题解 CF1305C 【Kuroni and Impossible Calculation】
发现直接做不容易并且
int main(){
cin >> n >> m ;
for (int i = 1 ; i <= n ; ++ i)
scanf("%lld", &base[i]) ;
if (n > m) return puts("0"), 0 ;
for (int i = 1 ; i <= n ; ++ i)
for (int j = i + 1 ; j <= n ; ++ j)
(ans *= (LL)abs(base[i] - base[j])) %= m ;
cout << ans << endl ;
}