POJ 2689 Prime Distance（大區間素數篩選）

Prime Distance

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9322		Accepted: 2531

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

 

 

/*
 * POJ 2689 Prime Distance
 * 給出一個區間[L,U]，找出區間內容、相鄰的距離最近的兩個素數和
 * 距離最遠的兩個素數。
 * 1<=L<U<=2,147,483,647 區間長度不超過1,000,000
 * 就是要篩選出[L,U]之間的素數
 */

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;

const int MAXN=100010;
int prime[MAXN+1];
void getPrime()
{
    memset(prime,0,sizeof(prime));
    for(int i=2;i<=MAXN;i++)
    {
        if(!prime[i])prime[++prime[0]]=i;
        for(int j=1;j<=prime[0]&&prime[j]<=MAXN/i;j++)
        {
            prime[prime[j]*i]=1;
            if(i%prime[j]==0)break;
        }
    }
}
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=1;i<=prime[0]&&(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()
{
    getPrime();
    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);
        }
    }
}