Prime Distance

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18731		Accepted: 5006

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).

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.

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.

2 17
14 17

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

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string.h>
 4 #include <stdlib.h>
 5 #include <iostream>
 6 #include <sstream>
 7 #include <queue>
 8 #include <vector>
 9 #include <algorithm>
10 #define maxn 50010
11 #define maxm 1000010
12 #define inf  0x3f3f3f3f
13 using namespace std;
14 typedef long long ll;
15 int vis[maxm],f[maxm];
16 int prime[maxn],prime1[maxm];
17 int mind,maxd,minl,minr,maxl,maxr;
18 int t,t1;
19 ll L,R;
20 void init()                          //筛选出50000以内的素数
21 {
22     t=0;
23     memset(vis,0,sizeof(vis));
24     for(int i=2;i<maxn;i++)
25     {
26         if(!vis[i])
27         {
28             prime[t++]=i;
29             for(int j=i+i;j<maxn;j+=i)
30                 vis[j]=1;
31         }
32     }
33 }
34 void selet()
35 {
36     memset(vis,0,sizeof(vis));                   //*特别考虑1
37     for(int i=0;i<t;i++)                         //标记int范围内的合数
38     { 
39         ll b=L/prime[i];
40         while(b*prime[i]<L||b<=1)
41             b++;
42         for(ll j=b*prime[i];j<=R;j+=prime[i])     
43             vis[j-L]=1;                           //节约空间，降低空间复杂度
44     }
45     if(L==1)                                   
46         vis[0]=1;
47     t1=0;
48     for(ll i=L;i<=R;i++)                       //找出区间内的素数存进数组
49     {
50         if(!vis[i-L])
51             prime1[t1++]=i;
52     }
53 }
54 void solve()
55 {
56     selet();
57     mind=inf,maxd=-inf;
58     minl=minr=maxl=maxr=-1;
59     for(int i=1;i<t1;i++)                  //枚举区间内的素数，更新要输出的值
60     {
61         int d=prime1[i]-prime1[i-1];
62         if(d<mind)
63         {
64             mind=d;
65             minl=prime1[i-1];
66             minr=prime1[i];
67         }
68         if(d>maxd)
69         {
70             maxd=d;
71             maxl=prime1[i-1];
72             maxr=prime1[i];
73         }
74     }
75 }
76 int main(int argc, char const *argv[])
77 {
78     init();
79     while(cin>>L>>R)
80     {
81         solve();
82         if(t1<2) 
83             printf("There are no adjacent primes.\n");  
84         else 
85             printf("%d,%d are closest, %d,%d are most distant.\n",minl,minr,maxl,maxr);
86     }
87     return 0;
88 }