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 }