poj 2689 大范围内素数筛选

 1 /**
 2 给定一定范围求其内的素数
 3 
 4 注意：
 5 **/
 6 
 7 #include <iostream>
 8 #include <math.h>
 9 #include <cstring>
10 using namespace std;
11 #define maxn 1000000
12 long long  prime[500000];
13 long long cprime[500000];
14 long long  isprime[maxn+100];
15 long long  vis[maxn+100];
16 long long  q;
17 void getprime(){
18     //memset(isprime,0,sizeof(isprime));
19     q =-1;
20     long long  i,j;
21     isprime[0] = isprime[1] =1;
22     for(i=2;i<maxn;i++){
23         if(!isprime[i])
24             prime[++q] = i;
25         for(j=0;j<=q&&prime[j]*i<maxn;j++){
26             isprime[prime[j]*i] =1;
27             if(i%prime[j]==0)
28                 break;
29         }
30     }
31 }
32 
33 long long  qt;
34 void getprime1(long long  a, long long  b){
35     qt =-1;
36     if(b<maxn){
37         for(long long  i=a;i<=b;i++){
38             if(!isprime[i])
39                 cprime[++qt] = i;
40         }
41     }else{
42         memset(vis,0,sizeof(vis));
43         long long  i,k;
44         for(i=0;i<=b-a;i++){
45             vis[i] =1;
46         }
47         for(i=0;prime[i]*prime[i]<=b&&i<=q;i++){
48             k = a/prime[i];
49             if(k*prime[i]<a) k++;
50             if(k<=1) k++;
51             while(k*prime[i]<=b){
52                 vis[k*prime[i]-a] =0;
53                 k++;
54             }
55         }
56         for(i=0;i<=b-a;i++){
57             if(vis[i]==1)
58                 cprime[++qt] = i+a;
59         }
60     }
61 
62 }
63 int main()
64 {
65     getprime();
66     //for(int i=0;i<10;i++)
67     //    cout<<prime[i]<<endl;
68     long long  a,b;
69     while(cin>>a>>b){
70         long long  maxnum =-1;
71         long long  minnum =0x3f3f3f3f;
72         getprime1(a,b);
73         long long  max1,max2,min1,min2;
74        // for(int i=0;i<=qt;i++)
75         //    cout<<cprime[i]<<endl;
76         for(long long  i=0;i<qt;i++){
77             long long  temp = cprime[i+1]-cprime[i];
78             if(temp>maxnum){
79                 maxnum =temp;
80                 max1 = cprime[i];
81                 max2 = cprime[i+1];
82             }
83             if(temp<minnum){
84                 minnum =temp;
85                 min1 = cprime[i];
86                 min2 = cprime[i+1];
87             }
88         }
89         if((qt+1)<=1)
90             cout<<"There are no adjacent primes."<<endl;
91         else
92             cout<<min1<<","<<min2<<" are closest, "<<max1<<","<<max2<<" are most distant."<<endl;
93     }
94     return 0;
95 }