cp/codeforces/827/d.cc
2025-05-16 22:04:03 -05:00

123 lines
2.2 KiB
C++

#include <bits/stdc++.h> // {{{
// https://codeforces.com/blog/entry/96344
#pragma GCC optimize("O2,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
using namespace std;
using i32 = int32_t;
using u32 = uint32_t;
using i64 = int64_t;
using u64 = uint64_t;
using f64 = double;
using f128 = long double;
#if __cplusplus >= 202002L
template <typename T>
constexpr T MIN = std::numeric_limits<T>::min();
template <typename T>
constexpr T MAX = std::numeric_limits<T>::max();
template <typename T>
[[nodiscard]] static T sc(auto&& x) {
return static_cast<T>(x);
}
template <typename T>
[[nodiscard]] static T sz(auto&& x) {
return static_cast<T>(x.size());
}
#endif
static void NO() {
std::cout << "NO\n";
}
static void YES() {
std::cout << "YES\n";
}
template <typename T>
using vec = std::vector<T>;
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define ff first
#define ss second
#ifdef LOCAL
#define db(...) std::print(__VA_ARGS__)
#define dbln(...) std::println(__VA_ARGS__)
#else
#define db(...)
#define dbln(...)
#endif
// }}}
void solve() {
/*
a[i] has factors S, |S| < = log2(MAX_A[i] = 1000)
want to find rightmost j > i s.t. no common factors
factor -> [a[i] divisible by]
iter thru factors of a[i]
if i have log2(a[i]) factors;
there are 10^6 factors total
there are 10 factors here
so, factor each a[i], and manually compute
*/
u32 n;
cin >> n;
vector<u32> a(n);
for (auto& e : a)
cin >> e;
vector<bitset<1001>> coprime(1001);
for (u32 i = 1; i <= 1000; ++i) {
for (u32 j = 1; j <= 1000; ++j) {
coprime[i][j] = gcd(i, j) == 1;
}
}
vector<u32> right(1001, -1);
for (u32 i = 0; i < n; ++i) {
right[a[i]] = i;
}
u32 ans = 0;
for (i32 i = 1; i <= 1000; ++i) {
for (u32 j = 1; j <= 1000; ++j) {
if (coprime[i][j] && right[i] != -1 && right[j] != -1) {
ans = max(ans, right[i] + right[j] + 2);
}
}
}
if (ans == 0)
cout << -1;
else
cout << ans;
cout << '\n';
}
int main() { // {{{
cin.tie(nullptr)->sync_with_stdio(false);
cin.exceptions(cin.failbit);
u32 tc = 1;
cin >> tc;
for (u32 t = 0; t < tc; ++t) {
solve();
}
return 0;
}
// }}}