Prime Distance POJ - 2689 （数学 素数）

The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers.
Your program is given 2 numbers: L and U (1<=L< U<=2,147,483,647), and you are to find the two adjacent primes C1 and C2 (L<=C1< C2<=U) that are closest (i.e. C2-C1 is the minimum). If there are other pairs that are the same distance apart, use the first pair. You are also to find the two adjacent primes D1 and D2 (L<=D1< D2<=U) where D1 and D2 are as distant from each other as possible (again choosing the first pair if there is a tie).

Input

Each line of input will contain two positive integers, L and U, with L < U. The difference between L and U will not exceed 1,000,000.

Output

For each L and U, the output will either be the statement that there are no adjacent primes (because there are less than two primes between the two given numbers) or a line giving the two pairs of adjacent primes.

Sample Input

2 17
14 17

Sample Output

2,3 are closest, 7,11 are most distant.
There are no adjacent primes.

题意：找到l、r区间内，差值最大和最小的相邻质数。
思路：因为n很大，所以不可能直接找出所有的质数后遍历
对于区间1~n，我们只需要找出√n 范围内的素数，倍增标记剩余区间的合数，就可以得到区间的所有质数。

 1 #include<cstring>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cmath>
 5 using namespace std;
 6 
 7 const int maxn = 1e5;
 8 int prime[maxn];
 9 int tot;
10 void get_pri(int n)
11 {
12     bool v[n+5];
13     memset(v,0,sizeof(v));
14     for(int i=2; i<=n; i++)
15     {
16         if(!v[i])
17             prime[++tot] = i;
18         for(int j=i; j<=n/i; j++)
19         {
20             v[j*i] = 1;
21         }
22     }
23 }
24 int main()
25 {
26 
27     int l,r;
28     while(~scanf("%d%d",&l,&r))
29     {
30         tot = 0;
31         get_pri(sqrt(r));
32         bool vis[r-l+6];
33         memset(vis,0,sizeof(vis));
34         for(int i=1; i<=tot; i++)
35         {
36             for(int j=ceil(l*1.0/prime[i]); j<=r/prime[i]; j++)
37             {
38                 if(j == 1)continue;
39                 vis[prime[i]*j-l] = 1;
40             }
41         }
42         int ans[r-l+5];
43         int cnt = 0;
44         for(int i=0; i<=r-l; i++)
45         {
46             if(!vis[i])
47             {
48                 if(i+l == 1)continue;
49                 ans[++cnt] = i+l;
50             }
51         }
52         if(cnt < 2)
53             printf("There are no adjacent primes.\n");
54         else
55         {
56             int minn = 0x3f3f3f3f;
57             int maxx = 0;
58             int id1;
59             int id2;
60             for(int i=1; i<cnt; i++)
61             {
62                 int tmp = ans[i+1]-ans[i];
63                 if(tmp < minn)
64                 {
65                     minn = tmp;
66                     id1 = i;
67                 }
68                 if(tmp > maxx)
69                 {
70                     maxx = tmp;
71                     id2 = i;
72                 }
73             }
74             printf("%d,%d are closest, %d,%d are most distant.\n",ans[id1],ans[id1+1],ans[id2],ans[id2+1]);
75         }
76     }
77 }

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