poj 2689 区间筛素数

由于区间的右端点非常大（INT_MAX），而区间长度相对小（100W），所以考虑区间筛法，左端点为1的情况需要特判一下。

 1 #include <cstring>
 2 #include <cstdio>
 3 #include <cmath>
 4 using namespace std;
 5 
 6 typedef long long ll;
 7 const int MAX = 9999999;
 8 const int MIN = -1;
 9 const int N = 50000;
10 const int M = 5000;
11 const int K = 1000001;
12 bool visit[N];
13 int prime[M];
14 bool big_visit[K];
15 int big_prime[N];
16 int pn;
17 
18 void sieve( int n )
19 {
20     int r = sqrt( n + 0.5 );
21     memset( visit, 0, sizeof(visit) );
22     visit[0] = visit[1] = 1;
23     for ( int i = 2; i <= r; i++ )
24     {
25         if ( !visit[i] )
26         {
27             for ( int j = i * i; j <= n; j += i )
28             {
29                 visit[j] = 1;
30             }
31         }
32     }
33 }
34 
35 void get_prime( int n )
36 {
37     sieve(n);
38     pn = 0;
39     for ( int i = 0; i <= n; i++ )
40     {
41         if ( !visit[i] )
42         {
43             prime[pn++] = i;
44         }
45     }
46 }
47 
48 int main ()
49 {
50     get_prime( N - 1 );
51     int l, u;
52     while ( scanf("%d%d", &l, &u) != EOF )
53     {
54         memset( big_visit, 0, sizeof(big_visit) );
55         if ( l == 1 ) big_visit[0] = 1;
56         int bound = sqrt( u + 0.5 );
57         for ( int i = 0; prime[i] <= bound; i++ )
58         {
59             int r = ( ( ll ) l + prime[i] - 1 ) / prime[i];
60             if ( r < 2 ) r = 2;
61             for ( ll j = ( ll ) r * prime[i]; j <= ( ll ) u; j += prime[i] )
62             {
63                 big_visit[j - l] = 1;
64             }
65         }
66         int cnt = 0;
67         for ( int i = 0; i <= u - l; i++ )
68         {
69             if ( !big_visit[i] )
70             {
71                 big_prime[cnt++] = i;
72             }
73         }
74         if ( cnt == 1 )
75         {
76             printf("There are no adjacent primes.\n");
77             continue;
78         }
79         int minn = MAX, maxn = MIN, pma = -1, pmi = -1;
80         for ( int i = 0; i < cnt - 1; i++ )
81         {
82             int d = big_prime[i + 1] - big_prime[i];
83             if ( d > maxn )
84             {
85                 maxn = d;
86                 pma = i;
87             }
88             if ( d < minn )
89             {
90                 minn = d;
91                 pmi = i;
92             }
93         }
94         int o1 = big_prime[pmi] + l, o2 = big_prime[pmi + 1] + l;
95         int o3 = big_prime[pma] + l, o4 = big_prime[pma + 1] + l;
96         printf("%d,%d are closest, %d,%d are most distant.\n", o1, o2, o3, o4);
97     }
98     return 0;
99 }

 

posted @ 2015-07-27 12:28  hxy_has_been_used  阅读(189)  评论(0)  编辑  收藏  举报