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