Lucky number of tube (linear sieve)

Idea: This question can also be violent, but I first filter out prime numbers, and then determine whether n is a prime number for each inquiry, and if not, find a prime number smaller and larger than it. Finally found the nearest one.

#include <bits/stdc++.h>
using namespace std;
int prime[100010];
void is_prime(int x){
    
    
    for(int i=2;i<=sqrt(x);i++){
    
    
        if(prime[i]==0){
    
    
            for(int j=i*i;j<=x;j+=i){
    
    
                prime[j] = 1;
            }
        }
    }
}

int main()
{
    
    
    int t; cin>>t;
    is_prime(100000);
    while(t--){
    
    
        int n; cin>>n;
        if(prime[n]==0) cout<<"YES"<<endl;
        else {
    
    
            int cnt1=0,cnt2=0;
            for(int i=n;;i++){
    
    
                if(prime[i]==0){
    
    
                    cnt1 = i;
                    break;
                }
            }
            for(int i=n;;i--){
    
    
                if(prime[i]==0){
    
    
                    cnt2 = i;
                    break;
                }
            }
            int ans = min(abs(n-cnt1),abs(n-cnt2));
            cout<<ans<<endl;
        }
    }
    return 0;
}

At the beginning, the array was opened at 10005wa, and it was passed by increasing it later. Be sure to pay attention to the range next time.