POJ 2689 Prime Distance 大区间素数筛选

Prime Distance

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.

Source

​​Waterloo local 1998.10.17​​

算法分析：

题意：

出一个区间[L,U]，找出区间内容、相邻的距离最近的两个素数和
 * 距离最远的两个素数。
 * 1<=L<U<=2,147,483,647 区间长度不超过1,000,000

分析：

一开始，想把数据范围内的素数全部打出来，结果范围不够，打不到那么多。

原来有一个区间判断素数的问题，先限定范围，题目的L,U虽然很大，但L-U<=1e6，这就是在提示我们。该如何求出这个区间的素数呢？可以想到这个区间里的非素数的素数因子最大不会超过1e6，不然就会1e6*1e6超过U的上限，那么那些非素数的数的因子都小于等于1e6。那么先求出1-1e6之间的素数，然后用这些素数作为因子去删掉[L,U]之间的合数。剩下来的就是素数。

代码实现：

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
const int N=100010;
long long prime[N] = {0},num_prime = 0;    
long long  isNotPrime[N] = {1, 1};   
int Prime()    
{     
      for(long i = 2 ; i < N ; i ++)       
         {            
      if(! isNotPrime[i])               
         prime[num_prime ++]=i;  
           
      for(long j = 0 ; j < num_prime && i * prime[j] <  N ; j ++)
         {               
               isNotPrime[i * prime[j]] = 1;  
         if( !(i % prime[j] ) )  //¹Ø¼ü´¦2                  
            break;           
      }        
   }        
   return 0;   
}  
bool notprime[1000010];
int prime2[1000010];
void getPrime2(int L,int R)
{
    memset(notprime,false,sizeof(notprime));
    if(L<2)L=2;
    for(int i=0;i<num_prime&&(long long)prime[i]*prime[i]<=R;i++)
    {
        int s=L/prime[i]+(L%prime[i]>0);
        if(s==1)s=2;
        for(int j=s;(long long)j*prime[i]<=R;j++)
            if((long long)j*prime[i]>=L)
                notprime[j*prime[i]-L]=true;
    }
    prime2[0]=0;
    for(int i=0;i<=R-L;i++)
        if(!notprime[i])
            prime2[++prime2[0]]=i+L;
}
int main()
{
   Prime();
    int L,U;
    while(scanf("%d%d",&L,&U)==2)
    {
        getPrime2(L,U);
        if(prime2[0]<2)printf("There are no adjacent primes.\n");
        else
        {
            int x1=0,x2=100000000,y1=0,y2=0;
            for(int i=1;i<prime2[0];i++)
            {
                if(prime2[i+1]-prime2[i]<x2-x1)
                {
                    x1=prime2[i];
                    x2=prime2[i+1];
                }
                if(prime2[i+1]-prime2[i]>y2-y1)
                {
                    y1=prime2[i];
                    y2=prime2[i+1];
                }
            }
            printf("%d,%d are closest, %d,%d are most distant.\n",x1,x2,y1,y2);
        }
    }
}