/ SeriousOJ /

Record Detail

Time Exceeded


  
# Status Time Cost Memory Cost
#1 Accepted 254ms 32.75 MiB
#2 Accepted 295ms 32.957 MiB
#3 Accepted 752ms 32.965 MiB
#4 Accepted 740ms 32.961 MiB
#5 Accepted 826ms 32.961 MiB
#6 Accepted 885ms 32.953 MiB
#7 Accepted 848ms 32.957 MiB
#8 Accepted 697ms 32.961 MiB
#9 Accepted 690ms 32.723 MiB
#10 Accepted 561ms 32.965 MiB
#11 Accepted 541ms 32.961 MiB
#12 Time Exceeded ≥1087ms ≥32.945 MiB
#13 Time Exceeded ≥1099ms ≥32.867 MiB

Code

// #pragma GCC optimize("O3,unroll-loops,Ofast")
// #pragma GCC target("avx2")
#include<bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
#define int long long
#define endl '\n'

using namespace std;
using pii = pair<int, int>;
using tup = tuple<int, int, int>;

// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// template <class T> using ordered_set = tree<T, null_type,
//                          less<T>, rb_tree_tag, tree_order_statistics_node_update>;

const int inf = (1e15)+5;
const int mod = 1000000007;
const double eps = 1e-6;
const int N = 200005;

vector<int> divs[N];
void preprocess() {
	for (int i = 1; i < N; i++) {
		for (int j = i; j < N; j += i) {
			divs[j].push_back(i);
		}
	}
}

int n, a[N];
int cnt[N];


int calc(int x) {
	int tot = 0, out = 0;
	for(int i=1; i<=n; i++) {
		if(a[i] % x == 0) tot++;
		else out = gcd(out, a[i]);
	}

	int onno = n - tot;
	int lage;

	if(n % 2 == 0) {
		if(tot < n/2) return 0;
		lage = n / 2;
	}

	else {
		if(tot >= n/2+1) lage = n/2;
		else if(tot == n/2) lage = n/2+1;
		else return 0;
	}

	lage -= onno;

	for(int i=1; i<=n; i++) {
		if(a[i] % x) continue;
		for(int j : divs[a[i]])
			cnt[j]++;
	}

	// cout << tot << ' ' << onno << ' ' << lage << ' ' << out << endl;
	int ret = 0;
	for(int i=1; i<N; i++) {
		// cout << i << ' ' << cnt[i] << endl;
		if(out % i or cnt[i] < lage) continue;
		ret = max(ret, i);
	}

	for(int i=1; i<=n; i++) {
		if(a[i] % x) continue;
		for(int j : divs[a[i]])
			cnt[j]--;
	}
	// if(out % x == 0) ret *= x;
	
	// cout << x << ": " << ret + x << endl;
	return ret + x;
}

void solve(int tc) {
	cin >> n;
	for(int i=1; i<=n; i++) cin >> a[i];

	int ans = 0;
	for(int i : divs[a[1]]) ans = max(ans, calc(i));

	cout << ans << endl;
}
    
int32_t main() {
    cin.tie(NULL)->sync_with_stdio(false);
    cout.precision(10);

    preprocess();

    int T = 1;
    cin >> T;

    for (int i = 1; i <= T; i++) {
        solve(i);
    }

    return 0;
}

Information

Submit By
Type
Submission
Problem
P1076 Even Odd GCD (Easy Version)
Contest
Bangladesh 2.0
Language
C++20 (G++ 13.2.0)
Submit At
2024-08-16 17:06:08
Judged At
2024-11-11 03:12:43
Judged By
Score
35
Total Time
≥1099ms
Peak Memory
≥32.965 MiB