poj 2689(素数)

这道题是学习素数筛法的经典，应用到了区间筛素数。具体思路是先筛出1到sqrt(2147483647)之间的所有素数，然后再通过已经晒好素数筛出给定区间的素数。

PS:如果用的int存low,high(输入)。注意high=2147483647.加1就变成负数了。

 1 // File Name: 2689.cpp
 2 // Author: Missa_Chen
 3 // Created Time: 2013年05月28日 星期二 23时16分51秒
 4 
 5 #include<iostream>
 6 #include<string>
 7 #include<algorithm>
 8 #include<cstdio>
 9 #include<cstring>
10 #include<cmath>
11 #include<queue>
12 #include<map>
13 #include<stack>
14 #include<set>
15 #include<cstdlib>
16 
17 using namespace std;
18 
19 #define LL long long
20 const int inf = 0x3f3f3f3f;
21 const int maxn = 1e6 + 5;
22 int low, high;
23 int prim[maxn], p[maxn], pn;
24 void prime()
25 {
26     pn = 0;
27     memset(p,0,sizeof(p));
28     p[0] = p[1] = 1;
29     for (int i = 2; i < maxn; ++i)
30     {
31         if (p[i]) continue;
32         for (int j = i << 1; j < maxn; j += i)
33             p[j] = 1;
34         prim[pn++] = i;
35     }
36 }
37 void prime2()
38 {
39     for (int i = 0; i < maxn; ++i) p[i] = 1;
40     if (low == 1)
41         p[0] = 0;
42     for (int i = 0; i < pn; ++i)
43     {
44         if (prim[i] *(LL)prim[i] > high) break;
45         LL st = low / prim[i];
46         for (LL j = st * prim[i]; j <= high; j += prim[i])
47         {
48             if (j < low) continue;
49             if (j == prim[i]) continue;
50             p[j - low] = 0;
51         }
52     }
53     int min_ans = inf, max_ans = -1, pre = -1;
54     int res1,res2,res3,res4;
55     for (int i = low; i <= high; ++i)
56     {
57         if (i < 0) break;//坑爹。以后还是用LL吧
58         if (p[i - low])
59         {
60             if (pre == -1)
61                 pre = i;
62             else
63             {
64                 if (i - pre < min_ans)
65                 {
66                     min_ans = i - pre;
67                     res1 = pre;
68                     res2 = i;
69                 }
70                 if (i - pre > max_ans)
71                 {
72                     max_ans = i - pre;
73                     res3 = pre;
74                     res4 = i;
75                 }
76                 pre = i;
77             }
78         }
79     }
80     if (max_ans == -1)
81     {
82         printf("There are no adjacent primes.\n");
83     }
84     else
85     {
86         printf("%d,%d are closest, %d,%d are most distant.\n",res1,res2,res3,res4);
87     }
88 
89 }
90 int main()
91 {
92     prime();
93     while (~scanf("%d%d",&low,&high))
94     {
95         prime2();
96     }
97     return 0;
98 }

 

posted @ 2013-05-29 00:18  Missa  阅读(382)  评论(0)  编辑  收藏  举报