思路：
先筛一遍质数，对于n<0,需要先将其连加到刚好为正，答案为2n+1+刚好为正数时的那个答案；
对于n>0,每一次n++，看n或者n2+1是否为质数。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<math.h>
using namespace std;
#define ll long long
#define mod 1000000007ll
int primes[30000100];
bool st[30000100];
int cnt;
void prime(int n)
{
for (int i = 2; i <= n; i ++ )
{
if (st[i] == 0) primes[cnt ++ ] = i;
for (int j = 0; j<cnt && i*primes[j] <= 20000050; j ++ )
{
st[primes[j] * i] = 1;
if (i % primes[j] == 0) break;
}
}
}
int main()
{
int t;
cin>>t;
prime(10000000);
while(t--)
{
int n;
cin>>n;
if(n<0)
{
if(!st[-n+1])cout<<1+(-n)*2+1<<endl;
else if(!st[3-2*n])cout<<2+2*(-n)+1<<endl;
else
{
ll ans = -n*2+1;
n = -n+1;
while(st[n] && st[n*2+1])
{
ans+=2;
n++;
}
if(!st[n])ans++;
else ans+=2;
printf("%lld\n",ans);
}
}
else if(n==0)cout<<3<<endl;
else if(n == 1)cout<<2<<endl;
else
{
if(!st[n])cout<<1<<endl;
else if(!st[n+n+1])cout<<2<<endl;
else if(!st[n+n-1])cout<<2<<endl;
else
{
ll ans = n*2+1;
n++;
while(st[n] && st[n*2+1])
{
ans+=2;
n++;
}
if(!st[n])ans++;
else ans+=2;
printf("%lld\n",ans);
}
}
}
}