ACwing 196. 质数距离

#include <bits/stdc++.h>

using namespace std;
const int N = , M = ;
int v[M];
long long prime[N],prim[N];
int cun=;
int inint(int n)
{
    for(int i=;i<=n;i++)
    {
        if(!v[i])
        {
            v[i]=i;
            prim[cun++]=i;
        }
        for(int j=;j<cun;j++)
        {
            if(prim[j]>n/i||prim[j]>v[i])break;
            v[i*prim[j]]= prim[j];
        }
    }
}
int main()
{
    long long l, r;
    inint();
    while(cin >> l>> r)
   {
       memset(v,,sizeof(v));
       for(int i=;i<cun;i++)
       {

           long long p=prim[i];
           for(long long j = max(*p,(l+p-)/p*p);j<=r;j+=p)
           {
               v[j-l]=;
           }
       }
       cun =;
       for(int i= ;i<=r-l;i++)
       {
           if(v[i]==&&i+l>)prime[cun++] = i;
       }
       if(cun <)
       {
           cout <<"There are no adjacent primes."<<endl;
       }
       else{
           int mind=,maxd=;
           for(int i=;i+<cun;i++)
           {
               int d = prime[i+] - prime[i];
               if(d>prime[maxd+]-prime[maxd])maxd=i;
               if(d<prime[mind+]-prime[mind])mind = i;
           }
           printf("%d,%d are closest, %d,%d are most distant.\n",prime[mind]+l,prime[mind+]+l,prime[maxd]+l,prime[maxd+]+l);
       }
   }

    return ;
}