UVA10140 Prime Distance

#include<iostream>
#include<cstring>
using namespace std;
const int maxn=1e6+10;
 
long long l,r,prime[maxn],vis[maxn],p[maxn],a[maxn];
int tot=0;

int pre(long long n)
{
    for(int i=2;i<=n;i++)
    {
        if(!vis[i])
        {
            vis[i]=i;
            prime[++tot]=i;
        }
        for(int j=1;j<=tot;j++)
        {
            if(i*prime[j]>n)
            {
                break;
            }
            if(prime[j]>vis[i])
            {
                break;
            }
            vis[i*prime[j]]=prime[j];
        }
    }
}
int main()
{
    pre(5e4+10);
    while(cin>>l>>r)
    {
        memset(p,0,sizeof(p));
        if(l==1)
        {
            l=2;
        }
        long long cnt=0,min1,min2,max1,max2,minn=0x3f3f3f3f,maxx=-0x3f3f3f3f;
        for(int j=1;j<=tot;j++)
        {
            for(int i=l/prime[j];i<=r/prime[j];i++)
            {
                while(i<=1)
                {
                    i++;
                }
                if(prime[j]*i<l||prime[j]*i>r)
                {
                    continue;
                }
                p[prime[j]*i-l]=1;
            }  
        }
        for(int i=l;i<=r;i++)
        {
            if(p[i-l]==0)
            {
                a[++cnt]=i;
            }
        }
        if(cnt<=1)
        {
            //cout<<cnt<<endl;
            cout<<"There are no adjacent primes."<<endl;
            continue;
        }
        for(int i=1;i<=cnt-1;i++)
        {
            if(a[i+1]-a[i]<minn)
            {
                minn=a[i+1]-a[i];
                min1=a[i+1];
                min2=a[i];
            }
            if(a[i+1]-a[i]>maxx)
            {
                maxx=a[i+1]-a[i];
                max1=a[i+1];
                max2=a[i];
            }
        }
        //cout<<cnt<<endl;
        cout<<min2<<","<<min1<<" are closest, "<<max2<<","<<max1<<" are most distant."<<endl;
    }
    return 0;
}