POJ 2689 - Prime Distance - [筛法求素数]

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题目链接：http://poj.org/problem?id=2689

Time Limit: 1000MS Memory Limit: 65536K

Description

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：

给出st与ed（1 <= st < ed <= 2147483647 且 ed - st <= 1000000），求[st,ed]区间内，相邻的两个素数中，差最小的和差最大的（若存在差同样大的一对素数，则有先给出最小的一对素数）；

 

题解：

显然，不可能直接去筛2147483647以内的素数；

不难观察到，ed - st <= 1000000，可以直接去筛每个test case的[st,ed]区间内的素数；

那么，我们不妨先用欧拉筛法筛出[0,46341]区间内的素数（46341≈sqrt(2147483647)）；

然后对于每个test case的[st,ed]区间，我们采用埃筛的方法，用已经筛选出的在[0,46341]区间内的素数，去筛[st,ed]区间内的素数；

 

AC代码：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 #define MAX 46345
 5 typedef long long ll;
 6 
 7 bool isPrime[1000005];
 8 
 9 //欧拉筛法求[0,46345]内的素数 - begin
10 int Prime1[MAX],cnt1;
11 void screen1()
12 {
13     memset(isPrime,1,sizeof(isPrime));
14     cnt1=0;
15     isPrime[0]=isPrime[1]=0;
16     for(int i=2;i<=MAX;i++)
17     {
18         if(isPrime[i]) Prime1[cnt1++]=i;
19         for(int j=0;j<cnt1;j++)
20         {
21             if(i*Prime1[j]>MAX) break;
22             isPrime[(i*Prime1[j])]=0;
23             if(i%Prime1[j]==0) break;
24         }
25     }
26 }
27 //欧拉筛法求[0,46345]内的素数 - end
28 
29 //埃筛求[st,ed]内的素数 - begin
30 ll st,ed;
31 ll Prime2[1000005];
32 int cnt2;
33 void screen2()
34 {
35     memset(isPrime,1,sizeof(isPrime));
36     for(int i=0;i<cnt1;i++)
37     {
38         if(Prime1[i]>ed) break;//这种情况下，已经不可能再筛了，跳出
39 
40         ll tmp=st/Prime1[i];
41         while(tmp*Prime1[i]<st) tmp++;
42         while(tmp<=1) tmp++;
43         
44         for(ll j=tmp*Prime1[i];j<=ed;j+=Prime1[i]) isPrime[j-st]=0;
45     }
46     if(st==1) isPrime[0]=0;
47         //得到了[st,ed]内的素数的标记数组
48         
49     cnt2=0;
50     for(ll i=st;i<=ed;i++)
51     {
52         if(isPrime[i-st]) Prime2[cnt2++]=i;
53     }
54         //根据标记数组，得到存储[st,ed]内所有素数的数组
55 }
56 //埃筛求[st,ed]内的素数 - end
57 
58 int main()
59 {
60     screen1();
61     while(scanf("%lld%lld",&st,&ed)!=EOF)
62     {
63         screen2();
64         if(cnt2<=1)
65         {
66             printf("There are no adjacent primes.\n");
67             continue;
68         }
69         else
70         {
71             ll min_dist=0x3f3f3f3f,max_dist=0;
72             ll c1,c2,d1,d2;
73             for(int i=0;i<cnt2-1;i++)
74             {
75                 if(Prime2[i+1]-Prime2[i]<min_dist)
76                 {
77                     min_dist=Prime2[i+1]-Prime2[i];
78                     c1=Prime2[i], c2=Prime2[i+1];
79                 }
80                 if(Prime2[i+1]-Prime2[i]>max_dist)
81                 {
82                     max_dist=Prime2[i+1]-Prime2[i];
83                     d1=Prime2[i], d2=Prime2[i+1];
84                 }
85             }
86             printf("%I64d,%I64d are closest, %I64d,%I64d are most distant.\n",c1,c2,d1,d2);
87         }
88     }
89 }

 

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