Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18707		Accepted: 4998

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.

 #include<cstdio>

 #include<cstring>

 #include<iostream>

 #include<algorithm>

 using namespace std;

 typedef long long ll;

 const int N=1e6+;

 const ll INF=0x3f3f3f3f;

 bool prime1[N];

 bool prime2[N];

 ll ss[N];

 ll num=;

 void findprime(ll n,ll m){       //区间筛素数，筛的是从[n,m)的，下面传的参数是n,m+1,所以保证是求的闭区间的。好好想想。

     for(ll i=;i*i<m;i++) prime1[i]=true;                     //这是对[2，sqrt(m))的初始化。

     for(ll i=;i<m-n;i++) prime2[i]=true;                     //对下标偏移之后的[n,m）进行初始化。

     for(ll i=;i*i<m;i++){

         if(prime1[i]){

             for(ll j=*i;j*j<m;j+=i) prime1[j]=false;                    //找出来[2，sqrt(m))中不是素数的筛出去。

             for(ll j=max((ll),(n+i-)/i)*i;j<m;j+=i) prime2[j-n]=false; //将偏移后的[n,m)中的不是素数的筛出去。

         }            //j=max((ll)2,(n+i-1)/i)*i;j<m;j+=i)意思就是从最接近n的数开始，将那些合数的倍数，从2倍开始，筛掉。

     }

 }

 int main(){

     ll n,m,h;

     ll minn,maxx;

     while(~scanf("%lld%lld",&n,&m)){

         memset(prime2,,sizeof(prime2));

         memset(ss,,sizeof(ss));

         findprime(n,m+);                                                     //这样保证是求的[n,m]这个闭区间的

         h=;

         for(int i=;i<m-n+;i++){

             if(prime2[i]&&i+n!=)//如果是true就是成立的，因为上面区间筛素数没有筛掉1(既不是素数也不是合数)，所以判断一下。

             ss[h++]=i+n;                                                  //将下标偏移量再加回来。

         }

         if(h<) printf("There are no adjacent primes.\n");                //保证必须有>=2个素数。

         else{

             minn=INF;maxx=-;

         for(int i=;i<h-;i++){                                            //我太菜，只会这样找。

             minn=min(minn,ss[i+]-ss[i]);                                  //找出距离最近的值是多少。

             maxx=max(maxx,ss[i+]-ss[i]);                                  //找出距离最远的值是多少。

         }

         for(int i=;i<h-;i++){

             if(ss[i+]-ss[i]==minn){                          //将距离最近的2个素数找出来，输出来最先找到的一组就可以。

                 printf("%lld,%lld are closest, ",ss[i],ss[i+]);

                 break;

             }

         }

         for(int i=;i<h-;i++){                                             //意思同上。

             if(ss[i+]-ss[i]==maxx){

                 printf("%lld,%lld are most distant.\n",ss[i],ss[i+]);

                 break;

             }

         }

         }

     }

     return ;

 }