POJ 2689 - Prime Distance - [筛法求素数]

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 #define MAX 46345
 5 typedef long long ll;
 6 
 7 bool isPrime[1000005];
 8 
 9 //欧拉筛法求[0,46345]内的素数 - begin
10 int Prime1[MAX],cnt1;
11 void screen1()
12 {
13     memset(isPrime,1,sizeof(isPrime));
14     cnt1=0;
15     isPrime[0]=isPrime[1]=0;
16     for(int i=2;i<=MAX;i++)
17     {
18         if(isPrime[i]) Prime1[cnt1++]=i;
19         for(int j=0;j<cnt1;j++)
20         {
21             if(i*Prime1[j]>MAX) break;
22             isPrime[(i*Prime1[j])]=0;
23             if(i%Prime1[j]==0) break;
24         }
25     }
26 }
27 //欧拉筛法求[0,46345]内的素数 - end
28 
29 //埃筛求[st,ed]内的素数 - begin
30 ll st,ed;
31 ll Prime2[1000005];
32 int cnt2;
33 void screen2()
34 {
35     memset(isPrime,1,sizeof(isPrime));
36     for(int i=0;i<cnt1;i++)
37     {
38         if(Prime1[i]>ed) break;//这种情况下，已经不可能再筛了，跳出
39 
40         ll tmp=st/Prime1[i];
41         while(tmp*Prime1[i]<st) tmp++;
42         while(tmp<=1) tmp++;
43         
44         for(ll j=tmp*Prime1[i];j<=ed;j+=Prime1[i]) isPrime[j-st]=0;
45     }
46     if(st==1) isPrime[0]=0;
47         //得到了[st,ed]内的素数的标记数组
48         
49     cnt2=0;
50     for(ll i=st;i<=ed;i++)
51     {
52         if(isPrime[i-st]) Prime2[cnt2++]=i;
53     }
54         //根据标记数组，得到存储[st,ed]内所有素数的数组
55 }
56 //埃筛求[st,ed]内的素数 - end
57 
58 int main()
59 {
60     screen1();
61     while(scanf("%lld%lld",&st,&ed)!=EOF)
62     {
63         screen2();
64         if(cnt2<=1)
65         {
66             printf("There are no adjacent primes.\n");
67             continue;
68         }
69         else
70         {
71             ll min_dist=0x3f3f3f3f,max_dist=0;
72             ll c1,c2,d1,d2;
73             for(int i=0;i<cnt2-1;i++)
74             {
75                 if(Prime2[i+1]-Prime2[i]<min_dist)
76                 {
77                     min_dist=Prime2[i+1]-Prime2[i];
78                     c1=Prime2[i], c2=Prime2[i+1];
79                 }
80                 if(Prime2[i+1]-Prime2[i]>max_dist)
81                 {
82                     max_dist=Prime2[i+1]-Prime2[i];
83                     d1=Prime2[i], d2=Prime2[i+1];
84                 }
85             }
86             printf("%I64d,%I64d are closest, %I64d,%I64d are most distant.\n",c1,c2,d1,d2);
87         }
88     }
89 }