C. Koxia and Number Theory （线性同余）https://codeforces.com/contest/1770/problem/C

https://codeforces.com/contest/1770/attachments/download/18470/editorial.pdf

这个pdf都写得很明白了，这个c题终于懂了，麻了à$a_i^j \leq b^j$

本来只需要枚举n/2之内的质数，但是因为n很小，所以直接枚举n/2也可以

• 对于给定的 $x$，让我们记 $b_i = a_i + x$。条件「 $gcd(b_i,b_j) = 1$对于 恒成 $1 \leq i < j \leq n$立」等价于「每个质数 $p$ 只能整除至多一个 $b_i$」。那么，对于一个质数  $p$，我们能否判断 $p$， 是否总是整除至少两个 $b_i$（无论$x$的取值）

点通我的就是这句话

#include<bits/stdc++.h>
#define debug1(a) cout<<#a<<'='<< a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<endl;
#define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<"  "<<#e<<" = "<<e<<endl;
#define debug0(x) cout << "debug0: " << x << endl
#define fr(t, i, n)for (long long i = t; i < n; i++)
#define YES cout<<"Yes"<<endl
#define nO cout<<"no"<<endl
#define fi first
#define se second
#define int long long
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;

//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize(2)

const int N = 110;

int a[N];
bool solve() 
{
    int n;cin >> n;
    for(int i = 0;i < n;i ++)cin >> a[i];
    sort(a,a+n);

    for(int i = 1;i < n;i ++)
        if(a[i] == a[i-1])
        {
            return 0;
        }
    
    
    bool acc = 1;
    for(int i = 2;i <= n;i ++)
    {
        vector<int> num(i,0);
        for(int j = 0;j < n;j ++)num[a[j] % i]++;
        if(*min_element(num.begin(),num.end()) >= 2)acc = 0;
    }
    return acc;
    
}

signed main()
{
    /*
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    */
    int T = 1;cin >> T;
	
    while(T--){
        puts(solve()?"YES":"NO");
    }
    return 0;
}
/*

*/