思想：

用到了欧拉筛，然后对数组的内存的巧妙处理。

推荐的博客：

https://blog.csdn.net/Feynman1999/article/details/79533130?depth_1-utm_source=distribute.pc_relevant.none-task&utm_source=distribute.pc_relevant.none-task

推荐的b站：

https://www.bilibili.com/video/av89230175?from=search&seid=1935489020745550394

代码：

#include<iostream>
#include<stdio.h>
#include<vector>
using namespace std;
const int maxn = (1<<16)+10;
bool valid[maxn];
int ans[maxn];
bool vis[maxn<<4];
int tot;
void get_prime(int n){
    tot = 0;
    for(int i=2;i<=n;i++)
        valid[i] = true;
    for(int i=2;i<=n;i++){
        if(valid[i])
            ans[++tot] =i;
        for(int j=1;j<=tot&&ans[j]*i<=n;j++){
            valid[ans[j]*i] = false;
            if(i%ans[j]==0)
                break;
        }
    }
}

int main(){
    get_prime(1<<16);
    int L,R;
    while(cin>>L>>R){
        for(int i=1;i<=(R-L+1);i++)
            vis[i] = true;
        if(L==1) vis[1] = false;
        for(int i=1;i<=tot;i++){
            for(int j=max(ans[i],(L-1)/ans[i]+1);j<=R/ans[i];j++){
                vis[ans[i]*j-L+1] = false;
            }
        }
        int num = 0;
        for(int i=1;i<=(R-L+1);i++)
            if(vis[i]) num++;
        if(num>=2){
            vector<int>tep;
            for(int i=1;i<=(R-L+1);i++){
                if(vis[i]) tep.push_back(L+i-1);
            }
            int ans1 = 0,ans2 = 0;
            int maxx = tep[1] - tep[0],minn = tep[1]-tep[0];
            for(int i=2;i<tep.size();i++){
                if(tep[i]-tep[i-1]>maxx){
                    ans2 = i-1;
                    maxx = tep[i] -tep[i-1];
                }
                if(tep[i]-tep[i-1]<minn){
                    ans1 =i-1;
                    minn = tep[i] - tep[i-1];
                }
            }
            cout<<tep[ans1]<<","<<tep[ans1+1]<<" are closest, "<<tep[ans2]<<","<<tep[ans2+1]<<" are most distant."<<endl;
        }
        else{
            cout<<"There are no adjacent primes."<<endl;
        }
    }
    return 0;
}