(p==-1){p=i;continue;}
31                 if(max_ans<i-p){max_ans=i-p;x1=p+L;y1=i+L;}
32                 if(min_ans>i-p){min_ans=i-p;x2=p+L;y2=i+L;}
33                 p=i;
34             }
35         }
36         if(max_ans==-1)cout<<"There are no adjacent primes."<<endl;
37         else cout<<x2<<","<<y2<<" are closest, "<<x1<<","<<y1<<" are most distant."<<endl;
38     }
39     return 0;
40 }