题目

题解思路

大合数N必然存在一个质因子小于等于根号N，不然怎么组合出他。

所以我们只需筛出2到根号R的素数，再用这个区间的素数来筛出区间里的素数。
将区间的每个素数减去l来存入数组寻找差值。（类似离散化吧）

时间复杂度Nloglogn（因为我用的埃式筛 感觉和欧拉筛差不了多少）约等于On

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#include <map>
#include <string>
using namespace std;

const  int  INF =  0x3f3f3f3f;

const  int   N = 1e6 + 10 ;

bool pre[N] ;

int vis[N];
long long  l ,r ;

void in()
{
    pre[1] = 1 ;
    for (long long  i = 2 ; i <= 50000 ; i++ )
    {
        if ( ! pre[i] )
        {
            for (int j = i*2 ; j <= 50000 ; j+=i )
                pre[j] = 1 ;
            for (long long  k = max ((l + i - 1 ) / i * i , 2*i) ; k <= r ; k += i )
            {
                vis[k-l] = 1 ;
            }
        }
    }
}

int main ()
{
    ios::sync_with_stdio(false);
    while( cin >> l >> r)
    {
        memset(vis, 0 , sizeof(vis) );
        int mi = INF , ma = -1 ;
        int t1 = INF , t2 = -1 ;
        int mit1 , mit2  , mat1 , mat2   ;
        in();
        for (int i = 0 ; i <= r - l  ;  )
        {
            if ( i == 0 && l == 1 )
            {
                i++;
                continue;
            }
            if ( vis[i] == 0 )
            {
                t1 = i + l ;
                i++;
                while( i <= r - l && vis[i] )
                    i++;
                if ( i <= r - l )
                {
                    t2 = i + l ;
                    if ( t2 - t1  > ma )
                    {
                        mat1 = t1;
                        mat2 = t2;
                        ma = t2 - t1 ;
                    }
                    if ( t2 - t1 < mi )
                    {
                        mi = t2 - t1 ;
                        mit1 = t1 ;
                        mit2 = t2 ;
                    }
                }
            }else
             i++;
        }
        if ( ma == -1 )
        {
            cout<<"There are no adjacent primes.\n";
        }else
        {
            cout<<mit1<<","<<mit2<<" are closest, " <<mat1<<","<<mat2<<" are most distant.\n";
        }
    }
    return 0 ;
}