D) K-good

原创已于 2022-03-25 16:43:17 修改 · 1k 阅读

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于 2022-03-25 11:17:39 首次发布

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$n=\frac{(k + 1) * k }{2}+ k * s-> n * 2= k * (k + 2 * s + 1) 观察可以发现一定是偶数乘奇数的情况，如果n是2的幂，那么结果偶数乘偶数，一定不合法，所以最终答案就是这个$

// Problem: C. Make Equal With Mod
// Contest: Codeforces - CodeTON Round 1 (Div. 1 + Div. 2, Rated, Prizes!)
// URL: https://codeforces.com/contest/1656/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 2022-03-24 22:59:03
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a;
		a = a * a;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}

void solve() {
	ll n;
	cin >> n;
	ll c = log(n) / log(2);
	
	
	
		ll ans = 1;
		while (n %2==0) {
			ans *= 2;
			n >>= 1;
		}
	if (n == 1) {
		puts("-1");
	}
	else cout << min (n, ans * 2) << endl;
	
}
int main () {
	// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}